Design of Steel and Timber Structure — Chapter 6: Compression Members

Most repeated · 4 of 21 exams
Asked 2 times
2075 Bhadra · 14 marks
2072 Magh · 12 marks

Design a built up column 10 m long to carry a factored axial compressive load of 1000 kN. The column is restrained in position but not in direction at both ends. Design the column with connecting system as lacing (single lacing) with bolted (or welded) connection. Use two channel back to back. Assume steel of grade Fe 410, E250 C and bolts grade 4.6.

Similar questions: Built-up column with battens, 10 m, 1080 kN (2073 Magh) · Built-up column, lacing, 8 m, 1100 kN (2079 Chaitra)

Answer

Given: factored axial load $P=1000$ kN, length 10 m, held in position but not restrained in direction at both ends → $K=1.0$ , $KL=10$ m. Two channels back to back with single bolted lacing; E250 ( $f_y=250$ , $f_u=410$ ); bolts M20 grade 4.6. Built-up members belong to buckling class c (IS 800:2007 Table 10, $\alpha=0.49$ ).

1. Trial section: 2 ISMC 300

$A=2\times4564=9128$ mm², $I_{xx}=2\times6362.6\times10^4$ mm⁴, $r_{xx}=118.1$ mm, $C_{yy}=23.6$ mm, $b_f=90$ mm.

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{10000}{118.1}=84.7$
Effective slenderness of laced column $=1.05\lambda=88.9$ (cl. 7.6.1.4)
$f_{cc}=\dfrac{\pi^2E}{\lambda^2}$ , $\lambda_n=\sqrt{f_y/f_{cc}}$ , $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$ , $f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}=122.6$ N/mm²
$P_d=A f_{cd}=9128\times122.6\times10^{-3}=1119$ kN $>1000$ ✓

2. Spacing of channels (so that $I_{yy}\ge I_{xx}$ )

\left(C_{yy}+\frac{s}{2}\right)^2\ge\frac{I_{xx}-I_{yy}}{A_{ch}}\Rightarrow C_{yy}+\frac{s}{2}\ge115.2\Rightarrow s\ge183\ \text{mm}

Provide a clear gap $s=190$ mm between webs. Then $I_{yy}=2[I_{yy,ch}+A_{ch}(C_{yy}+s/2)^2]=13461\times10^4$ mm⁴, $r_{yy}=121.4$ mm, $\lambda_{yy}=82.3\ (<\lambda_{xx}),$ 1.05\lambda_{yy}=86.5 $,$ P_{dy}=1153$ kN ✓.

3. Lacing (single, bolted)

   |\     /|
   | \   / |   theta = 45 deg to axis
   |  \ /  |   S = distance between bolt lines
   |  / \  |
   | /   \ |

Distance between bolt lines (at middle of flanges) $S=s+b_f=190+90=280$ mm
Transverse shear $V=2.5\%P=25.0$ kN; two lacing planes: $V_p=12.50$ kN per plane
Lacing at $\theta=45°$ (limits 40°-70°): force $F=\dfrac{V_p}{\sin\theta}=17.68$ kN
Length of bar between end bolts $l=\dfrac{S}{\sin\theta}=396$ mm
Thickness $t\ge\dfrac{l}{40}=10$ mm (single lacing); width $\ge3d=60$ mm → flat 60 × 10 mm
Slenderness $\lambda=\dfrac{l}{t/\sqrt{12}}=137<145$ ✓; $f_{cd}=68$ N/mm² → compressive strength $=41.0$ kN $>F$ ✓; tension $T_{dn}=\dfrac{0.9(60-22)t f_u}{1.25}=112.2$ kN $>F$ ✓
Bolt connection: $V_{dsb}=45.3$ kN (single shear), bearing on 10 mm $=99.4$ kN → 1 M20 bolt at each end
Spacing of lacing points on one channel $=\dfrac{2S}{\tan\theta}=560$ mm; slenderness of the channel between them $=\dfrac{560}{26.1}=21.5\le\min(50,\,0.7\times89)$ ✓ (cl. 7.6.3).

4. Tie plates

At the ends, lacing is ended with tie (batten) plates, length $\ge S$ + edges $=360$ mm, depth $\ge S=280$ mm for a plate directly at the end, thickness $\ge S/50=6$ mm: tie plate 360 × 280 × 6 mm, connected by 2 M20 bolts to each channel.

Answer: 2 ISMC 300 back to back, clear gap 190 mm, single lacing 60 × 10 mm flats at 45° with M20 bolts, tie plates at the ends.

Most repeated · 4 of 21 exams
2079 Chaitra · 14 marks

Design a built-up column 8 m long to carry a factored axial compressive load of 1100 kN. The column is restrained in position but not in direction at both ends. Design the column with connecting system as lacing with bolted connections. Use two channel sections back to back. Use steel of grade Fe 410.

Similar questions: Built-up laced column, 10 m, 1000 kN (2075 Bhadra) · Built-up column with battens, 10 m, 1080 kN (2073 Magh)

Answer

Given/assumptions: factored load $P_u=1100$ kN; length 8 m, restrained in position but not in direction at both ends → $K=1.0$ , $KL=8$ m. Two channels back to back, single bolted lacing (M20, grade 4.6); Fe410 / E250 ( $f_y=250$ , $f_u=410$ ).

1. Section: 2 ISMC 300 back to back (webs facing, flanges outward)

$A=2\times4564=9128$ mm², $I_{xx}=2\times6362.6\times10^4$ mm⁴, $r_{xx}=118.1$ mm, $C_{yy}=23.6$ mm, $b_f=90$ mm. Built-up members use buckling class c (IS 800:2007 Table 10, $\alpha=0.49$ ).

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{8000}{118.1}=67.8$
Laced column: $\lambda_e=1.05\lambda=71.1$ (cl. 7.6.1.4)
$f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$ , $\lambda_n=\sqrt{f_y/f_{cc}}$ , $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=150.4$ N/mm²
$P_d=A f_{cd}=1373$ kN $\ge P_u=1100$ kN ✓

2. Clear gap for equal stiffness

Centroid of each section from the column axis $x_c=23.6+s/2$ . For $I_{yy}\ge I_{xx}$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=115.2\ \text{mm}\ \Rightarrow\ s\ge183\ \text{mm}

Provide $s=190$ mm. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=13461\times10^4$ mm⁴, $r_{yy}=121.4$ mm, $\lambda_{yy}=65.9$ , $1.05\lambda_{yy}=69.2$ , $P_{dy}=1402$ kN $\ge P_u$ ✓.

3. Design of lacing (single lacing, two planes)

   |\     /|
   | \   / |    theta = 45 deg to the axis
   |  \ /  |    S = distance between connection lines
   |  / \  |
   | /   \ |

Distance between connection lines $S=s+b_f=190+90=280$ mm
Transverse shear $V=2.5\%\,P_u=27.5$ kN; per lacing plane $V_p=13.75$ kN
Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=19.45$ kN
Length of bar $l=\dfrac{S}{\sin\theta}=396$ mm; effective length = $l$ (between inner end bolts)
$t\ge l/40=10$ mm; width $\ge3d=60$ mm → flat 60 × 10 mm
$\lambda=\dfrac{l}{t/\sqrt{12}}=137<145$ ✓ (cl. 7.6.3); $f_{cd}=68$ N/mm², $P_c=41.0$ kN $>F$ ✓
Tension: $T_{dn}=\dfrac{0.9(w-d_0)tf_u}{\gamma_{m1}}=112.2$ kN $>F$ ✓
Connection: M20 (4.6) $V_{dsb}=45.3$ kN, bearing on 10 mm $=99.4$ kN → 1 bolt(s) at each end
Lacing angle $\theta=45°$ lies between 40° and 70° ✓
Spacing of lacing points on one section $=\dfrac{2S}{\tan\theta}=560$ mm; slenderness of the section between them $=\dfrac{560}{26.1}=21.5\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

4. Tie plates

Tie (end) plates at both ends of the lacing system: length $\ge S+$ edge $=360$ mm, depth $\ge S=280$ mm, thickness $\ge S/50=6$ mm → 360 × 280 × 6 mm at each end, connected to each section by 2 bolts.

Answer: 2 ISMC 300 back to back (webs facing, flanges outward), clear gap 190 mm, single lacing of flats 60 × 10 mm at 45°, tie plates 360 × 280 × 6 mm at the ends.

Most repeated · 4 of 21 exams
2073 Magh · 15 marks

Design a built up column 10 m long to carry a factored axial compressive load of 1080 kN. The column is restrained in position but not in direction at both ends. Design the column with connecting system as battens with bolted connection. Use two channel back to back assume steel of grade Fe 410, E250A and bolts grade 4.6.

Similar questions: Built-up laced column, 10 m, 1000 kN (2075 Bhadra) · Built-up column, lacing, 8 m, 1100 kN (2079 Chaitra)

Answer

Approach. Battened column of two channels, designed to IS 800:2007 (cl. 7.1.2 for compressive strength and cl. 7.7 for battened columns). Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm²; bolts M16 grade 4.6 ( $f_{ub} = 400$ N/mm²), $\gamma_{m0} = 1.10$ , $\gamma_{mb} = 1.25$ .

Data

$P = 1080$ kN (factored). Ends restrained in position but not in direction, so $KL = 1.0 \times 10$ m $= 10000$ mm (Table 11).
Buckling curve c, $\alpha = 0.49$ (built-up member, Table 10). Effective slenderness of a battened column $= 1.1 \times$ actual maximum slenderness (cl. 7.7.1.4).

Step 1: Trial section

Try 2 × ISMC 350: $A = 54.4$ cm² each, $I_{xx} = 10000$ cm⁴, $I_{yy} = 434$ cm⁴, $C_{yy} = 2.44$ cm, $B = 100$ mm, $t_w = 8.3$ mm, $t_f = 13.5$ mm, $r_{y} = 2.82$ cm.

\begin{aligned} r_{zz} &= \sqrt{I_{xx}/A} = 13.56\ \text{cm} \\ \lambda_{zz} &= 10000/135.6 = 73.8 \\ \lambda_{eff} &= 1.1 \times 73.8 = 81.1 \end{aligned}

Curve c at $\lambda_{eff} = 81.1$ : $f_{cc} = 300$ N/mm², $\lambda_n = 0.913$ , $\phi = 1.092$ , $\chi = 0.592$ .

\begin{aligned} f_{cd} &= 0.592 \times 250/1.1 = 134.5\ \text{N/mm}^2 \\ P_d &= 2A f_{cd} = 10880 \times 134.5/10^3 = 1463\ \text{kN} > 1080\ \text{kN} \end{aligned}

Step 2: Spacing of channels

For $I_{yy} \ge I_{zz}$ : $h \ge 2\sqrt{(I_x - I_y)/A} = 2\sqrt{(10000 - 434) \times 10^4/5440} = 265$ mm.

Gap between webs $g = h - 2C_{yy} = 265 - 48.8 = 216$ mm; provide g = 220 mm, so $h = 269$ mm.

$I_{yy} = 2[434 + 54.4(13.44)^2] = 20521$ cm⁴, $r_{yy} = 13.73$ cm, $\lambda_{yy} = 72.8$ . This is not more than $\lambda_{zz} = 73.8$ , so the capacity $P_d = 1463$ kN is safe. Overall width $= 2B + g = 420$ mm.

Step 3: Spacing of battens (cl. 7.7.3)

The slenderness of one channel between batten connections must not exceed 50 or $0.7\lambda_{max} = 0.7 \times 73.8 = 51.6$ . So $\lambda_1 \le 50.0$ and with $r_{min} = r_y = 28.2$ mm:

\begin{aligned} C &\le 50.0 \times 28.2 = 1412\ \text{mm} \\ \text{Bays} &\ge 10000/1412 \to 8\ (\ge 3) \end{aligned}

The main-member check of Step 7 (bending from the batten moment) needs a smaller spacing, so adopt 11 bays: $C = 10000/11 = 909$ mm, $\lambda_1 = 909/28.2 = 32.2$ .

Step 4: Forces on battens (cl. 7.7.2.1)

Transverse shear $V_t = 0.025 \times 1080 = 27.0$ kN, $N = 2$ planes. Bolt lines are at mid flange, so $S = g + B = 220 + 100 = 320$ mm.

\begin{aligned} V_b &= \frac{V_t C}{N S} = \frac{27.0 \times 909}{2 \times 320} = 38.4\ \text{kN} \\ M &= \frac{V_t C}{2N} = \frac{27.0 \times 909}{4} = 6136\ \text{kN·mm} \end{aligned}

Step 5: Size of battens (cl. 7.7.2.3)

End battens: effective depth $\ge h = 269$ mm; intermediate $\ge 0.75h = 202$ mm; in no case $< 2B = 200$ mm. Use the same size throughout: effective depth 270 mm (between outer bolts), overall depth $= 270 + 2 \times 30 = 330$ mm.

Thickness $\ge S/50 = 320/50 = 6.4$ mm; provide 8 mm. Batten plate: 420 mm × 330 mm × 8 mm.

Plate check: $\sigma_b = M/Z = 6136364/145200 = 42.3$ N/mm² $< f_y/\gamma_{m0} = 227$ ; $\tau = V_b/(t\,d) = 14.5$ N/mm² $< f_y/(\sqrt3\gamma_{m0}) = 131$ . OK.

Step 6: Bolts (M16, grade 4.6)

$V_{dsb} = 29.0$ kN (single shear, threaded), $V_{dpb} = 58.3$ kN ( $t = 8$ mm, $k_b = 0.556$ ). Bolt value $= 29.0$ kN.

Try 5 bolts in a line at 50 mm pitch on each channel (extent 200 mm $\le$ 270 mm). Bolt force due to $M$ (farthest bolt) $= M r_{max}/\Sigma r^2 = 24.5$ kN; due to shear $= V_b/n = 7.7$ kN.

R = \sqrt{24.5^2 + 7.7^2} = 25.7\ \text{kN} \le 29.0\ \text{kN}

Provide 5 bolts M16 per batten per channel. Edge distance 30 mm, pitch 50 mm ( $\ge 2.5d = 40$ mm).

Step 7: Check of the main member (cl. 7.7.2.1)

Each channel carries $P/2 = 540$ kN with local moment $M = 6.14$ kN·m about its own $y$ -axis. $\lambda_1 = 32.2$ , $f_{cd} = 208.4$ N/mm², $P_{d1} = 5440 \times 208.4/10^3 = 1134$ kN. $M_d = Z_y f_y/\gamma_{m0} = 57.3 \times 10^3 \times 250/1.1 = 13.02$ kN·m ( $Z_y = 57.3$ cm³).

\frac{P/2}{P_{d1}} + \frac{M}{M_d} = \frac{540}{1134} + \frac{6.14}{13.02} = 0.95 < 1.0

The main member is safe.

Final design

2 × ISMC 350 back to back, gap 220 mm, $P_d = 1463$ kN $> 1080$ kN.
11 equal bays of 909 mm; battens 420 × 330 × 8 mm at each bay point and at both ends 5 M16 bolts per channel per batten.

  |==|            |==|
  |  |------------|  |  <- end batten
  |  |            |  |
  |  |------------|  |  <- batten
  |  |            |  |  (bay C = 909 mm)
  |  |------------|  |  <- batten
  |==|            |==|
  channel     channel

Most repeated · 3 of 21 exams
Asked 3 times
2078 Chaitra · 6 marks
2071 Magh · 8 marks
2068 Bhadra (old course) · 6 marks

Draw neat sketch of column slab base and write down design procedure (design steps of column bases / process to find thickness of slab base foundation).

Answer

Slab base

A slab base is a rectangular steel plate welded (or connected by cleats/angles) to the foot of the column. The column end is machined to bear on the plate, so most of the load goes by direct contact; the weld is nominal. The plate spreads the load over concrete and is fixed with anchor bolts.

        elevation                    plan
     |  |  column  |  |        +---------------+
     |  |          |  |        |  a            |
   ==|==|==========|==|==      |  +---------+  |
   base plate, ts, B x L        |  | column  |  | L
   ______________________       |  +---------+  |
   | concrete pedestal  |       |   b           |
   anchor bolts  o      o       +---------------+
                                       B

Design procedure (IS 800:2007 cl. 7.4)

Factored load $P$ and concrete grade; bearing strength of concrete $=0.45f_{ck}$ (cl. 7.4.1).
Required area $A_{req}=\dfrac{P}{0.45f_{ck}}$ .
Plate size: choose $B\times L$ so that $B\cdot L\ge A_{req}$ , with projections $\approx50$ mm or more beyond the column; the projections should be about equal to make the plate thickness economical.
Actual bearing pressure: $w=\dfrac{P}{B\,L}\le0.45f_{ck}$ .
Projections: $a$ = larger projection and $b$ = smaller projection of the plate beyond the column (flange tips and web faces).
Thickness (cl. 7.4.3.1):

t_s=\sqrt{\frac{2.5\,w\,(a^2-0.3b^2)\,\gamma_{m0}}{f_y}}\ \ge t_f

where $t_f$ is the column flange thickness, $\gamma_{m0}=1.10$ . 7. Connection: column to plate by nominal fillet weld (or full contact if milled) all round; provide at least 2 anchor bolts (usually 4) of 20 mm for fixing (design for any uplift or moment, if present). 8. Check: plate is thick enough ( $t_s\ge$ calculated, round up to 12, 16, 20, 25 mm...), concrete pressure not exceeded, and the foundation (pedestal or footing) is checked separately for bearing and size.

Asked 2 times
2073 Magh · 6 marks
2072 Asoj · 6 marks

Design a slab base for a column ISHB 350@710.2 N/m subjected to a factored axial compressive load of 1000 kN. Concrete pedestal of grade M20.

Answer

Data: ISHB 350 @ 710.2 N/m (72.4 kg/m): $h=350$ , $b_f=250$ , $t_f=11.6$ , $t_w=10.1$ mm. Factored load $P=1000$ kN; pedestal M20; steel E250 ( $f_y=250$ ).

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

$0.45f_{ck}=0.45\times20=9.0$ N/mm²

2. Plate size

A_{req}=\frac{1000\times10^3}{9.0}=111111\ \text{mm}^2

Provide equal projections of 50 mm: $L=350+2\times50=450$ mm, $B=250+2\times50=350$ mm. Area $=157500$ mm² $>A_{req}$ ✓

w=\frac{1000\times10^3}{450\times350}=6.35\ \text{N/mm}^2<9.0

3. Thickness (cl. 7.4.3.1)

Projections $a=b=50$ mm:

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times6.35\times(50^2-0.3\times50^2)\times1.10}{250}}=11.06\ \text{mm}

Also $t_s\ge t_f=11.6$ mm. Provide $t_s=12$ mm.

4. Connections

Column to plate: 6 mm fillet weld all round (nominal; column end machined for bearing).
Anchor bolts: 4 nos. M20 (minimum), embedded 450 mm, with washer plate, placed outside the flanges.

Answer: base plate 450 × 350 × 12 mm, 4 anchor bolts M20, 6 mm weld.

2069 Bhadra · 10 marks

Design the base plate for a ISHB 350 column to carry factored load of 1200KN. Take E250 grade of steel and M20 grade of concrete.

Similar questions: Base plate for ISHB 450, 1400 kN (2070 Bhadra)

Answer

Data: ISHB 350 @ 72.4 kg/m: $h=350$ , $b_f=250$ , $t_f=11.6$ , $t_w=10.1$ mm. Factored load $P=1200$ kN; steel E250; concrete M20.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

$0.45f_{ck}=0.45\times20=9.00$ N/mm²

2. Plate size

A_{req}=\frac{P}{0.45f_{ck}}=\frac{1200\times10^3}{9.00}=133333\ \text{mm}^2

Projection of about 50 mm all round the column (not less than 50 mm): $L=350+2\times50=450$ mm, $B=250+2\times50=350$ mm. Area provided $=157500$ mm² $>A_{req}$ ✓

w=\frac{P}{BL}=\frac{1200\times10^3}{450\times350}=7.62\ \text{N/mm}^2<9.00\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections $a=50$ mm and $b=50$ mm (larger and smaller):

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.62\times(50^2-0.3\times50^2)\times1.10}{250}}=12.11\ \text{mm}

It must also be not less than the flange thickness $t_f=11.6$ mm. Provide $t_s=14$ mm.

4. Connections

The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (load is transferred mainly by contact).
Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

Answer: base plate 450 × 350 × 14 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M20 concrete.

2070 Bhadra · 8 marks

Design the base plate for the column ISHB 450 to carry a factored load of 1400 kN. Take E250 grade of steel and M20 grade of concrete.

Similar questions: Base plate for ISHB 350, 1200 kN (2069 Bhadra)

Answer

Data: ISHB 450 @ 87.2 kg/m: $h=450$ , $b_f=250$ , $t_f=13.7$ , $t_w=9.8$ mm. Factored load $P=1400$ kN; steel E250; concrete M20.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

$0.45f_{ck}=0.45\times20=9.00$ N/mm²

2. Plate size

A_{req}=\frac{P}{0.45f_{ck}}=\frac{1400\times10^3}{9.00}=155556\ \text{mm}^2

Projection of about 50 mm all round the column (not less than 50 mm): $L=450+2\times50=550$ mm, $B=250+2\times50=350$ mm. Area provided $=192500$ mm² $>A_{req}$ ✓

w=\frac{P}{BL}=\frac{1400\times10^3}{550\times350}=7.27\ \text{N/mm}^2<9.00\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections $a=50$ mm and $b=50$ mm (larger and smaller):

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.27\times(50^2-0.3\times50^2)\times1.10}{250}}=11.83\ \text{mm}

It must also be not less than the flange thickness $t_f=13.7$ mm. Provide $t_s=14$ mm.

4. Connections

The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (load is transferred mainly by contact).
Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

Answer: base plate 550 × 350 × 14 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M20 concrete.

2080 Chaitra · 14 marks

Design a built-up column using double channel-section placed toe-to-toe, having unsupported length 4 m, subjected to factored axial compressive load of 1000 kN. Use batten as lattice member and bolted connection to connect batten and column. The ends of the columns are effectively held in position at both ends but not restrained against rotation.

Similar questions: Braced built-up column, 3 m, 1600 kN (2078 Chaitra)

Answer

Given/assumptions: factored load $P_u=1000$ kN; unsupported length 4 m; ends held in position but not restrained against rotation → $K=1.0$ , $KL=4$ m. Two channels toe to toe, battens, bolted connections (M20, 4.6); E250 ( $f_y=250$ , $f_u=410$ ).

1. Section: 2 ISMC 225 toe to toe (flanges facing)

$A=6734$ mm², $I_{xx}=2\times2442.8\times10^4$ mm⁴, $r_{xx}=85.2$ mm, $C_{yy}=23.0$ mm, $b_f=80$ mm, $r_{yy,ch}=23.6$ mm. Built-up members are buckling class c.

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{4000}{85.2}=47.0$
Battened column: $\lambda_e=1.1\lambda=51.7$ (cl. 7.6.1.5)
$f_{cd}=181.0$ N/mm² (curve c, $f_y=250$ , $\gamma_{m0}=1.10$ ) → $P_d=A f_{cd}=1219$ kN $\ge1000$ kN ✓

2. Clear gap (so that $I_{yy}\ge I_{xx}$ )

Centroid of each section from the column axis $x_c=57.0+s/2$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=81.8\ \text{mm}\Rightarrow s\ge50\ \text{mm}

Provide $s=50$ mm. $I_{yy}=4902\times10^4$ mm⁴, $r_{yy}=85.3$ mm, $1.1\lambda_{yy}=51.6$ , $P_{dy}=1220$ kN $\ge P_u$ ✓

3. Spacing of battens (cl. 7.7.1.4)

Slenderness of one section between battens $\le\min(50,\ 0.7\lambda_e)=36.2$ :

C\le 36.2\times23.6=853\ \text{mm}

For the length of 4000 mm provide 5 bays, i.e. 6 battens at $C=800$ mm c/c.

4. Size of battens (cl. 7.7.2)

Distance between the connection lines $a=S=s+b_f=130$ mm
End battens: depth $\ge a$ → 260 mm; intermediate battens: depth $\ge\max(0.75a,\ 2b_f)$ → 260 mm
Thickness $\ge a/50=130/50$ → 8 mm; length of batten $=a+80=210$ mm.

5. Design forces on a batten

Transverse shear $V_t=2.5\%P_u=25.0$ kN, number of batten planes $N=2$ :

V_b=\frac{V_tC}{NS}=\frac{25.0\times800}{2\times130}=76.92\ \text{kN},\qquad M=\frac{V_tC}{2N}=5.000\ \text{kN·m}

Plate in bending: $Z=\dfrac{t d^2}{6}=90133$ mm³ (intermediate batten); $\sigma=\dfrac{M}{Z}=55.5$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓ ( $M_d=20.48$ kN·m)
Shear: $\tau=\dfrac{V_b}{dt}=37.0$ N/mm² $<\dfrac{f_y}{\sqrt3\gamma_{m0}}=131$ N/mm² ✓

6. Bolts connecting battens to the sections (M20, 4.6)

4 bolts in a vertical line at each end (pitch 60 mm, end 40 mm): $V_{dsb}=45.3$ kN, bearing on 8 mm $=79.5$ kN → value $45.3$ kN.

Direct shear per bolt $=V_b/n=19.23$ kN; moment effect on extreme bolt $=\dfrac{M y_{max}}{\Sigma y^2}=25.00$ kN
Resultant $R=\sqrt{19.23^2+25.00^2}=31.54$ kN $<45.3$ kN ✓

Answer: 2 ISMC 225 toe to toe (flanges facing), clear gap 50 mm; battens 260 mm deep (end) and 260 mm deep (intermediate) × 8 mm thick at 800 mm c/c, with 4 M20 bolts at each end.

2078 Chaitra · 14 marks

Design of braced built-up column using double channel-section, having unsupported length 3.0m, subjected to factored axial load of 1600 kN. Use single lacing as lattice member and bolted connection to connect lacing and column. The ends of the columns are effectively held in position at both ends but not restrained against rotation.

Similar questions: Built-up column with battens, 4 m, 1000 kN (2080 Chaitra)

Answer

Given/assumptions: factored load $P_u=1600$ kN; unsupported length 3.0 m; ends held in position but not restrained in rotation → $K=1.0$ , $KL=3$ m. Two channels (back to back), single lacing, bolted (M20, grade 4.6); E250 ( $f_y=250$ , $f_u=410$ ).

1. Section: 2 ISMC 250 back to back (webs facing, flanges outward)

$A=2\times3867=7734$ mm², $I_{xx}=2\times3816.8\times10^4$ mm⁴, $r_{xx}=99.3$ mm, $C_{yy}=23.0$ mm, $b_f=80$ mm. Built-up members use buckling class c (IS 800:2007 Table 10, $\alpha=0.49$ ).

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{3000}{99.3}=30.2$
Laced column: $\lambda_e=1.05\lambda=31.7$ (cl. 7.6.1.4)
$f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$ , $\lambda_n=\sqrt{f_y/f_{cc}}$ , $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=209.1$ N/mm²
$P_d=A f_{cd}=1617$ kN $\ge P_u=1600$ kN ✓

2. Clear gap for equal stiffness

Centroid of each section from the column axis $x_c=23.0+s/2$ . For $I_{yy}\ge I_{xx}$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=96.5\ \text{mm}\ \Rightarrow\ s\ge147\ \text{mm}

Provide $s=150$ mm. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=7866\times10^4$ mm⁴, $r_{yy}=100.8$ mm, $\lambda_{yy}=29.7$ , $1.05\lambda_{yy}=31.2$ , $P_{dy}=1622$ kN $\ge P_u$ ✓.

3. Design of lacing (single lacing, two planes)

   |\     /|
   | \   / |    theta = 45 deg to the axis
   |  \ /  |    S = distance between connection lines
   |  / \  |
   | /   \ |

Distance between connection lines $S=s+b_f=150+80=230$ mm
Transverse shear $V=2.5\%\,P_u=40.0$ kN; per lacing plane $V_p=20.00$ kN
Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=28.28$ kN
Length of bar $l=\dfrac{S}{\sin\theta}=325$ mm; effective length = $l$ (between inner end bolts)
$t\ge l/40=10$ mm; width $\ge3d=60$ mm → flat 60 × 10 mm
$\lambda=\dfrac{l}{t/\sqrt{12}}=113<145$ ✓ (cl. 7.6.3); $f_{cd}=91$ N/mm², $P_c=54.9$ kN $>F$ ✓
Tension: $T_{dn}=\dfrac{0.9(w-d_0)tf_u}{\gamma_{m1}}=112.2$ kN $>F$ ✓
Connection: M20 (4.6) $V_{dsb}=45.3$ kN, bearing on 10 mm $=99.4$ kN → 1 bolt(s) at each end
Lacing angle $\theta=45°$ lies between 40° and 70° ✓
Spacing of lacing points on one section $=\dfrac{2S}{\tan\theta}=460$ mm; slenderness of the section between them $=\dfrac{460}{23.8}=19.3\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

4. Tie plates

Tie (end) plates at both ends of the lacing system: length $\ge S+$ edge $=310$ mm, depth $\ge S=230$ mm, thickness $\ge S/50=6$ mm → 310 × 230 × 6 mm at each end, connected to each section by 2 bolts.

Answer: 2 ISMC 250 back to back (webs facing, flanges outward), clear gap 150 mm, single lacing of flats 60 × 10 mm at 45°, tie plates 310 × 230 × 6 mm at the ends.

2077 Chaitra · 10 marks

Design slab base for a column SC300 carrying factor axial lead of 1200KN if concrete grade used is M20.

Answer

Assumption: the dimensions of "SC 300" are not in the section tables used here, so the column is taken as a stocky column section of overall size $h=300$ mm, $b_f=300$ mm and flange thickness $t_f\approx16$ mm (similar to a heavy column section); only $h$ , $b_f$ and $t_f$ matter here. Factored axial load $P=1200$ kN, pedestal M20, steel E250 ( $f_y=250$ ).

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

$0.45f_{ck}=0.45\times20=9.00$ N/mm²

2. Plate size

A_{req}=\frac{P}{0.45f_{ck}}=\frac{1200\times10^3}{9.00}=133333\ \text{mm}^2

Projection of about 50 mm all round the column (not less than 50 mm): $L=300+2\times50=400$ mm, $B=300+2\times50=400$ mm. Area provided $=160000$ mm² $>A_{req}$ ✓

w=\frac{P}{BL}=\frac{1200\times10^3}{400\times400}=7.50\ \text{N/mm}^2<9.00\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections $a=50$ mm and $b=50$ mm (larger and smaller):

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.50\times(50^2-0.3\times50^2)\times1.10}{250}}=12.02\ \text{mm}

It must also be not less than the flange thickness $t_f=16$ mm. Provide $t_s=16$ mm.

4. Connections

The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (load is transferred mainly by contact).
Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

Answer: base plate 400 × 400 × 16 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M20 concrete.

2081 Chaitra · 8 marks

Design a slab base for ISSC 200 which has to carry a design axial load of 1200 kN resting on a concrete pedestal with M 20 grade of concrete.

Answer

Assumption: the section "ISSC 200" (SC 200) is taken with overall size $h=200$ mm, $b_f=200$ mm and flange thickness $t_f\approx15$ mm (only these three values are needed for the base plate). Factored axial load $P=1200$ kN, pedestal M20, E250 steel.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

$0.45f_{ck}=0.45\times20=9.00$ N/mm²

2. Plate size

A_{req}=\frac{P}{0.45f_{ck}}=\frac{1200\times10^3}{9.00}=133333\ \text{mm}^2

Projection of about 85 mm all round the column (not less than 50 mm): $L=200+2\times85=370$ mm, $B=200+2\times85=370$ mm. Area provided $=136900$ mm² $>A_{req}$ ✓

w=\frac{P}{BL}=\frac{1200\times10^3}{370\times370}=8.77\ \text{N/mm}^2<9.00\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections $a=85$ mm and $b=85$ mm (larger and smaller):

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times8.77\times(85^2-0.3\times85^2)\times1.10}{250}}=22.08\ \text{mm}

It must also be not less than the flange thickness $t_f=15$ mm. Provide $t_s=25$ mm.

4. Connections

The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (load is transferred mainly by contact).
Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

Answer: base plate 370 × 370 × 25 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M20 concrete.

2079 Chaitra · 3 marks

Draw neat sketch of column gusseted base.

Answer

A gusseted base is used for heavy columns whose end is not machined. The column is connected through gusset plates and angle cleats (or channels) to the base plate, so the load spreads over a larger area of concrete.

   elevation                         plan
        | |  column |  |        +--------------------+
        | |         |  |        |  o              o  |
        | |  gusset |  |        |   +------------+   |
     ===|=|==plates=|==|===     |   |  column +  |   |
       /| |         |  |\       |   |  gussets   |   |
      / | |_________|  | \      |   +------------+   |
     /__|______________|__\     |  o              o  |
     angle cleats  base plate    +--------------------+
     ____________________________     anchor bolts o
          concrete pedestal

Main parts:

Column (rolled section, usually with flange cover plates) carrying the load.
Gusset plates welded to the column and to the base plate; they spread the load.
Gusset angles / cleats (two per gusset) connecting the gusset plate or column flange to the base plate by bolts or welds.
Base plate resting on the concrete pedestal, whose thickness is found from the bending of the plate between the stiffening plates.
Anchor bolts fixing the base to the foundation.
Grout between the base plate and the concrete.

2077 Chaitra · 10 marks

Design a gusseted base for a column ISHB 350 @ 710 N/m with two plates 450mm×20mm carrying factored load 2000KN. The column is to be supported on Concrete pedestal with M20 grade concrete.

Answer

Data and assumptions: ISHB 350 @ 710 N/m ( $h=350$ , $b_f=250$ , $t_f=11.6$ mm) with two 450 × 20 mm plates (one on each flange) forming the column shaft: overall shaft size $D=350+2\times20=390$ mm by $B_s=450$ mm. Factored load $P=2000$ kN; pedestal M20; E250 steel. Gusset angles (ISA 100×100×10) are used on the long sides.

        |==450x20 plate==|
        |    ISHB 350    |      shaft 390 x 450
        |==450x20 plate==|
     ISA cleats ---> |_|   |_| <--- ISA cleats
    ===================================
         base plate  L x B x t

1. Bearing and plate area

$0.45f_{ck}=9.0$ N/mm²; $A_{req}=\dfrac{2000\times10^3}{9.0}=222222$ mm². Provide base plate $L\times B=500\times520$ mm (area $260000$ mm² $>A_{req}$ ):

w=\frac{2000\times10^3}{500\times520}=7.69\ \text{N/mm}^2<9.0\ \text{N/mm}^2\ \checkmark

2. Thickness of base plate

Projections beyond the shaft: $a=\dfrac{500-390}{2}=55$ mm, $b=\dfrac{520-450}{2}=35$ mm.

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.69\times(55^2-0.3\times35^2)\times1.10}{250}}=15.0\ \text{mm}

Because the plate is also loaded between the gusset angles, provide $t_s=16$ mm (minimum 16 mm for a gusseted base).

3. Transfer of load from the shaft to the base plate

The column end is not machined, so the full 2000 kN is carried by fillet welds (shop welds, $s=10$ mm):

q=\frac{f_u}{\sqrt3\gamma_{mw}}\times0.7s=189.4\times0.7\times10=1326\ \text{N/mm},\qquad L_w=\frac{2000\times10^3}{1326}=1509\ \text{mm}

Weld all round the shaft and the cover plates: perimeter $2(390+450)=1680$ mm $\ge$ required; in addition provide the two gusset angles ISA 100×100×10, 450 mm long, welded to the cover plates and to the base plate with 8 mm fillets, which carry the balance. If the column is milled to bear, the welds need carry only a part.

4. Anchor bolts

4 nos. M24 (or more) near the corners, embedded 600 mm with plate washers.

Answer: base plate 500 × 520 × 16 mm; shaft ISHB 350 + 2 PL 450×20; ISA 100×100×10 gusset angles; 10 mm fillet weld; 4 M24 anchor bolts.

2071 Bhadra · 6 marks

Design a slab base for a column SC220 to transfer an axial load of 1000 kN. Take Fe410 grade steel and M30 for concrete.

Answer

Assumption: SC 220 taken as $h=220$ mm, $b_f=220$ mm, $t_f\approx14$ mm. The load of 1000 kN is the factored axial load. Concrete M30 ( $f_{ck}=30$ ), steel Fe410/E250 ( $f_y=250$ ).

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

$0.45f_{ck}=0.45\times30=13.50$ N/mm²

2. Plate size

A_{req}=\frac{P}{0.45f_{ck}}=\frac{1000\times10^3}{13.50}=74074\ \text{mm}^2

Projection of about 50 mm all round the column (not less than 50 mm): $L=220+2\times50=320$ mm, $B=220+2\times50=320$ mm. Area provided $=102400$ mm² $>A_{req}$ ✓

w=\frac{P}{BL}=\frac{1000\times10^3}{320\times320}=9.77\ \text{N/mm}^2<13.50\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections $a=50$ mm and $b=50$ mm (larger and smaller):

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times9.77\times(50^2-0.3\times50^2)\times1.10}{250}}=13.71\ \text{mm}

It must also be not less than the flange thickness $t_f=14$ mm. Provide $t_s=14$ mm.

4. Connections

The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (load is transferred mainly by contact).
Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

Answer: base plate 320 × 320 × 14 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M30 concrete.

2072 Magh · 8 marks

Design a slab base for a column ISMB 350 @ 52.4 kg/m to carry a service load of 850 KN. Assume Fe410 grade steel and M25 concrete.

Answer

Data: ISMB 350 @ 52.4 kg/m: $h=350$ , $b_f=140$ , $t_f=14.2$ , $t_w=8.1$ mm. Service load $850$ kN → factored load $P=1.5\times850=1275$ kN. Steel Fe410 ( $f_y=250$ ); concrete M25.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

$0.45f_{ck}=0.45\times25=11.25$ N/mm²

2. Plate size

A_{req}=\frac{P}{0.45f_{ck}}=\frac{1275\times10^3}{11.25}=113333\ \text{mm}^2

Projection of about 55 mm all round the column (not less than 50 mm): $L=350+2\times55=460$ mm, $B=140+2\times55=250$ mm. Area provided $=115000$ mm² $>A_{req}$ ✓

w=\frac{P}{BL}=\frac{1275\times10^3}{460\times250}=11.09\ \text{N/mm}^2<11.25\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections $a=55$ mm and $b=55$ mm (larger and smaller):

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times11.09\times(55^2-0.3\times55^2)\times1.10}{250}}=16.07\ \text{mm}

It must also be not less than the flange thickness $t_f=14.2$ mm. Provide $t_s=18$ mm.

4. Connections

The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (load is transferred mainly by contact).
Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

Answer: base plate 460 × 250 × 18 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M25 concrete.

2070 Magh · 10 marks

Design the foundation base for an ISHB 350 column to carry factored load of 120[?] KN. Assume steel and M20 concrete.

Answer

Reading of the doubtful value: the load is printed as "120[?] kN"; it is read as 1200 kN (factored), consistent with the similar ISHB 350 base-plate questions. If the load is really 120 kN, the minimum 50 mm projection and thickness $\ge t_f$ govern: a plate $350\times450\times12$ mm would do. Data: ISHB 350: $h=350$ , $b_f=250$ , $t_f=11.6$ mm; steel E250; concrete M20.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

$0.45f_{ck}=0.45\times20=9.00$ N/mm²

2. Plate size

A_{req}=\frac{P}{0.45f_{ck}}=\frac{1200\times10^3}{9.00}=133333\ \text{mm}^2

Projection of about 50 mm all round the column (not less than 50 mm): $L=350+2\times50=450$ mm, $B=250+2\times50=350$ mm. Area provided $=157500$ mm² $>A_{req}$ ✓

w=\frac{P}{BL}=\frac{1200\times10^3}{450\times350}=7.62\ \text{N/mm}^2<9.00\ \text{N/mm}^2

3. Thickness (cl. 7.4.3.1)

Projections $a=50$ mm and $b=50$ mm (larger and smaller):

t_s=\sqrt{\frac{2.5w(a^2-0.3b^2)\gamma_{m0}}{f_y}}=\sqrt{\frac{2.5\times7.62\times(50^2-0.3\times50^2)\times1.10}{250}}=12.11\ \text{mm}

It must also be not less than the flange thickness $t_f=11.6$ mm. Provide $t_s=14$ mm.

4. Connections

The column end is machined to bear on the plate; connect with a nominal 6 mm fillet weld all round (load is transferred mainly by contact).
Anchor bolts: 4 nos. M20, embedded about 450 mm with a washer plate (for erection and to resist any uplift).

Answer: base plate 450 × 350 × 14 mm, 4 anchor bolts M20, 6 mm weld, pedestal of M20 concrete.

2081 Chaitra · 14 marks

Design a built-up column consisting of two channels placed toe-to-toe. The column carries an axial factored load of 1500 kN. The effective length of column is 10 m. Also, design the lacing system using welded connection. Take Fe 415 grade steel.

Answer

Given/assumptions: factored load $P_u=1500$ kN; effective length $KL=10$ m (given); two channels toe to toe with single lacing and welded connection. "Fe 415" is read as ordinary structural steel with $f_y=250$ N/mm², $f_u=410$ N/mm² (IS 2062 E250); if $f_y=415$ were intended, redo with that value. Shop fillet welds.

1. Section: 2 ISMC 350 toe to toe (flanges facing, webs outward)

$A=2\times5366=10732$ mm², $I_{xx}=2\times10008.0\times10^4$ mm⁴, $r_{xx}=136.6$ mm, $C_{yy}=24.4$ mm, $b_f=100$ mm. Built-up members use buckling class c (IS 800:2007 Table 10, $\alpha=0.49$ ).

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{10000}{136.6}=73.2$
Laced column: $\lambda_e=1.05\lambda=76.9$ (cl. 7.6.1.4)
$f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$ , $\lambda_n=\sqrt{f_y/f_{cc}}$ , $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=141.2$ N/mm²
$P_d=A f_{cd}=1515$ kN $\ge P_u=1500$ kN ✓

2. Clear gap for equal stiffness

Centroid of each section from the column axis $x_c=75.6+s/2$ . For $I_{yy}\ge I_{xx}$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=133.6\ \text{mm}\ \Rightarrow\ s\ge116\ \text{mm}

Provide $s=120$ mm. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=20595\times10^4$ mm⁴, $r_{yy}=138.5$ mm, $\lambda_{yy}=72.2$ , $1.05\lambda_{yy}=75.8$ , $P_{dy}=1534$ kN $\ge P_u$ ✓.

3. Design of lacing (single lacing, two planes)

   |\     /|
   | \   / |    theta = 45 deg to the axis
   |  \ /  |    S = distance between connection lines
   |  / \  |
   | /   \ |

Distance between connection lines $S=s+b_f=120+100=220$ mm
Transverse shear $V=2.5\%\,P_u=37.5$ kN; per lacing plane $V_p=18.75$ kN
Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=26.52$ kN
Length of bar $l=\dfrac{S}{\sin\theta}=311$ mm; effective length for welded lacing $=0.7l=218$ mm
$t\ge l_e/40=8$ mm; width 50 mm → flat 50 × 8 mm
$\lambda=\dfrac{l_e}{t/\sqrt{12}}=94<145$ ✓; $f_{cd}=115$ N/mm², $P_c=45.9$ kN $>F$ ✓
Weld: size 5 mm shop fillet; strength $=\dfrac{410}{\sqrt3\times1.25}\times0.7\times5=662.8$ N/mm → length required $=\dfrac{F}{q}=40$ mm; provide 55 mm of weld at each end (not less than the bar width and $4s$ ), returned round the ends
Lacing angle $\theta=45°$ lies between 40° and 70° ✓
Spacing of lacing points on one section $=\dfrac{2S}{\tan\theta}=440$ mm; slenderness of the section between them $=\dfrac{440}{28.3}=15.5\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

4. Tie plates

Tie (end) plates at both ends of the lacing system: length $\ge S+$ edge $=300$ mm, depth $\ge S=220$ mm, thickness $\ge S/50=6$ mm → 300 × 220 × 6 mm at each end, connected to each section by 2 bolts.

Answer: 2 ISMC 350 toe to toe (flanges facing, webs outward), clear gap 120 mm, single lacing of flats 50 × 8 mm at 45°, tie plates 300 × 220 × 6 mm at the ends.

2080 Chaitra · 8 marks

Design splice plate and the connection for a column section ISHB 350 to support a factored axial load of 1000 kN. Use M20 bolts of grade 4.6 and steel of grade E250. Assume the column ends are machined. The connection is to be spliced at a height of 3 meters.

Answer

Data: ISHB 350 @ 72.4 kg/m ( $h=350$ , $b_f=250$ , $t_f=11.6$ , $t_w=10.1$ mm, $A=9221$ mm²), factored axial load $1000$ kN at the splice, ends machined for bearing. M20 bolts grade 4.6 ( $d_0=22$ mm); E250.

Design basis

Because the column ends are machined and in full contact, most of the load goes by direct bearing. The splice plates and bolts are designed to carry at least 50 % of the factored load ( $P_s$ ), and to hold the two lengths in line (bending from accidental eccentricity and erection loads).

P_s=0.5\times1000=500\ \text{kN}

Shared in proportion to the area: flanges $A_f=2\times250\times11.6=5800$ mm², web $A_w=(350-2\times11.6)\times10.1=3301$ mm².

Flanges: $P_f=P_s\dfrac{A_f}{A}=314.5$ kN → 157.2 kN per flange
Web: $P_w=P_s\dfrac{A_w}{A}=179.0$ kN

Bolt values (M20, 4.6)

Single shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN; double shear: $103.3$ kN
Bearing ( $e=40$ , $p=60$ ): $k_b=0.606$ ; on flange 11.6 mm $=115.3$ kN, on web 10.1 mm $=100.4$ kN
Flange bolt value $45.26$ kN (single shear, outer splice plate); web bolt value $100.4$ kN (double shear, two web plates).

Flange splice

$n=\dfrac{157.2}{45.26}=3.5$ → 4 bolts on each side of the joint per flange (2 rows of 2, pitch 60, gauge 120, end 40 mm). Flange splice plate $250\times10$ mm, length $=2\times(40+60+40)+$ gap $\approx300$ mm. Stress in the plate $=\dfrac{157.2\times10^3}{250\times10}=62.9$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓

Web splice

$n=\dfrac{179.0}{100.4}$ → 2 bolts per side in double shear; web plates (two, one on each face) $200\times8$ mm, length 250 mm, with 2 rows of bolts.

Notes

The splice is at 3 m height, so bending from lateral loads is small; the plates and bolts also hold the two lengths in line and resist erection loads.
Packing plates are provided if the two lengths differ in size.

Answer: flange splice plates 250 × 10 mm with 4 M20 bolts each side per flange; web splice plates 2 × (200 × 8 mm) with 2 M20 bolts each side.

2079 Chaitra · 8 marks

Determine design axial compressive load carrying capacity of a double angle IS 150×75×12 provided at opposite side of gusset plate, length of member is 3.5 m.

Answer

Data and assumptions: 2 ISA 150×75×12 on opposite faces of a gusset plate (taken as 10 mm thick), long legs (150 mm) in contact with the gusset, length $L=3.5$ m, steel E250 ( $f_y=250$ ). The angles are connected at each end by at least two bolts (or welds), so (IS 800:2007 cl. 7.5.2) the effective length is taken as $KL=0.85L$ (between 0.7L and 1.0L depending on end restraint). Buckling class c for angles (Table 10).

Section properties (from the geometry of the leg-thickness rectangles)

Area of one angle $A_a=2556$ mm² (SP 6: about 2540 mm²), $c_x=17.1$ mm (from the back of the 150 mm leg)
One angle: $I_{xx}=591.9\times10^4$ mm⁴ (axis parallel to the 75 mm leg), $I_{yy}=102.0\times10^4$ mm⁴ (axis parallel to the 150 mm leg)
Pair: $A=2A_a=5112$ mm²
About the axis parallel to the gusset: $I_y=2[I_{yy}+A_a(t_g/2+c_x)^2]=453.5\times10^4$ mm⁴, $r_y=29.8$ mm
About the axis perpendicular to the gusset: $I_x=2I_{xx}=1183.8\times10^4$ mm⁴, $r_x=48.1$ mm
$r_{min}=29.8$ mm

Slenderness and design stress

KL=0.85\times3500=2975\ \text{mm},\qquad \lambda=\frac{KL}{r_{min}}=\frac{2975}{29.8}=99.9\ (<180)

Using curve c ( $\alpha=0.49$ ): $f_{cc}=\dfrac{\pi^2E}{\lambda^2}$ , $\lambda_n=\sqrt{f_y/f_{cc}}$ , $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$ ,

f_{cd}=\frac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}=107.2\ \text{N/mm}^2

Capacity

P_d=A\,f_{cd}=5112\times107.2\times10^{-3}=548\ \text{kN}

Tack bolts (stitching) must be provided at spacing so that the slenderness of a single angle between them is not more than $0.7\lambda$ and the pair acts together.

Answer: design axial compressive capacity $\approx548$ kN.

2076 Baisakh · 12 marks

Design a built-up column to carry an axial load of 12 00kN and composed of two channels placed back to back. The effective length of the member is 6m. Design the column using battens system and Fe410 grade of steel.

Answer

Given/assumptions: axial load $P_u=1200$ kN (taken as factored); effective length 6 m ( $KL=6$ m); two channels back to back with battens, bolted (M20, grade 4.6); Fe410 / E250 ( $f_y=250$ , $f_u=410$ ).

1. Section: 2 ISMC 250 back to back (flanges outward)

$A=7734$ mm², $I_{xx}=2\times3816.8\times10^4$ mm⁴, $r_{xx}=99.3$ mm, $C_{yy}=23.0$ mm, $b_f=80$ mm, $r_{yy,ch}=23.8$ mm. Built-up members are buckling class c.

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{6000}{99.3}=60.4$
Battened column: $\lambda_e=1.1\lambda=66.4$ (cl. 7.6.1.5)
$f_{cd}=158.0$ N/mm² (curve c, $f_y=250$ , $\gamma_{m0}=1.10$ ) → $P_d=A f_{cd}=1222$ kN $\ge1200$ kN ✓

2. Clear gap (so that $I_{yy}\ge I_{xx}$ )

Centroid of each section from the column axis $x_c=23.0+s/2$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=96.5\ \text{mm}\Rightarrow s\ge147\ \text{mm}

Provide $s=150$ mm. $I_{yy}=7866\times10^4$ mm⁴, $r_{yy}=100.8$ mm, $1.1\lambda_{yy}=65.4$ , $P_{dy}=1234$ kN $\ge P_u$ ✓

3. Spacing of battens (cl. 7.7.1.4)

Slenderness of one section between battens $\le\min(50,\ 0.7\lambda_e)=46.5$ :

C\le 46.5\times23.8=1107\ \text{mm}

For the length of 6000 mm provide 6 bays, i.e. 7 battens at $C=1000$ mm c/c.

4. Size of battens (cl. 7.7.2)

Distance between the connection lines $a=S=s+b_f=230$ mm
End battens: depth $\ge a$ → 260 mm; intermediate battens: depth $\ge\max(0.75a,\ 2b_f)$ → 260 mm
Thickness $\ge a/50=230/50$ → 8 mm; length of batten $=a+80=310$ mm.

5. Design forces on a batten

Transverse shear $V_t=2.5\%P_u=30.0$ kN, number of batten planes $N=2$ :

V_b=\frac{V_tC}{NS}=\frac{30.0\times1000}{2\times230}=65.22\ \text{kN},\qquad M=\frac{V_tC}{2N}=7.500\ \text{kN·m}

Plate in bending: $Z=\dfrac{t d^2}{6}=90133$ mm³ (intermediate batten); $\sigma=\dfrac{M}{Z}=83.2$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓ ( $M_d=20.48$ kN·m)
Shear: $\tau=\dfrac{V_b}{dt}=31.4$ N/mm² $<\dfrac{f_y}{\sqrt3\gamma_{m0}}=131$ N/mm² ✓

6. Bolts connecting battens to the sections (M20, 4.6)

4 bolts in a vertical line at each end (pitch 60 mm, end 40 mm): $V_{dsb}=45.3$ kN, bearing on 8 mm $=79.5$ kN → value $45.3$ kN.

Direct shear per bolt $=V_b/n=16.30$ kN; moment effect on extreme bolt $=\dfrac{M y_{max}}{\Sigma y^2}=37.50$ kN
Resultant $R=\sqrt{16.30^2+37.50^2}=40.89$ kN $<45.3$ kN ✓

Answer: 2 ISMC 250 back to back (flanges outward), clear gap 150 mm; battens 260 mm deep (end) and 260 mm deep (intermediate) × 8 mm thick at 1000 mm c/c, with 4 M20 bolts at each end.

2076 Bhadra · 10 marks

Design a built up column using double channel section. Unsupported length of column is 6m, both end of column are restrained against lateral displacement but free in rotation. Column is subjected to design axial load of 1200 kN. Use battens.

Answer

Given/assumptions: design axial load $P_u=1200$ kN; unsupported length 6 m; both ends restrained against lateral displacement but free to rotate (hinged) → $K=1.0$ , $KL=6$ m. Two channels back to back with battens, bolted (M20, grade 4.6); E250 ( $f_y=250$ , $f_u=410$ ).

1. Section: 2 ISMC 250 back to back (flanges outward)

$A=7734$ mm², $I_{xx}=2\times3816.8\times10^4$ mm⁴, $r_{xx}=99.3$ mm, $C_{yy}=23.0$ mm, $b_f=80$ mm, $r_{yy,ch}=23.8$ mm. Built-up members are buckling class c.

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{6000}{99.3}=60.4$
Battened column: $\lambda_e=1.1\lambda=66.4$ (cl. 7.6.1.5)
$f_{cd}=158.0$ N/mm² (curve c, $f_y=250$ , $\gamma_{m0}=1.10$ ) → $P_d=A f_{cd}=1222$ kN $\ge1200$ kN ✓

2. Clear gap (so that $I_{yy}\ge I_{xx}$ )

Centroid of each section from the column axis $x_c=23.0+s/2$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=96.5\ \text{mm}\Rightarrow s\ge147\ \text{mm}

Provide $s=150$ mm. $I_{yy}=7866\times10^4$ mm⁴, $r_{yy}=100.8$ mm, $1.1\lambda_{yy}=65.4$ , $P_{dy}=1234$ kN $\ge P_u$ ✓

3. Spacing of battens (cl. 7.7.1.4)

Slenderness of one section between battens $\le\min(50,\ 0.7\lambda_e)=46.5$ :

C\le 46.5\times23.8=1107\ \text{mm}

For the length of 6000 mm provide 6 bays, i.e. 7 battens at $C=1000$ mm c/c.

4. Size of battens (cl. 7.7.2)

Distance between the connection lines $a=S=s+b_f=230$ mm
End battens: depth $\ge a$ → 260 mm; intermediate battens: depth $\ge\max(0.75a,\ 2b_f)$ → 260 mm
Thickness $\ge a/50=230/50$ → 8 mm; length of batten $=a+80=310$ mm.

5. Design forces on a batten

Transverse shear $V_t=2.5\%P_u=30.0$ kN, number of batten planes $N=2$ :

V_b=\frac{V_tC}{NS}=\frac{30.0\times1000}{2\times230}=65.22\ \text{kN},\qquad M=\frac{V_tC}{2N}=7.500\ \text{kN·m}

Plate in bending: $Z=\dfrac{t d^2}{6}=90133$ mm³ (intermediate batten); $\sigma=\dfrac{M}{Z}=83.2$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓ ( $M_d=20.48$ kN·m)
Shear: $\tau=\dfrac{V_b}{dt}=31.4$ N/mm² $<\dfrac{f_y}{\sqrt3\gamma_{m0}}=131$ N/mm² ✓

6. Bolts connecting battens to the sections (M20, 4.6)

4 bolts in a vertical line at each end (pitch 60 mm, end 40 mm): $V_{dsb}=45.3$ kN, bearing on 8 mm $=79.5$ kN → value $45.3$ kN.

Direct shear per bolt $=V_b/n=16.30$ kN; moment effect on extreme bolt $=\dfrac{M y_{max}}{\Sigma y^2}=37.50$ kN
Resultant $R=\sqrt{16.30^2+37.50^2}=40.89$ kN $<45.3$ kN ✓

Answer: 2 ISMC 250 back to back (flanges outward), clear gap 150 mm; battens 260 mm deep (end) and 260 mm deep (intermediate) × 8 mm thick at 1000 mm c/c, with 4 M20 bolts at each end.

2076 Bhadra · 6 marks

Explain about Buckling Behaviour of column.

Answer

Buckling of columns

A column under axial compression stays straight until a critical load is reached; beyond this it bends sideways (buckles) suddenly with a large lateral deflection, even though the stress may be well below the yield stress. Buckling is a loss of stability, not a material failure.

Euler's behaviour of an ideal (perfect) column

Euler's critical load for a perfectly straight, elastic, pin-ended column:

P_{cr}=\frac{\pi^2EI}{(KL)^2},\qquad f_{cc}=\frac{\pi^2E}{\lambda^2},\ \ \lambda=\frac{KL}{r}

The load is directly proportional to $EI$ and inversely to the square of the effective length $KL$ .
A slender column (large $\lambda$ ) buckles elastically at $f_{cc}<f_y$ ; a stocky column (small $\lambda$ ) reaches yield before buckling.
Buckling occurs about the axis of least resistance (largest slenderness).

 load P          Euler hyperbola
  fy |--------.
     |         \.
     |   real    \.   Euler curve
     |   curve ..  `---.____  fcc = pi^2 E/lambda^2
     +----------------------------> lambda

Behaviour of real columns

Real columns differ from Euler's ideal column because of:

Initial crookedness (out-of-straightness), usually $L/1000$ .
Eccentricity of load and non-uniform end restraint.
Residual stresses from rolling/welding, which cause early yielding of parts of the section.
Material non-linearity, yielding in the inelastic range for intermediate slenderness.
Local buckling of thin flanges or webs, and torsional / flexural-torsional buckling of open sections (angles, channels, tees).

For intermediate $\lambda$ the strength falls below both the yield load and the Euler load, so a smooth curve is used (Perry-Robertson):

f_{cd}=\frac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}},\quad \phi=0.5[1+\alpha(\lambda_n-0.2)+\lambda_n^2],\ \ \lambda_n=\sqrt{f_y/f_{cc}}

The imperfection factor $\alpha$ depends on the buckling class: a (0.21), b (0.34), c (0.49), d (0.76), chosen from IS 800:2007 Table 10 according to section type, axis and flange thickness.

Effective length

Support conditions change the buckled shape: pinned-pinned $K=1.0$ ; fixed-fixed $0.65$ (design 0.8); fixed-pinned $0.8$ (design 0.8-0.85); fixed-free $2.0$ (design 2.0). A column that is free to sway has larger $KL$ .

Remedies

Reduce $KL$ (bracing, intermediate restraints), increase $r$ (use a deeper or hollow/built-up section), use stocky compact sections, and provide adequate stiffness of end connections.

2075 Baisakh · 20 marks

Design a column section subjected to axial load of 1500 KN using rolled steel 'I' sections. Height of the column is 8 m. Both ends of column are fixed and exits sway condition. Also design lateral bracing of column using battening system.

Answer

Given/assumptions: factored load $P_u=1500$ kN; height 8 m; both ends fixed against rotation but the frame is free to sway → effective length factor $K=1.2$ (IS 800:2007 Table 11, recommended value), so $KL=1.2\times8000=9600$ mm. The column is built up of two rolled I-sections (ISMB) placed side by side and tied by battens (bolted, M20 4.6); E250 ( $f_y=250$ , $f_u=410$ ). Battens are on the flange faces, so the bolt lines are at the middle of the flanges, $S=s+b_f$ .

1. Section: 2 ISMB 350 placed side by side with a gap (webs parallel)

$A=13342$ mm², $I_{xx}=2\times13630.3\times10^4$ mm⁴, $r_{xx}=142.9$ mm, $C_{yy}=70$ mm, $b_f=140$ mm, $r_{yy,ch}=28.4$ mm. Built-up members are buckling class c.

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{9600}{142.9}=67.2$
Battened column: $\lambda_e=1.1\lambda=73.9$ (cl. 7.6.1.5)
$f_{cd}=146.0$ N/mm² (curve c, $f_y=250$ , $\gamma_{m0}=1.10$ ) → $P_d=A f_{cd}=1948$ kN $\ge1500$ kN ✓

2. Clear gap (so that $I_{yy}\ge I_{xx}$ )

Centroid of each section from the column axis $x_c=70.0+s/2$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=140.1\ \text{mm}\Rightarrow s\ge140\ \text{mm}

Provide $s=150$ mm. $I_{yy}=29127\times10^4$ mm⁴, $r_{yy}=147.8$ mm, $1.1\lambda_{yy}=71.5$ , $P_{dy}=2000$ kN $\ge P_u$ ✓

3. Spacing of battens (cl. 7.7.1.4)

Slenderness of one section between battens $\le\min(50,\ 0.7\lambda_e)=50.0$ :

C\le 50.0\times28.4=1420\ \text{mm}

For the length of 8000 mm provide 6 bays, i.e. 7 battens at $C=1333$ mm c/c.

4. Size of battens (cl. 7.7.2)

Distance between the connection lines $a=S=s+b_f=290$ mm
End battens: depth $\ge a$ → 320 mm; intermediate battens: depth $\ge\max(0.75a,\ 2b_f)$ → 320 mm
Thickness $\ge a/50=290/50$ → 8 mm; length of batten $=a+80=370$ mm.

5. Design forces on a batten

Transverse shear $V_t=2.5\%P_u=37.5$ kN, number of batten planes $N=2$ :

V_b=\frac{V_tC}{NS}=\frac{37.5\times1333}{2\times290}=86.21\ \text{kN},\qquad M=\frac{V_tC}{2N}=12.500\ \text{kN·m}

Plate in bending: $Z=\dfrac{t d^2}{6}=136533$ mm³ (intermediate batten); $\sigma=\dfrac{M}{Z}=91.6$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓ ( $M_d=31.03$ kN·m)
Shear: $\tau=\dfrac{V_b}{dt}=33.7$ N/mm² $<\dfrac{f_y}{\sqrt3\gamma_{m0}}=131$ N/mm² ✓

6. Bolts connecting battens to the sections (M20, 4.6)

5 bolts in a vertical line at each end (pitch 60 mm, end 40 mm): $V_{dsb}=45.3$ kN, bearing on 8 mm $=79.5$ kN → value $45.3$ kN.

Direct shear per bolt $=V_b/n=17.24$ kN; moment effect on extreme bolt $=\dfrac{M y_{max}}{\Sigma y^2}=41.67$ kN
Resultant $R=\sqrt{17.24^2+41.67^2}=45.09$ kN $<45.3$ kN ✓

Answer: 2 ISMB 350 placed side by side with a gap (webs parallel), clear gap 150 mm; battens 320 mm deep (end) and 320 mm deep (intermediate) × 8 mm thick at 1333 mm c/c, with 5 M20 bolts at each end.

2074 Bhadra · 12 marks

Design a built up column to carry an axial load of 1100 KN. The length of column is 8m and is effectively held in position at both ends but not restrained against rotation. Use single lacing system with bolted connection. Grade of steel E250, M10 [?] Bolt, 4.6 grade. The built up column should be consists of double channel back to back.

Answer

Reading of the doubtful value: the bolt size is printed "M10 [?]". M10 is too small for the plate and flange used here, so M20 is adopted (grade 4.6); with M10 two bolts per end would be needed.

Given/assumptions: factored load $P_u=1100$ kN; length 8 m, held in position at both ends but not restrained against rotation → $K=1.0$ , $KL=8$ m. Two channels back to back, single bolted lacing; E250 ( $f_y=250$ , $f_u=410$ ).

1. Section: 2 ISMC 300 back to back (webs facing, flanges outward)

$A=2\times4564=9128$ mm², $I_{xx}=2\times6362.6\times10^4$ mm⁴, $r_{xx}=118.1$ mm, $C_{yy}=23.6$ mm, $b_f=90$ mm. Built-up members use buckling class c (IS 800:2007 Table 10, $\alpha=0.49$ ).

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{8000}{118.1}=67.8$
Laced column: $\lambda_e=1.05\lambda=71.1$ (cl. 7.6.1.4)
$f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$ , $\lambda_n=\sqrt{f_y/f_{cc}}$ , $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=150.4$ N/mm²
$P_d=A f_{cd}=1373$ kN $\ge P_u=1100$ kN ✓

2. Clear gap for equal stiffness

Centroid of each section from the column axis $x_c=23.6+s/2$ . For $I_{yy}\ge I_{xx}$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=115.2\ \text{mm}\ \Rightarrow\ s\ge183\ \text{mm}

Provide $s=190$ mm. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=13461\times10^4$ mm⁴, $r_{yy}=121.4$ mm, $\lambda_{yy}=65.9$ , $1.05\lambda_{yy}=69.2$ , $P_{dy}=1402$ kN $\ge P_u$ ✓.

3. Design of lacing (single lacing, two planes)

   |\     /|
   | \   / |    theta = 45 deg to the axis
   |  \ /  |    S = distance between connection lines
   |  / \  |
   | /   \ |

Distance between connection lines $S=s+b_f=190+90=280$ mm
Transverse shear $V=2.5\%\,P_u=27.5$ kN; per lacing plane $V_p=13.75$ kN
Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=19.45$ kN
Length of bar $l=\dfrac{S}{\sin\theta}=396$ mm; effective length = $l$ (between inner end bolts)
$t\ge l/40=10$ mm; width $\ge3d=60$ mm → flat 60 × 10 mm
$\lambda=\dfrac{l}{t/\sqrt{12}}=137<145$ ✓ (cl. 7.6.3); $f_{cd}=68$ N/mm², $P_c=41.0$ kN $>F$ ✓
Tension: $T_{dn}=\dfrac{0.9(w-d_0)tf_u}{\gamma_{m1}}=112.2$ kN $>F$ ✓
Connection: M20 (4.6) $V_{dsb}=45.3$ kN, bearing on 10 mm $=99.4$ kN → 1 bolt(s) at each end
Lacing angle $\theta=45°$ lies between 40° and 70° ✓
Spacing of lacing points on one section $=\dfrac{2S}{\tan\theta}=560$ mm; slenderness of the section between them $=\dfrac{560}{26.1}=21.5\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

4. Tie plates

Tie (end) plates at both ends of the lacing system: length $\ge S+$ edge $=360$ mm, depth $\ge S=280$ mm, thickness $\ge S/50=6$ mm → 360 × 280 × 6 mm at each end, connected to each section by 2 bolts.

Answer: 2 ISMC 300 back to back (webs facing, flanges outward), clear gap 190 mm, single lacing of flats 60 × 10 mm at 45°, tie plates 360 × 280 × 6 mm at the ends.

2074 Bhadra · 8 marks

An ISHB 250 @ 536 N/m column carrying a factored axial load of 900KN. The column ends machined. Design the splice connection. Use M16 bolts.

Answer

Data: ISHB 250 @ 536 N/m ( $h=250$ , $b_f=250$ , $t_f=9.7$ , $t_w=8.8$ mm, $A=6971$ mm²), factored axial load 900 kN, ends machined for bearing; M16 bolts (grade 4.6, $d_0=18$ mm); E250.

Design basis

With machined ends in contact, the load is mostly transferred by bearing; the splice plates and bolts are designed for 50 % of the factored load, shared by area:

$P_s=0.5\times900=450$ kN
Flanges $A_f=2\times250\times9.7=4850$ mm²; web $A_w=(250-2\times9.7)\times8.8=2029$ mm²
Flange share $P_f=P_s\dfrac{A_f}{A}=313.1$ kN (156.5 kN per flange); web share $P_w=131.0$ kN

Bolt values (M16, 4.6)

Single shear $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=29.01$ kN; double shear $=66.1$ kN
Bearing ( $e=35$ , $p=50$ ): $k_b=0.648$ ; flange 9.7 mm: $82.5$ kN; web 8.8 mm: $74.8$ kN
Flange bolt value $29.01$ kN; web bolt value $66.1$ kN

Flange splice

$n=\dfrac{156.5}{29.01}=5.4$ → 6 bolts each side per flange (2 rows of 3, pitch 50 mm, gauge 100 mm, end 35 mm). Outer flange plate $250\times10$ mm × about 340 mm long. Plate stress $=\dfrac{156.5\times10^3}{250\times10}=62.6$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓

Web splice

Two web plates (one each face), double shear: $n=\dfrac{131.0}{66.1}$ → 2 bolts each side; web plates $170\times8$ mm × 200 mm long.

Answer: flange plates 250 × 10 mm with 6 M16 bolts each side per flange; two web plates 170 × 8 mm with 2 M16 bolts each side.

2073 Bhadra · 15 marks

A 7.5m long built-up and laced column has to carry a factored axial load of 1250KN. The column is restrained in position but not in direction at each end. Design the column with single lacing system. Connection shall consist of two channels placed back to back at a suitable spacing.

Answer

Given/assumptions: factored load $P_u=1250$ kN; length 7.5 m, restrained in position but not in direction at each end → $K=1.0$ , $KL=7.5$ m. Two channels back to back at a suitable spacing with single lacing; bolted connections (M20, grade 4.6); steel E250 ( $f_y=250$ , $f_u=410$ ).

1. Section: 2 ISMC 300 back to back (webs facing, flanges outward)

$A=2\times4564=9128$ mm², $I_{xx}=2\times6362.6\times10^4$ mm⁴, $r_{xx}=118.1$ mm, $C_{yy}=23.6$ mm, $b_f=90$ mm. Built-up members use buckling class c (IS 800:2007 Table 10, $\alpha=0.49$ ).

$\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{7500}{118.1}=63.5$
Laced column: $\lambda_e=1.05\lambda=66.7$ (cl. 7.6.1.4)
$f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$ , $\lambda_n=\sqrt{f_y/f_{cc}}$ , $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=157.5$ N/mm²
$P_d=A f_{cd}=1438$ kN $\ge P_u=1250$ kN ✓

2. Clear gap for equal stiffness

Centroid of each section from the column axis $x_c=23.6+s/2$ . For $I_{yy}\ge I_{xx}$ :

x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=115.2\ \text{mm}\ \Rightarrow\ s\ge183\ \text{mm}

Provide $s=190$ mm. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=13461\times10^4$ mm⁴, $r_{yy}=121.4$ mm, $\lambda_{yy}=61.8$ , $1.05\lambda_{yy}=64.8$ , $P_{dy}=1465$ kN $\ge P_u$ ✓.

3. Design of lacing (single lacing, two planes)

   |\     /|
   | \   / |    theta = 45 deg to the axis
   |  \ /  |    S = distance between connection lines
   |  / \  |
   | /   \ |

Distance between connection lines $S=s+b_f=190+90=280$ mm
Transverse shear $V=2.5\%\,P_u=31.2$ kN; per lacing plane $V_p=15.62$ kN
Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=22.10$ kN
Length of bar $l=\dfrac{S}{\sin\theta}=396$ mm; effective length = $l$ (between inner end bolts)
$t\ge l/40=10$ mm; width $\ge3d=60$ mm → flat 60 × 10 mm
$\lambda=\dfrac{l}{t/\sqrt{12}}=137<145$ ✓ (cl. 7.6.3); $f_{cd}=68$ N/mm², $P_c=41.0$ kN $>F$ ✓
Tension: $T_{dn}=\dfrac{0.9(w-d_0)tf_u}{\gamma_{m1}}=112.2$ kN $>F$ ✓
Connection: M20 (4.6) $V_{dsb}=45.3$ kN, bearing on 10 mm $=99.4$ kN → 1 bolt(s) at each end
Lacing angle $\theta=45°$ lies between 40° and 70° ✓
Spacing of lacing points on one section $=\dfrac{2S}{\tan\theta}=560$ mm; slenderness of the section between them $=\dfrac{560}{26.1}=21.5\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

4. Tie plates

Tie (end) plates at both ends of the lacing system: length $\ge S+$ edge $=360$ mm, depth $\ge S=280$ mm, thickness $\ge S/50=6$ mm → 360 × 280 × 6 mm at each end, connected to each section by 2 bolts.

Answer: 2 ISMC 300 back to back (webs facing, flanges outward), clear gap 190 mm, single lacing of flats 60 × 10 mm at 45°, tie plates 360 × 280 × 6 mm at the ends.

2072 Asoj · 16 marks

Design column to carry an axial load of 1200 kN. The column is effectively held in position but not restrained against rotation at both ends. Design the column using two channels placed toe to toe if center to center distance between connections is 6 m. Design the column using lacing and Fe 410 steel.

Answer

Approach. Laced column of two channels (toe to toe), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$ .

Data

Factored axial load $P = 1200$ kN; effective length $KL = 6.0$ m (both axes, ends held in position, not restrained against rotation, $K = 1.0$ , Table 11).
Built-up member: buckling curve c, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5).

Step 1: Trial section

Try 2 × ISMC 250 (IS 808): $A = 39.0$ cm² each, $I_{xx} = 3880$ cm⁴, $I_{yy} = 211$ cm⁴, $C_{yy} = 2.30$ cm, $B = 80$ mm, $t_w = 7.2$ mm, $t_f = 14.1$ mm.

Total area $A_e = 2 \times 3900 = 7800$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 9.97$ cm.

\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{6000}{99.7} = 60.2 \\ \lambda_{eff} &= 1.05 \times 60.2 = 63.2 \end{aligned}

For $\lambda_{eff} = 63.2$ , curve c: $f_{cc} = \pi^2E/\lambda^2 = 495$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.711$ , $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.878$ , $\chi = 0.718$ .

\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.718 \times 250/1.1 = 163.2\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 7800 \times 163.2/10^3 = 1273\ \text{kN} > 1200\ \text{kN} \end{aligned}

Step 2: Spacing of channels

To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels:

\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(3880 - 211) \times 10^4}{3900}} = 194\ \text{mm} \end{aligned}

Gap between toes $g = h - 2(B - C_{yy}) = 194 - 114.0 = 80$ mm. Provide g = 80 mm (centroid distance $h = 194$ mm).

\begin{aligned} I_{yy} &= 2[211 + 39.0(9.70)^2] = 7761\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 9.97\ \text{cm}, \quad \lambda_{yy} = 6000/99.7 = 60.2 \end{aligned}

$\lambda_{yy} = 60.2 \le \lambda_{zz} = 60.2$ , so the zz-axis governs and the capacity $P_d = 1273$ kN $\ge 1200$ kN. Safe. Overall width of column $= 2B + g = 240$ mm.

  +-------+   gap g   +-------+
  | ISMC  |<--------->| ISMC  |
  +-------+           +-------+
  flanges point towards each other (webs outside)
  lacing bolted to the flange tips (top and bottom)

Step 3: Lacing system (single lacing, two planes)

Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 1200 = 30.0$ kN; shared by two lacing planes: $15.00$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 80 + 80 = 160$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 160/\sin 45^\circ = 226\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{15.00}{\sin 45^\circ} = 21.21\ \text{kN} \end{aligned}

Spacing of connections along one channel $a_1 = 2s\cot\theta = 320$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 320/23.3 = 13.8$ , which is less than the smaller of 50 and $0.7\lambda_{max} = 42.1$ . OK.

Flat size. Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 5.7$ mm (cl. 7.6.3) and for $KL/r \le 145$ , $t \ge l\sqrt{12}/145 = 5.4$ mm. Provide ISF 60 × 6 mm.

$r = t/\sqrt{12} = 1.73$ mm, $l/r = 130.6 < 145$ OK. $f_{cd}$ (curve c) $= 73.7$ N/mm², $P_{d,lace} = 360 \times 73.7/10^3 = 26.5$ kN $> F = 21.21$ kN. OK.

Step 4: Connection of lacing (M16, grade 4.6 bolts)

\begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 6 \times 410}{1.25 \times 10^3} = 43.7\ \text{kN} \end{aligned}

( $k_b = e/3d_0 = 30/54 = 0.556$ with edge distance 30 mm $\ge 1.5d_0$ .) Bolt value $= 29.0$ kN. Bolts needed $= 21.21/29.0 = 0.73$ , so provide 1 bolt(s) M16 at each end of every bar.

Net section of flat in tension: $A_n = (60 - 18) \times 6 = 252$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 74.4$ kN $> 21.21$ kN. OK.

Step 5: End tie plates

Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth $\ge h = 194$ mm; thickness $\ge s/50 = 3.2$ mm. Provide 240 mm wide × 200 mm deep × 6 mm thick.

With $C = 320$ mm, $N=2$ : $V_b = V_tC/(NS) = 30.0$ kN and $M = V_tC/2N = 2400$ kN·mm. Using 3 M16 bolts at 50 mm pitch on each channel: max bolt force from $M$ $= 24.0$ kN, from shear $= 10.0$ kN, resultant $= 26.0$ kN $\le 29.0$ kN. OK.

Final design

Column: 2 × ISMC 250 toe to toe, gap 80 mm, $P_d = 1273$ kN $> 1200$ kN.
Lacing: single, flats 60 × 6 mm at 45°, M16 bolts, 1 per end.
End tie plates 240 × 200 × 6 mm.

  +--------------------+   <- end tie plate
  | \    /\    /\    /|
  |  \  /  \  /  \  / |   single lacing
  |   \/    \/    \/  |   (same direction
  |   /\    /\    /\  |    on both faces)
  +--------------------+   <- end tie plate

2071 Bhadra · 14 marks

Design a built-up column of the effective length of 5 m to carry an axial load of 900 kN using two channels and single lacing. Design the connections using bolt. The grade of the steel is Fe410.

Answer

Approach. Laced column of two channels (back to back), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$ .

Data

Factored axial load $P = 900$ kN; effective length $KL = 5.0$ m (given, same for both axes).
Built-up member: buckling curve c, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5).

Step 1: Trial section

Try 2 × ISMC 200 (IS 808): $A = 28.5$ cm² each, $I_{xx} = 1830$ cm⁴, $I_{yy} = 141$ cm⁴, $C_{yy} = 2.20$ cm, $B = 75$ mm, $t_w = 6.2$ mm, $t_f = 11.4$ mm.

Total area $A_e = 2 \times 2850 = 5700$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 8.01$ cm.

\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{5000}{80.1} = 62.4 \\ \lambda_{eff} &= 1.05 \times 62.4 = 65.5 \end{aligned}

For $\lambda_{eff} = 65.5$ , curve c: $f_{cc} = \pi^2E/\lambda^2 = 460$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.737$ , $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.903$ , $\chi = 0.701$ .

\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.701 \times 250/1.1 = 159.4\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 5700 \times 159.4/10^3 = 909\ \text{kN} > 900\ \text{kN} \end{aligned}

Step 2: Spacing of channels

To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels:

\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(1830 - 141) \times 10^4}{2850}} = 154\ \text{mm} \end{aligned}

Gap between webs $g = h - 2C_{yy} = 154 - 44.0 = 110$ mm. Provide g = 110 mm (centroid distance $h = 154$ mm).

\begin{aligned} I_{yy} &= 2[141 + 28.5(7.70)^2] = 3662\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 8.01\ \text{cm}, \quad \lambda_{yy} = 5000/80.1 = 62.4 \end{aligned}

$\lambda_{yy} = 62.4 \le \lambda_{zz} = 62.4$ , so the zz-axis governs and the capacity $P_d = 909$ kN $\ge 900$ kN. Safe. Overall width of column $= 2B + g = 260$ mm.

  +-------+   gap g   +-------+
  | ISMC  |<--------->| ISMC  |
  +-------+           +-------+
  webs face each other, flanges point outward
  lacing bolted to the flanges (top and bottom)

Step 3: Lacing system (single lacing, two planes)

Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 900 = 22.5$ kN; shared by two lacing planes: $11.25$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 110 + 75 = 185$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 185/\sin 45^\circ = 262\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{11.25}{\sin 45^\circ} = 15.91\ \text{kN} \end{aligned}

Spacing of connections along one channel $a_1 = 2s\cot\theta = 370$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 370/22.2 = 16.6$ , which is less than the smaller of 50 and $0.7\lambda_{max} = 43.7$ . OK.

Flat size. Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 6.5$ mm (cl. 7.6.3) and for $KL/r \le 145$ , $t \ge l\sqrt{12}/145 = 6.3$ mm. Provide ISF 60 × 8 mm.

$r = t/\sqrt{12} = 2.31$ mm, $l/r = 113.3 < 145$ OK. $f_{cd}$ (curve c) $= 90.8$ N/mm², $P_{d,lace} = 480 \times 90.8/10^3 = 43.6$ kN $> F = 15.91$ kN. OK.

Step 4: Connection of lacing (M16, grade 4.6 bolts)

\begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 8 \times 410}{1.25 \times 10^3} = 58.3\ \text{kN} \end{aligned}

( $k_b = e/3d_0 = 30/54 = 0.556$ with edge distance 30 mm $\ge 1.5d_0$ .) Bolt value $= 29.0$ kN. Bolts needed $= 15.91/29.0 = 0.55$ , so provide 1 bolt(s) M16 at each end of every bar.

Net section of flat in tension: $A_n = (60 - 18) \times 8 = 336$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 99.2$ kN $> 15.91$ kN. OK.

Step 5: End tie plates

Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth $\ge h = 154$ mm; thickness $\ge s/50 = 3.7$ mm. Provide 260 mm wide × 160 mm deep × 6 mm thick.

With $C = 370$ mm, $N=2$ : $V_b = V_tC/(NS) = 22.5$ kN and $M = V_tC/2N = 2081$ kN·mm. Using 3 M16 bolts at 50 mm pitch on each channel: max bolt force from $M$ $= 20.8$ kN, from shear $= 7.5$ kN, resultant $= 22.1$ kN $\le 29.0$ kN. OK.

Final design

Column: 2 × ISMC 200 back to back, gap 110 mm, $P_d = 909$ kN $> 900$ kN.
Lacing: single, flats 60 × 8 mm at 45°, M16 bolts, 1 per end.
End tie plates 260 × 160 × 6 mm.

  +--------------------+   <- end tie plate
  | \    /\    /\    /|
  |  \  /  \  /  \  / |   single lacing
  |   \/    \/    \/  |   (same direction
  |   /\    /\    /\  |    on both faces)
  +--------------------+   <- end tie plate

2069 Bhadra · 14 marks

Design a built-up column of the effective length of 6m to carry an axial load of 1000KN using two channels and laces. Design the connections using welds. The grade of the steel is E250C.

Answer

Approach. Laced column of two channels (back to back), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$ .

Data

Factored axial load $P = 1000$ kN; effective length $KL = 6.0$ m (given, same for both axes).
Built-up member: buckling curve c, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5).

Step 1: Trial section

Try 2 × ISMC 225 (IS 808): $A = 33.3$ cm² each, $I_{xx} = 2710$ cm⁴, $I_{yy} = 188$ cm⁴, $C_{yy} = 2.31$ cm, $B = 80$ mm, $t_w = 6.5$ mm, $t_f = 12.4$ mm.

Total area $A_e = 2 \times 3330 = 6660$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 9.02$ cm.

\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{6000}{90.2} = 66.5 \\ \lambda_{eff} &= 1.05 \times 66.5 = 69.8 \end{aligned}

For $\lambda_{eff} = 69.8$ , curve c: $f_{cc} = \pi^2E/\lambda^2 = 405$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.786$ , $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.952$ , $\chi = 0.671$ .

\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.671 \times 250/1.1 = 152.5\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 6660 \times 152.5/10^3 = 1016\ \text{kN} > 1000\ \text{kN} \end{aligned}

Step 2: Spacing of channels

To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels:

\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(2710 - 188) \times 10^4}{3330}} = 174\ \text{mm} \end{aligned}

Gap between webs $g = h - 2C_{yy} = 174 - 46.2 = 128$ mm. Provide g = 130 mm (centroid distance $h = 176$ mm).

\begin{aligned} I_{yy} &= 2[188 + 33.3(8.81)^2] = 5545\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 9.12\ \text{cm}, \quad \lambda_{yy} = 6000/91.2 = 65.8 \end{aligned}

$\lambda_{yy} = 65.8 \le \lambda_{zz} = 66.5$ , so the zz-axis governs and the capacity $P_d = 1016$ kN $\ge 1000$ kN. Safe. Overall width of column $= 2B + g = 290$ mm.

  +-------+   gap g   +-------+
  | ISMC  |<--------->| ISMC  |
  +-------+           +-------+
  webs face each other, flanges point outward
  lacing bolted to the flanges (top and bottom)

Step 3: Lacing system (single lacing, two planes)

Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 1000 = 25.0$ kN; shared by two lacing planes: $12.50$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 130 + 80 = 210$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 210/\sin 45^\circ = 297\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{12.50}{\sin 45^\circ} = 17.68\ \text{kN} \end{aligned}

Spacing of connections along one channel $a_1 = 2s\cot\theta = 420$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 420/23.8 = 17.7$ , which is less than the smaller of 50 and $0.7\lambda_{max} = 46.6$ . OK.

Flat size. Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 7.4$ mm (cl. 7.6.3) and for $KL/r \le 145$ , $t \ge l\sqrt{12}/145 = 7.1$ mm. Provide ISF 60 × 8 mm.

$r = t/\sqrt{12} = 2.31$ mm, $l/r = 128.6 < 145$ OK. $f_{cd}$ (curve c) $= 75.5$ N/mm², $P_{d,lace} = 480 \times 75.5/10^3 = 36.2$ kN $> F = 17.68$ kN. OK.

Step 4: Welded connection of lacing

For a welded lacing the effective length is $0.7\times$ the distance between the inner ends of the welds (cl. 7.6.6.3); the check above was done on $l = 297$ mm, which is conservative.

Fillet weld size $s = 4$ mm (min 3 mm for 6 to 10 mm plates, max $t - 1.5$ ). Design strength per mm:

\begin{aligned} f_w &= \frac{0.7 s f_u}{\sqrt3\,\gamma_{mw}} = \frac{0.7 \times 4 \times 410}{\sqrt3 \times 1.25} = 530\ \text{N/mm} \\ L_w &= \frac{F}{f_w} = \frac{17678}{530} = 33\ \text{mm} \end{aligned}

Provide a weld of length $35$ mm on each side of the bar at each end (total 70 mm > 33 mm). Lap on the channel flange $\ge 4t = 32$ mm (cl. 7.6.7). Weld along both sides for the full lap length.

Step 5: End tie plates

Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth $\ge h = 176$ mm; thickness $\ge s/50 = 4.2$ mm. Provide 290 mm wide × 180 mm deep × 6 mm thick.

The tie plate is fillet welded (size 5 mm) to the flange of each channel along two lines of length 180 mm. With $C = 420$ mm, $N=2$ : $V_b = V_tC/(NS) = 25.0$ kN and $M = V_tC/2N = 2625$ kN·mm. Weld modulus per unit throat $Z_w = d^2/3 = 10800$ mm²: $q_m = M/Z_w = 243$ N/mm, $q_v = V_b/2d = 69$ N/mm, resultant $= 253$ N/mm $<$ weld strength $663$ N/mm. OK.

Final design

Column: 2 × ISMC 225 back to back, gap 130 mm, $P_d = 1016$ kN $> 1000$ kN.
Lacing: single, flats 60 × 8 mm at 45°, 4 mm fillet welds.
End tie plates 290 × 180 × 6 mm.

  +--------------------+   <- end tie plate
  | \    /\    /\    /|
  |  \  /  \  /  \  / |   single lacing
  |   \/    \/    \/  |   (same direction
  |   /\    /\    /\  |    on both faces)
  +--------------------+   <- end tie plate

2068 Bhadra (old course) · 14 marks

Design a column to carry an axial load of 800KN using two channels laced together. The length of the column is 6m and is effectively held in position at both ends but not restrained against rotation.

Answer

Approach. Laced column of two channels (back to back), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$ .

Data

Factored axial load $P = 800$ kN; effective length $KL = 6.0$ m (both axes, ends held in position, not restrained against rotation, $K = 1.0$ , Table 11).
Built-up member: buckling curve c, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5).

Step 1: Trial section

Try 2 × ISMC 225 (IS 808): $A = 33.3$ cm² each, $I_{xx} = 2710$ cm⁴, $I_{yy} = 188$ cm⁴, $C_{yy} = 2.31$ cm, $B = 80$ mm, $t_w = 6.5$ mm, $t_f = 12.4$ mm.

Total area $A_e = 2 \times 3330 = 6660$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 9.02$ cm.

\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{6000}{90.2} = 66.5 \\ \lambda_{eff} &= 1.05 \times 66.5 = 69.8 \end{aligned}

For $\lambda_{eff} = 69.8$ , curve c: $f_{cc} = \pi^2E/\lambda^2 = 405$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.786$ , $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.952$ , $\chi = 0.671$ .

\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.671 \times 250/1.1 = 152.5\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 6660 \times 152.5/10^3 = 1016\ \text{kN} > 800\ \text{kN} \end{aligned}

Step 2: Spacing of channels

To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels:

\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(2710 - 188) \times 10^4}{3330}} = 174\ \text{mm} \end{aligned}

Gap between webs $g = h - 2C_{yy} = 174 - 46.2 = 128$ mm. Provide g = 130 mm (centroid distance $h = 176$ mm).

\begin{aligned} I_{yy} &= 2[188 + 33.3(8.81)^2] = 5545\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 9.12\ \text{cm}, \quad \lambda_{yy} = 6000/91.2 = 65.8 \end{aligned}

$\lambda_{yy} = 65.8 \le \lambda_{zz} = 66.5$ , so the zz-axis governs and the capacity $P_d = 1016$ kN $\ge 800$ kN. Safe. Overall width of column $= 2B + g = 290$ mm.

  +-------+   gap g   +-------+
  | ISMC  |<--------->| ISMC  |
  +-------+           +-------+
  webs face each other, flanges point outward
  lacing bolted to the flanges (top and bottom)

Step 3: Lacing system (single lacing, two planes)

Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 800 = 20.0$ kN; shared by two lacing planes: $10.00$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 130 + 80 = 210$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 210/\sin 45^\circ = 297\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{10.00}{\sin 45^\circ} = 14.14\ \text{kN} \end{aligned}

Spacing of connections along one channel $a_1 = 2s\cot\theta = 420$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 420/23.8 = 17.7$ , which is less than the smaller of 50 and $0.7\lambda_{max} = 46.6$ . OK.

Flat size. Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 7.4$ mm (cl. 7.6.3) and for $KL/r \le 145$ , $t \ge l\sqrt{12}/145 = 7.1$ mm. Provide ISF 60 × 8 mm.

$r = t/\sqrt{12} = 2.31$ mm, $l/r = 128.6 < 145$ OK. $f_{cd}$ (curve c) $= 75.5$ N/mm², $P_{d,lace} = 480 \times 75.5/10^3 = 36.2$ kN $> F = 14.14$ kN. OK.

Step 4: Connection of lacing (M16, grade 4.6 bolts)

\begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 8 \times 410}{1.25 \times 10^3} = 58.3\ \text{kN} \end{aligned}

( $k_b = e/3d_0 = 30/54 = 0.556$ with edge distance 30 mm $\ge 1.5d_0$ .) Bolt value $= 29.0$ kN. Bolts needed $= 14.14/29.0 = 0.49$ , so provide 1 bolt(s) M16 at each end of every bar.

Net section of flat in tension: $A_n = (60 - 18) \times 8 = 336$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 99.2$ kN $> 14.14$ kN. OK.

Step 5: End tie plates

Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth $\ge h = 176$ mm; thickness $\ge s/50 = 4.2$ mm. Provide 290 mm wide × 180 mm deep × 6 mm thick.

With $C = 420$ mm, $N=2$ : $V_b = V_tC/(NS) = 20.0$ kN and $M = V_tC/2N = 2100$ kN·mm. Using 3 M16 bolts at 50 mm pitch on each channel: max bolt force from $M$ $= 21.0$ kN, from shear $= 6.7$ kN, resultant $= 22.0$ kN $\le 29.0$ kN. OK.

Final design

Column: 2 × ISMC 225 back to back, gap 130 mm, $P_d = 1016$ kN $> 800$ kN.
Lacing: single, flats 60 × 8 mm at 45°, M16 bolts, 1 per end.
End tie plates 290 × 180 × 6 mm.

  +--------------------+   <- end tie plate
  | \    /\    /\    /|
  |  \  /  \  /  \  / |   single lacing
  |   \/    \/    \/  |   (same direction
  |   /\    /\    /\  |    on both faces)
  +--------------------+   <- end tie plate

2070 Bhadra · 12 marks

A bridge compression member is built using two channels ISLC 400 @ 45.8 kg/m placed toe to toe. The effective length of the member is 8.0 m. The width over the backs of two channels is 40 cm. The channels are properly connected by lacings. i) Calculate the safe load for the member. ii) Design the lacing systems using M16 properly class 4.6 grade bolts.

Answer

Approach. Two channels toe to toe, laced; IS 800:2007 cl. 7.1.2 (compressive strength) and cl. 7.6 (laced columns). $f_y = 250$ N/mm² (E250), $\gamma_{m0} = 1.10$ . Section properties of ISLC 400 @ 45.8 kg/m (IS 808): $A = 58.3$ cm², $I_{xx} = 14000$ cm⁴, $I_{yy} = 462$ cm⁴, $C_{yy} = 2.37$ cm, $b = 100$ mm, $t_w = 8.0$ mm, $t_f = 14.0$ mm (my reading of the table).

(i) Safe load of the member

Width over the backs $= 400$ mm, so the distance between the centroids of the channels is $h = 400 - 2C_{yy} = 400 - 47.4 = 352.6$ mm and the gap between toes is $400 - 2\times100 = 200$ mm.

\begin{aligned} I_{zz} &= 2\times14000 = 28000\ \text{cm}^4,\quad r_{zz} = 15.50\ \text{cm} \\ I_{yy} &= 2[462 + 58.3\,(17.63)^2] = 37165\ \text{cm}^4,\quad r_{yy} = 17.85\ \text{cm} \\ \lambda_{zz} &= 8000/155.0 = 51.6,\quad \lambda_{yy} = 8000/178.5 = 44.8 \end{aligned}

Maximum slenderness $= 51.6$ (zz-axis). Laced column: $\lambda_{eff} = 1.05\times51.6 = 54.2$ (cl. 7.6.1.5). Buckling curve c ( $\alpha = 0.49$ , Table 10):

\begin{aligned} f_{cc} &= \frac{\pi^2E}{\lambda^2} = 672\ \text{N/mm}^2,\ \lambda_n = 0.610,\ \phi = 0.787,\ \chi = 0.779 \\ f_{cd} &= \chi f_y/\gamma_{m0} = 177.1\ \text{N/mm}^2 \\ P_d &= 2A f_{cd} = 2\times5830\times177.1/10^3 = 2065\ \text{kN} \end{aligned}

Answer (i): safe (design) load $\approx 2065$ kN (factored). Working load $= 2065/1.5 \approx 1377$ kN.

(ii) Design of lacing (M16, grade 4.6 bolts, single lacing)

Transverse shear $V_t = 2.5\%\,P = 0.025\times2065 = 51.6$ kN; per lacing plane $= 25.8$ kN (cl. 7.6.6.1).

Bolt lines at mid-flange, so $s = 200 + 100 = 300$ mm. Angle of lacing $\theta = 45^\circ$ (between 40° and 70°).

\begin{aligned} l &= s/\sin\theta = 424\ \text{mm} \\ F &= \frac{V_t/2}{\sin\theta} = 36.5\ \text{kN} \\ a_1 &= 2s\cot\theta = 600\ \text{mm} \end{aligned}

Check of main member between lacing points: $a_1/r_{y} = 600/28.2 = 21.3$ , below the smaller of 50 and $0.7\lambda_{max} = 36.1$ . OK (cl. 7.6.5.1).

Flat: width 60 mm ( $\ge 3d = 48$ mm); thickness $\ge l/40 = 10.6$ mm and $\ge l\sqrt{12}/145 = 10.1$ mm, so use ISF 60 × 12 mm. $l/r = 122.5 \le 145$ ; $f_{cd} = 81.2$ N/mm², $P_{d,bar} = 58.5$ kN $> 36.5$ kN. OK.

Bolts M16, 4.6: $V_{dsb} = \frac{400}{\sqrt3}\times\frac{0.78\times\pi\times16^2/4}{1.25\times10^3} = 29.0$ kN; $V_{dpb} = 2.5k_bdtf_u/\gamma_{mb} = 87.5$ kN ( $k_b = 0.556$ , $t = 12$ mm). Bolt value $= 29.0$ kN, so number of bolts $= 36.5/29.0$ , provide 2 bolt(s) M16 at each end of each bar.

End tie plates (cl. 7.6.8, 7.7.2): size $400\times360\times8$ mm (depth $\ge$ distance between centroids $= 353$ mm; thickness $\ge s/50$ ).

2071 Magh · 5+7 marks

Design a bridge compression member using two channels placed back to back to carry a factorial load of 1200 kN, if effective length of column is 8.5 m. Also design the single lacing system using tie bar.

Answer

Approach. Laced column of two channels (back to back), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$ .

Data

Factored axial load $P = 1200$ kN; effective length $KL = 8.5$ m (given, same for both axes).
Built-up member: buckling curve c, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5).

Step 1: Trial section

Try 2 × ISMC 300 (IS 808): $A = 46.3$ cm² each, $I_{xx} = 6420$ cm⁴, $I_{yy} = 313$ cm⁴, $C_{yy} = 2.35$ cm, $B = 90$ mm, $t_w = 7.8$ mm, $t_f = 13.6$ mm.

Total area $A_e = 2 \times 4630 = 9260$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 11.78$ cm.

\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{8500}{117.8} = 72.2 \\ \lambda_{eff} &= 1.05 \times 72.2 = 75.8 \end{aligned}

For $\lambda_{eff} = 75.8$ , curve c: $f_{cc} = \pi^2E/\lambda^2 = 344$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.853$ , $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 1.024$ , $\chi = 0.629$ .

\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.629 \times 250/1.1 = 142.9\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 9260 \times 142.9/10^3 = 1324\ \text{kN} > 1200\ \text{kN} \end{aligned}

Step 2: Spacing of channels

To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels:

\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(6420 - 313) \times 10^4}{4630}} = 230\ \text{mm} \end{aligned}

Gap between webs $g = h - 2C_{yy} = 230 - 47.0 = 183$ mm. Provide g = 190 mm (centroid distance $h = 237$ mm).

\begin{aligned} I_{yy} &= 2[313 + 46.3(11.85)^2] = 13629\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 12.13\ \text{cm}, \quad \lambda_{yy} = 8500/121.3 = 70.1 \end{aligned}

$\lambda_{yy} = 70.1 \le \lambda_{zz} = 72.2$ , so the zz-axis governs and the capacity $P_d = 1324$ kN $\ge 1200$ kN. Safe. Overall width of column $= 2B + g = 370$ mm.

  +-------+   gap g   +-------+
  | ISMC  |<--------->| ISMC  |
  +-------+           +-------+
  webs face each other, flanges point outward
  lacing bolted to the flanges (top and bottom)

Step 3: Single lacing system using flat tie bars (two planes)

Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 1200 = 30.0$ kN; shared by two lacing planes: $15.00$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 190 + 90 = 280$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 280/\sin 45^\circ = 396\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{15.00}{\sin 45^\circ} = 21.21\ \text{kN} \end{aligned}

Spacing of connections along one channel $a_1 = 2s\cot\theta = 560$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 560/26.0 = 21.5$ , which is less than the smaller of 50 and $0.7\lambda_{max} = 50.5$ . OK.

Tie bar (flat) size. Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 9.9$ mm (cl. 7.6.3) and for $KL/r \le 145$ , $t \ge l\sqrt{12}/145 = 9.5$ mm. Provide ISF 60 × 10 mm.

$r = t/\sqrt{12} = 2.89$ mm, $l/r = 137.2 < 145$ OK. $f_{cd}$ (curve c) $= 68.4$ N/mm², $P_{d,lace} = 600 \times 68.4/10^3 = 41.0$ kN $> F = 21.21$ kN. OK.

Step 4: Connection of lacing (M16, grade 4.6 bolts)

\begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 10 \times 410}{1.25 \times 10^3} = 72.9\ \text{kN} \end{aligned}

( $k_b = e/3d_0 = 30/54 = 0.556$ with edge distance 30 mm $\ge 1.5d_0$ .) Bolt value $= 29.0$ kN. Bolts needed $= 21.21/29.0 = 0.73$ , so provide 1 bolt(s) M16 at each end of every bar.

Net section of flat in tension: $A_n = (60 - 18) \times 10 = 420$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 124.0$ kN $> 21.21$ kN. OK.

Step 5: End tie plates

Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth $\ge h = 237$ mm; thickness $\ge s/50 = 5.6$ mm. Provide 370 mm wide × 240 mm deep × 6 mm thick.

With $C = 560$ mm, $N=2$ : $V_b = V_tC/(NS) = 30.0$ kN and $M = V_tC/2N = 4200$ kN·mm. Using 4 M16 bolts at 50 mm pitch on each channel: max bolt force from $M$ $= 25.2$ kN, from shear $= 7.5$ kN, resultant $= 26.3$ kN $\le 29.0$ kN. OK.

Final design

Column: 2 × ISMC 300 back to back, gap 190 mm, $P_d = 1324$ kN $> 1200$ kN.
Lacing: single, flats 60 × 10 mm at 45°, M16 bolts, 1 per end.
End tie plates 370 × 240 × 6 mm.

  +--------------------+   <- end tie plate
  | \    /\    /\    /|
  |  \  /  \  /  \  / |   single lacing
  |   \/    \/    \/  |   (same direction
  |   /\    /\    /\  |    on both faces)
  +--------------------+   <- end tie plate

2069 Bhadra · 10 marks

The center to center distance between the end connections of a discontinuous strut consisting of two L75 75×8 is 3.0m. Calculate the design load carrying capacity in compression if angles are connected to the same side of a gusset by more than one bolt in each angle. The grade of the steel is E250.

Answer

Approach. Two angles ISA 75×75×8 connected back to back on the same side of a gusset by more than one bolt in each angle. For such a discontinuous double-angle strut, IS 800:2007 cl. 7.5.2.2 says it is to be designed as for angles loaded through one leg (cl. 7.5.1.2), using the equivalent slenderness ratio from Table 12. The angles are stitched together (cl. 7.8, 10.2.5), so the capacity is twice that of one angle. $f_y = 250$ N/mm² (E250), $\gamma_{m0} = 1.10$ .

Data (IS 808)

ISA 75×75×8: $A = 11.4$ cm² = 1140 mm², minimum radius of gyration $r_{vv} = 1.45$ cm = 14.5 mm, $b_1 = b_2 = 75$ mm, $t = 8$ mm. Length between intersections $L = 3000$ mm. $\varepsilon = 1$ .

Equivalent slenderness ratio (cl. 7.5.1.2)

Constants for two or more bolts at each end (Table 12). The end fixity is not stated, so take the hinged gusset connection (conservative): $k_1 = 0.7$ , $k_2 = 0.6$ , $k_3 = 5$ .

\begin{aligned} \sqrt{\pi^2E/250} &= 88.86 \\ \lambda_{vv} &= \frac{L/r_{vv}}{\varepsilon\sqrt{\pi^2E/250}} = \frac{3000/14.5}{88.86} = 2.328 \\ \lambda_{\phi} &= \frac{(b_1+b_2)/2t}{\varepsilon\sqrt{\pi^2E/250}} = \frac{150/16}{88.86} = 0.106 \\ \lambda_e &= \sqrt{k_1 + k_2\lambda_{vv}^2 + k_3\lambda_\phi^2} = \sqrt{0.7 + 0.6(2.328)^2 + 5(0.106)^2} = 2.002 \end{aligned}

Equivalent slenderness ratio $KL/r = \lambda_e\times88.86 = 177.9$ , which is less than 180 (Table 3), so it is permitted.

Compressive strength of one angle (cl. 7.1.2.1, curve c, $\alpha = 0.49$ )

\begin{aligned} f_{cc} &= \pi^2E/\lambda^2 = 62.4\ \text{N/mm}^2 \\ \lambda_n &= \sqrt{f_y/f_{cc}} = 2.002 \\ \phi &= 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 2.946 \\ \chi &= 1/(\phi + \sqrt{\phi^2 - \lambda_n^2}) = 0.196 \\ f_{cd} &= \chi f_y/\gamma_{m0} = 44.5\ \text{N/mm}^2 \\ P_{d1} &= A f_{cd} = 1140\times44.5/10^3 = 50.7\ \text{kN} \end{aligned}

Capacity of the strut

P_d = 2\times50.7 = 101.5\ \text{kN}

Answer: design compressive strength $\approx 101$ kN (about 68 kN service load).

If the gusset connection is taken as fixed against in-plane rotation ( $k_1 = 0.20$ , $k_2 = 0.35$ , $k_3 = 20$ ), the same working gives $P_d = 2\times79.6 = 159$ kN. The true value lies between these; the hinged value is the safe one.

2070 Magh · 10 marks

Design a single equal angle to carry a compression of 50 KN. The centre to centre distance between the end connections is 2.0 M. Assume that at least two bolts are used for the end connections.

Answer

Approach. A single angle loaded through one leg and connected by at least two bolts at each end is designed with the equivalent slenderness ratio of IS 800:2007 cl. 7.5.1.2 (Table 12). $f_y = 250$ N/mm², $\gamma_{m0} = 1.10$ , $E = 2\times10^5$ N/mm².

Data

Factored load $P = 50$ kN; length between end connections $L = 2.0$ m; $\ge 2$ bolts at each end. End fixity is not stated, so a hinged connection is assumed (conservative): $k_1 = 0.70$ , $k_2 = 0.60$ , $k_3 = 5$ (Table 12).

Trial sections (equal angles from IS 808)

For each angle: $\lambda_{vv} = (L/r_{vv})/\sqrt{\pi^2E/250}$ , $\lambda_\phi = (b/t)/\sqrt{\pi^2E/250}$ (since $b_1 = b_2 = b$ ), $\lambda_e = \sqrt{k_1 + k_2\lambda_{vv}^2 + k_3\lambda_\phi^2}$ , then $KL/r = \lambda_e\sqrt{\pi^2E/250}$ and curve c.

Angle	A (cm²)	$r_{vv}$ (cm)	KL/r	$P_d$ (kN)	Result
ISA 60×60×6	6.84	1.15	155	38.2	Not safe
ISA 65×65×6	7.44	1.26	146	46.2	Not safe
ISA 70×70×6	8.06	1.35	139	53.8	Safe
ISA 75×75×5	7.27	1.46	134	51.6	Safe

The lightest safe angle is ISA 75×75×5 (mass 5.7 kg/m).

Detailed check of ISA 75×75×5

$A = 7.27$ cm² $= 727$ mm², $r_{vv} = 1.46$ cm.

\begin{aligned} \sqrt{\pi^2E/250} &= 88.86 \\ \lambda_{vv} &= \frac{2000/14.6}{88.86} = 1.542,\quad \lambda_\phi = \frac{75/5}{88.86} = 0.169 \\ \lambda_e &= \sqrt{0.70 + 0.60(1.542)^2 + 5(0.169)^2} = 1.506 \\ KL/r &= 1.506\times88.86 = 133.8 < 180 \end{aligned}

\begin{aligned} f_{cc} &= 110.2\ \text{N/mm}^2,\ \lambda_n = 1.506,\ \phi = 1.954,\ \chi = 0.313 \\ f_{cd} &= 71.0\ \text{N/mm}^2 \\ P_d &= A f_{cd} = 727\times71.0/10^3 = 51.6\ \text{kN} > 50\ \text{kN} \end{aligned}

Section class: $b/t = 75/5 = 15.0 \le 15.7\varepsilon = 15.7$ (Table 2, single angle, semi-compact limit). OK.

Answer: provide ISA 75×75×5, connected with at least two M16 bolts at each end; $P_d = 51.6$ kN $> 50$ kN.

Past questions and answers

1. Trial section: 2 ISMC 300

2. Spacing of channels (so that Iyy≥IxxI_{yy}\ge I_{xx}Iyy​≥Ixx​)

3. Lacing (single, bolted)

4. Tie plates

1. Section: 2 ISMC 300 back to back (webs facing, flanges outward)

2. Clear gap for equal stiffness

3. Design of lacing (single lacing, two planes)

4. Tie plates

Data

Step 1: Trial section

Step 2: Spacing of channels

Step 3: Spacing of battens (cl. 7.7.3)

Step 4: Forces on battens (cl. 7.7.2.1)

Step 5: Size of battens (cl. 7.7.2.3)

Step 6: Bolts (M16, grade 4.6)

Step 7: Check of the main member (cl. 7.7.2.1)

Final design

4.Draw neat sketch of column slab base and write down design procedure (design steps of column bases / process to find thickness of slab base foundation).

Slab base

Design procedure (IS 800:2007 cl. 7.4)

5.Design a slab base for a column ISHB 350@710.2 N/m subjected to a factored axial compressive load of 1000 kN. Concrete pedestal of grade M20.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

2. Plate size

3. Thickness (cl. 7.4.3.1)

4. Connections

6.Design the base plate for a ISHB 350 column to carry factored load of 1200KN. Take E250 grade of steel and M20 grade of concrete.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

2. Plate size

3. Thickness (cl. 7.4.3.1)

4. Connections

7.Design the base plate for the column ISHB 450 to carry a factored load of 1400 kN. Take E250 grade of steel and M20 grade of concrete.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

2. Plate size

3. Thickness (cl. 7.4.3.1)

4. Connections

1. Section: 2 ISMC 225 toe to toe (flanges facing)

2. Clear gap (so that Iyy≥IxxI_{yy}\ge I_{xx}Iyy​≥Ixx​)

3. Spacing of battens (cl. 7.7.1.4)

4. Size of battens (cl. 7.7.2)

5. Design forces on a batten

6. Bolts connecting battens to the sections (M20, 4.6)

1. Section: 2 ISMC 250 back to back (webs facing, flanges outward)

2. Clear gap for equal stiffness

3. Design of lacing (single lacing, two planes)

4. Tie plates

10.Design slab base for a column SC300 carrying factor axial lead of 1200KN if concrete grade used is M20.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

2. Plate size

3. Thickness (cl. 7.4.3.1)

4. Connections

11.Design a slab base for ISSC 200 which has to carry a design axial load of 1200 kN resting on a concrete pedestal with M 20 grade of concrete.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

2. Plate size

3. Thickness (cl. 7.4.3.1)

4. Connections

12.Draw neat sketch of column gusseted base.

13.Design a gusseted base for a column ISHB 350 @ 710 N/m with two plates 450mm×20mm carrying factored load 2000KN. The column is to be supported on Concrete pedestal with M20 grade concrete.

1. Bearing and plate area

2. Thickness of base plate

3. Transfer of load from the shaft to the base plate

4. Anchor bolts

14.Design a slab base for a column SC220 to transfer an axial load of 1000 kN. Take Fe410 grade steel and M30 for concrete.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

2. Plate size

3. Thickness (cl. 7.4.3.1)

4. Connections

15.Design a slab base for a column ISMB 350 @ 52.4 kg/m to carry a service load of 850 KN. Assume Fe410 grade steel and M25 concrete.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

2. Plate size

3. Thickness (cl. 7.4.3.1)

4. Connections

16.Design the foundation base for an ISHB 350 column to carry factored load of 120[?] KN. Assume steel and M20 concrete.

1. Bearing strength of concrete (IS 800:2007 cl. 7.4.1)

2. Plate size

3. Thickness (cl. 7.4.3.1)

4. Connections

17.Design a built-up column consisting of two channels placed toe-to-toe. The column carries an axial factored load of 1500 kN. The effective length of column is 10 m. Also, design the lacing system using welded connection. Take Fe 415 grade steel.

1. Section: 2 ISMC 350 toe to toe (flanges facing, webs outward)

2. Clear gap for equal stiffness

2. Spacing of channels (so that $I_{yy}\ge I_{xx}$ )

Draw neat sketch of column slab base and write down design procedure (design steps of column bases / process to find thickness of slab base foundation).

Design a slab base for a column ISHB 350@710.2 N/m subjected to a factored axial compressive load of 1000 kN. Concrete pedestal of grade M20.

Design the base plate for a ISHB 350 column to carry factored load of 1200KN. Take E250 grade of steel and M20 grade of concrete.

Design the base plate for the column ISHB 450 to carry a factored load of 1400 kN. Take E250 grade of steel and M20 grade of concrete.

2. Clear gap (so that $I_{yy}\ge I_{xx}$ )

Design slab base for a column SC300 carrying factor axial lead of 1200KN if concrete grade used is M20.

Design a slab base for ISSC 200 which has to carry a design axial load of 1200 kN resting on a concrete pedestal with M 20 grade of concrete.

Draw neat sketch of column gusseted base.

Design a gusseted base for a column ISHB 350 @ 710 N/m with two plates 450mm×20mm carrying factored load 2000KN. The column is to be supported on Concrete pedestal with M20 grade concrete.

Design a slab base for a column SC220 to transfer an axial load of 1000 kN. Take Fe410 grade steel and M30 for concrete.

Design a slab base for a column ISMB 350 @ 52.4 kg/m to carry a service load of 850 KN. Assume Fe410 grade steel and M25 concrete.

Design the foundation base for an ISHB 350 column to carry factored load of 120[?] KN. Assume steel and M20 concrete.

Design a built-up column consisting of two channels placed toe-to-toe. The column carries an axial factored load of 1500 kN. The effective length of column is 10 m. Also, design the lacing system using welded connection. Take Fe 415 grade steel.

Design splice plate and the connection for a column section ISHB 350 to support a factored axial load of 1000 kN. Use M20 bolts of grade 4.6 and steel of grade E250. Assume the column ends are machined. The connection is to be spliced at a height of 3 meters.

Determine design axial compressive load carrying capacity of a double angle IS 150×75×12 provided at opposite side of gusset plate, length of member is 3.5 m.

Design a built-up column to carry an axial load of 12 00kN and composed of two channels placed back to back. The effective length of the member is 6m. Design the column using battens system and Fe410 grade of steel.

2. Clear gap (so that $I_{yy}\ge I_{xx}$ )

Design a built up column using double channel section. Unsupported length of column is 6m, both end of column are restrained against lateral displacement but free in rotation. Column is subjected to design axial load of 1200 kN. Use battens.

2. Clear gap (so that $I_{yy}\ge I_{xx}$ )

Explain about Buckling Behaviour of column.

Design a column section subjected to axial load of 1500 KN using rolled steel 'I' sections. Height of the column is 8 m. Both ends of column are fixed and exits sway condition. Also design lateral bracing of column using battening system.

2. Clear gap (so that $I_{yy}\ge I_{xx}$ )

An ISHB 250 @ 536 N/m column carrying a factored axial load of 900KN. The column ends machined. Design the splice connection. Use M16 bolts.

A 7.5m long built-up and laced column has to carry a factored axial load of 1250KN. The column is restrained in position but not in direction at each end. Design the column with single lacing system. Connection shall consist of two channels placed back to back at a suitable spacing.

Design a built-up column of the effective length of 5 m to carry an axial load of 900 kN using two channels and single lacing. Design the connections using bolt. The grade of the steel is Fe410.