Tension Members

Given: axial pull (factored) $T=400$ kN, joint length available 300 mm, gusset 16 mm, M20 bolts grade 4.6, Fe410/E250 ($f_y=250$, $f_u=410$). A single angle connected by one leg would need more than 300 mm of bolting, so a lug angle is used at the end (IS 800:2007 cl. 10.12).

Section

Required $A_g\ge\dfrac{400\times10^3\times1.10}{250}=1760$ mm². Try ISA 100×100×12: $A_g=2257$ mm².

• Yielding: $T_{dg}=\dfrac{2257\times250}{1.10}=513.0$ kN $>400$ ✓
• Rupture: with a lug angle the whole angle is effectively connected, so no shear-lag reduction. $A_n=2257-12\times22=1993$ mm²; $T_{dn}=\dfrac{0.9\times1993\times410}{1.25}=588.3$ kN $>400$ ✓

Bolts (M20, 4.6)

• Shear $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing ($e=40$, $p=50$, $t=12$ mm): $k_b=0.508$, $V_{dpb}=99.9$ kN → bolt value $45.26$ kN

Forces with lug angle

Outstanding leg force $\approx T\dfrac{A_{out}}{A_g}$, $A_{out}\approx(100-6)\times12=1128$ mm²: $P_{out}=199.9$ kN.

• Connected leg force $=400-199.9=200.1$ kN → main angle to gusset: $n=200.1/45.26$ → 5 bolts
• Lug angle to gusset: $1.2\,P_{out}=239.9$ kN → 6 bolts
• Lug angle to main angle: $1.4\,P_{out}=279.9$ kN → 7 bolts

Layout

   gusset 16
  |==o==o==o==o==o==   main angle  (5 bolts @ 50)
  |==o==o==o==o==o==o  lug angle   (6 bolts @ 50)
        <- 300 mm ->

Pitch 50 mm ($\ge2.5d$), end distance 40 mm; length $40+5\times50=290\le300$ mm ✓. Lug angle: ISA 65×65×8 (width ≥ 0.35 times the main angle leg), connected to the other face of the gusset.

Answer: ISA 100×100×12 with lug angle ISA 65×65×8; 5 M20 bolts for the main angle, 6 for the lug angle to the gusset and 7 for the lug-to-angle connection.

Data: ISA 100×75×6 ($A_g=1010$ mm²), gusset 10 mm connected to the 100 mm leg by six 16 mm bolts ($d_0=18$ mm); $f_y=250$, $f_u=410$ N/mm². Assumed layout: single line of bolts, pitch $p=50$ mm, end distance 30 mm, gauge 45 mm from the toe (edge distance 45 mm). Bolts assumed grade 4.6.

1. Yielding of gross section (cl. 6.2)

$$T_{dg}=\frac{A_gf_y}{\gamma_{m0}}=\frac{1010\times250}{1.10}\times10^{-3}=229.5\ \text{kN}$$

2. Rupture of net section with shear lag (cl. 6.3.3)

• Connected leg: $A_{nc}=(100-\tfrac{6}{2}-18)\times6=474$ mm²
• Outstanding leg: $A_{go}=(75-\tfrac{6}{2})\times6=432$ mm²
• $w=75$ mm, $b_s=w+w_1-t=75+100-6=169$ mm, $L_c=5\times50=250$ mm

$$\beta=1.4-0.076\frac{w}{t}\frac{f_y}{f_u}\frac{b_s}{L_c}=1.4-0.076\times\frac{75}{6}\times\frac{250}{410}\times\frac{169}{250}=1.008\quad(0.7\le\beta\le1.443)$$ $$T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=238.9\ \text{kN}$$

3. Block shear (cl. 6.4)

$A_{vg}=1680$, $A_{vn}=1086$, $A_{tg}=270$, $A_{tn}=216$ mm²

• $T_{db1}=\dfrac{A_{vg}f_y}{\sqrt3\gamma_{m0}}+\dfrac{0.9A_{tn}f_u}{\gamma_{m1}}=284.2$ kN
• $T_{db2}=\dfrac{0.9A_{vn}f_u}{\sqrt3\gamma_{m1}}+\dfrac{A_{tg}f_y}{\gamma_{m0}}=246.5$ kN

Design tensile strength $T_d=\min(229.5,\ 238.9,\ 246.5)=229.5$ kN (yielding of gross section governs).

Design of connection for full capacity (16 mm, 4.6 bolts)

• $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=29.01$ kN
• Bearing on 6 mm angle ($e=30,\ p=50$): $k_b=0.556$, $V_{dpb}=43.7$ kN
• Number of bolts $=\dfrac{229.5}{29.01}=7.91$ → 8 bolts (the six bolts are not enough for the full capacity).

Answer: design tensile strength $\approx229.5$ kN; connect with 8 M16 bolts (4.6) at 50 mm pitch (more joint length) or use higher-grade bolts. With six bolts the connection strength would be only $6\times29.0=174.0$ kN.

Given: $T=280$ kN (design), $L=3.2$ m, gusset 10 mm, one line of M20 bolts (4.6), E250 ($f_y=250$, $f_u=410$ N/mm²), channel connected through its web.

1. Trial section

$A_g\ge\dfrac{T\gamma_{m0}}{f_y}=\dfrac{280\times10^3\times1.10}{250}=1232$ mm². Try ISMC 175: $A=2440$ mm², $h=175$, $b_f=75$, $t_f=10.2$, $t_w=5.7$ mm, $r_{yy}=22.3$ mm.

• Yielding: $T_{dg}=\dfrac{2440\times250}{1.10}=554.5$ kN $>280$ ✓
• Slenderness: $\dfrac{L}{r_{min}}=\dfrac{3200}{22.3}=143<400$ ✓ (IS 800 Table 3)

2. Bolts (M20, 4.6, single shear, $d_0=22$ mm)

• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing on 5.7 mm web ($e=40$, $p=50$): $k_b=0.508$, $V_{dpb}=47.4$ kN
• Bolt value $=45.26$ kN → $n=\dfrac{280}{45.26}=6.2$ → provide 7 bolts at 50 mm pitch, end distance 40 mm (joint length $380$ mm); capacity $316.8$ kN.

3. Rupture of net section with shear lag (cl. 6.3.3)

• $A_{nc}=(175-22)\times5.7=872$ mm² (connected web), $A_{go}=A-175\times5.7=1442$ mm² (outstanding flanges)
• $w=75$, $b_s=w+w_1-t_w=75+87.5-5.7=156.8$ mm, $L_c=6\times50=300$ mm

$$\beta=1.4-0.076\frac{w}{t_f}\frac{f_y}{f_u}\frac{b_s}{L_c}=1.222\ (\le1.443)$$ $$T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=658.0\ \text{kN}>280$$

4. Block shear (cl. 6.4)

$A_{vg}=1938$, $A_{vn}=1123$, $A_{tg}=499$, $A_{tn}=436$ mm² (failure along the bolt line and across to the web edge)

• $T_{db1}=\dfrac{A_{vg}f_y}{\sqrt3\gamma_{m0}}+\dfrac{0.9A_{tn}f_u}{\gamma_{m1}}=383.0$ kN
• $T_{db2}=\dfrac{0.9A_{vn}f_u}{\sqrt3\gamma_{m1}}+\dfrac{A_{tg}f_y}{\gamma_{m0}}=304.7$ kN → $T_{db}=304.7$ kN $>280$ ✓

Answer: ISMC 175 with 7 M20 (4.6) bolts at 50 mm pitch; design strength $=\min(554.5,\,658.0,\,304.7,\,316.8)=304.7$ kN $\ge280$ kN. Safe.

Given: design tension 200 kN; single unequal angle; HSFG bolts of property class 6.8 ($f_{ub}=600$ N/mm²); E250. Bolts are designed as friction-grip connection (no slip at the design load), M20, $d_0=22$ mm, $\mu_f=0.48$ (sand-blasted surface), $K_h=1.0$, $n_e=1$ (single shear), $\gamma_{mf}=1.25$ (IS 800:2007 cl. 10.4.3).

1. Section

Required $A_g\ge\dfrac{200\times10^3\times1.10}{250}=880$ mm². Try ISA 100×75×8, $A_g=1336$ mm², connected through the 100 mm leg.

2. HSFG bolts

• Proof load $F_o=0.7f_{ub}A_{nb}=0.7\times600\times245\times10^{-3}=102.9$ kN
• Slip resistance $V_{nsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times1\times1.0\times102.9}{1.25}=39.51$ kN
• Number of bolts $=\dfrac{200}{39.51}=5.1$ → 6 bolts, one line, pitch 50 mm ($\ge2.5d$), end distance 40 mm ($\ge1.5d_0=33$), gauge 55 mm from the heel. Capacity $237.1$ kN.

3. Member checks

• Yielding: $T_{dg}=\dfrac{1336\times250}{1.10}=303.6$ kN
• Net rupture with shear lag: $A_{nc}=(100-4-22)\times8=592$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=75+100-8=167$ mm, $L_c=250$ mm, $\beta=1.110$: $T_{dn}=318.0$ kN
• Block shear: $T_{db}=\min(384.7,312.2)=312.2$ kN

All exceed 200 kN. Slenderness limit (400) is checked with the member length (not given): $r_{min}=15.9$ mm allows $L\le6.3$ m.

Answer: ISA 100×75×8 connected by 6 M20 HSFG bolts (class 6.8) in a line at 50 mm pitch.

Shear lag is the non-uniform distribution of stress in a tension member when only part of its cross-section (e.g. one leg of an angle or the web of a channel) is connected to the gusset. The force enters through the connected element, so stress is highest near the connection and the outstanding element is less effective, especially over a short connection length.

Because the outstanding leg does not carry its full share, the net section reaches its strength earlier than the full area would suggest. IS 800:2007 (cl. 6.3.3) therefore reduces the net-section strength using a factor $\beta=1.4-0.076\dfrac{w}{t}\dfrac{f_y}{f_u}\dfrac{b_s}{L_c}$ for the outstanding leg:

$$T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}$$

Shear lag decreases with a longer connection length $L_c$ and by connecting the outstanding leg (e.g. with a lug angle).

Given: factored tension 250 kN, $L=3$ m, single row of M20 bolts (grade 4.6 assumed), $f_y=250$, $f_u=410$ N/mm².

1. Bolts

• $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN (single shear, thread in plane)
• Bearing on 8 mm angle ($e=35$, $p=50$): $k_b=0.508$; $V_{dpb}=66.6$ kN → bolt value $45.26$ kN
• $n=\dfrac{250}{45.26}=5.5$ → 6 bolts, pitch 50 mm, end distance 40 mm; capacity $271.6$ kN.

2. Section

$A_g\ge\dfrac{250\times1.10}{250}\times10^3=1100$ mm². Because the angle is connected through one leg, choose ISA 100×75×8 ($A_g=1336$ mm², $r_{vv}=16$ mm), long leg connected.

• Yielding: $T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN $>250$ ✓
• Rupture (cl. 6.3.3): $A_{nc}=(100-4-22)\times8=592$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=75+100-8=167$ mm, $L_c=250$ mm

$$\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{167}{250}=1.110,\qquad T_{dn}=\frac{0.9\times592\times410}{1.25}+\frac{1.110\times568\times250}{1.10}=318.0\ \text{kN}$$

• Block shear: $A_{vg}=2320$, $A_{vn}=1352$, $A_{tg}=360$, $A_{tn}=272$ mm² → $T_{db1}=384.7$, $T_{db2}=312.2$ kN
• Slenderness: $\dfrac{3000}{15.9}\approx189<400$ ✓

Design strength $=\min(303.6,318.0,312.2,271.6)=271.6$ kN $>250$ kN. Answer: ISA 100×75×8 (long leg connected) with 6 M20 bolts at 50 mm pitch.

Assumptions: the two angles are placed on the two faces of the gusset (the usual back-to-back arrangement), so M18 bolts (grade 4.6) are in double shear; gusset plate 10 mm; E250; $d_0=20$ mm.

1. Bolts

• $V_{dsb}=\dfrac{400}{\sqrt3\times1.25}(A_{nb}+A_{sb})=\dfrac{400}{\sqrt3\times1.25}(192+254)\times10^{-3}=82.40$ kN (one plane through threads)
• Bearing on the 10 mm gusset ($e=35$, $p=50$): $k_b=0.583$; $V_{dpb}=86.1$ kN
• Bolt value $=82.40$ kN → $n=300/82.40=3.6$ → 4 bolts, pitch 50 mm (min $2.5d=45$), end distance 35 mm ($\ge1.5d_0=30$); capacity $329.6$ kN.

2. Section

$A_g\ge\dfrac{300\times1.10}{250}\times10^3=1320$ mm². Try 2 ISA 75×75×6 ($A=866$ mm² each, total 1732 mm², $r_{vv}=14.6$ mm; $L=2.5$ m).

• Yielding: $T_{dg}=2\times\dfrac{866\times250}{1.10}=393.6$ kN $>300$ ✓
• Rupture per angle (carrying 150 kN): $A_{nc}=(75-3-20)\times6=312$ mm², $A_{go}=(75-3)\times6=432$ mm², $b_s=75+75-6=144$ mm, $L_c=150$ mm, $\beta=0.844$

$$T_{dn}=\frac{0.9\times312\times410}{1.25}+\frac{0.844\times432\times250}{1.10}=175.0\ \text{kN per angle}$$

• Block shear per angle: $\min(189.9,165.3)=165.3$ kN
• Strength of the pair $=2\times\min(196.8,175.0,165.3)=330.7$ kN $>300$ ✓

3. Slenderness and tacking

$r_{min}$ of the pair (about the axis parallel to the gusset) $\approx23$ mm so $L/r=2500/23=109<400$ ✓. Provide tack bolts at intervals not more than 1000 mm (and not less than 2 bolts) with packing pieces.

Answer: 2 ISA 75×75×6 on opposite faces of the gusset, connected with 4 M18 bolts (4.6) at 50 mm pitch.

Data: factored tension 400 kN; main flats 250×16 and 250×8; two cover plates 6 mm each ($2\times6=12>8$ mm); M20 HSFG bolts class 8.8 ($f_{ub}=800$); $d_0=22$ mm; $f_y=250$, $f_u=410$. Bolts are in double shear ($n_e=2$). The thinner flat (8 mm) governs the plate checks.

Plate checks (common)

• Gross section of 8 mm flat: $T_{dg}=\dfrac{250\times8\times250}{1.10}=454.5$ kN $>400$ ✓
• Net section (2 bolts in a line): $T_{dn}=\dfrac{0.9(250-2\times22)\times8\times410}{1.25}=486.5$ kN $>400$ ✓
• Covers: $2\times6\times250$: $T_{dg}=681.8$ kN ✓

(i) No slip at ultimate load (slip-critical, $\gamma_{mf}=1.25$)

• $F_o=0.7f_{ub}A_{nb}=0.7\times800\times245\times10^{-3}=137.2$ kN
• $V_{nsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times2\times1.0\times137.2}{1.25}=105.4$ kN per bolt ($\mu_f=0.48$, sand-blasted)
• $n=\dfrac{400}{105.4}=3.8$ → 4 bolts on each side (2 rows of 2), capacity $421.5$ kN.

(ii) Slip permitted at ultimate load (bearing type)

• Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{800}{\sqrt3\times1.25}(245+314)\times10^{-3}=206.6$ kN
• Bearing on 8 mm flat ($e=40$, $p=60$): $k_b=0.606$; $V_{dpb}=\dfrac{2.5k_bdtf_u}{1.25}=79.5$ kN → bolt value $79.5$ kN
• $n=\dfrac{400}{79.5}=5.03$ → 6 bolts on each side (2 rows of 3), capacity $477.1$ kN.

Detailing

Pitch 60 mm ($\ge2.5d$), gauge 80-100 mm, edge distance 40 mm; cover plates 250 wide. Surface preparation (clean, sand-blasted) is required in case (i).

Answer: (i) 4 bolts M20 (8.8) each side; (ii) 6 bolts M20 (8.8) each side.

Given: ISA 100×100×8 (centroid $c_x=27.5$ mm from the back), factored tension 200 kN, Fe410 ($f_u=410$ N/mm²), fillet welds at the toe and back, shop welding.

Weld size and strength (IS 800:2007 cl. 10.5)

• Thickness of angle 8 mm: minimum size 3-5 mm (Table 21, thicker part ≤ 10 mm → 3 mm; use at least 4 mm); maximum size at the edge $=t-1.5=6.5$ mm. Use $s=6$ mm.
• $f_{wd}=\dfrac{f_u}{\sqrt3\gamma_{mw}}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²
• Strength per mm $=f_{wd}\times0.7s=189.4\times4.2=795.4$ N/mm

Total length

$$L_{w}=\frac{200\times10^3}{795.4}=251.5\ \text{mm}$$

Distribution (so that the resultant coincides with the centroid)

Take moments about the back-weld (length $L_2$ at the heel) :

• Weld at toe: $L_1=L_w\dfrac{c_x}{100}=69.2$ mm (carries $200\times0.275=55$ kN)
• Weld at back (heel): $L_2=L_w-L_1=182.3$ mm (carries 145 kN)

Add $2s=12$ mm to each for end returns.

Answer: 6 mm fillet weld, toe weld 85 mm and back weld 195 mm long (total effective length about 251 mm).

Given: factored tension 400 kN; two angles on opposite faces of the gusset (bolts in double shear); M20 bolts, product grade C, class 5.6 ($f_{ub}=500$, $f_{yb}=300$); E250; gusset plate assumed 12 mm; $d_0=22$ mm.

1. Bolts

• $V_{dsb}=\dfrac{500}{\sqrt3\times1.25}(245+314)\times10^{-3}=129.1$ kN (one plane in threads)
• Bearing on gusset 12 mm ($e=40$, $p=60$): $k_b=\min(0.606,0.659,1.22,1)=0.606$; $V_{dpb}=119.3$ kN
• Bolt value $119.3$ kN → $n=\dfrac{400}{119.3}=3.35$ → 4 bolts, pitch 60 mm, end distance 40 mm; capacity $477.1$ kN.

2. Section

$A_g\ge\dfrac{400\times1.10}{250}\times10^3=1760$ mm². Try 2 ISA 75×75×8 ($A=1138$ mm² each, $2276$ mm²).

• Yielding: $T_{dg}=2\times\dfrac{1138\times250}{1.10}\times10^{-3}=517.3$ kN
• Rupture per angle (load 200 kN): $A_{nc}=(75-4-22)\times8=392$ mm², $A_{go}=(75-4)\times8=568$ mm², $b_s=142$ mm, $L_c=180$ mm, $\beta=1.057$ → $T_{dn}=252.2$ kN
• Block shear per angle: $\min(287.6,258.6)=258.6$ kN
• Strength of the pair $=2\times252.2=504.4$ kN $>400$ ✓ (member length not given; check $L/r\le400$, with $r_{min}\approx23$ mm allows $L\le9$ m).

Provide tack bolts at ≤ 1000 mm centres with packing.

Answer: 2 ISA 75×75×8 (one on each side of the gusset) with 4 M20 bolts (5.6) at 60 mm pitch.

Assumptions: factored tension 200 kN; Fe410 / E250; shop welding; the channel is welded to the gusset through its web (welds along the two web edges and across the web end).

1. Section

$A_g\ge\dfrac{200\times10^3\times1.10}{250}=880$ mm². Try ISMC 125: $A=1619$ mm², $h=125$, $b_f=65$, $t_f=8.1$, $t_w=5.0$ mm.

• $T_{dg}=\dfrac{1619\times250}{1.10}\times10^{-3}=368.0$ kN $>200$ ✓

2. Weld size and length

• Web thickness 5 mm: minimum size 3 mm (Table 21); use $s=4$ mm (below the maximum $t-1.5=3.5$ mm for a rounded edge, 4 mm is acceptable on the web face; 3 mm may also be used with longer weld).
• $f_{wd}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²; strength per mm $=189.4\times0.7\times4=530.2$ N/mm
• Required length $L_w=\dfrac{200\times10^3}{530.2}=377$ mm
• Provide: two longitudinal welds of 140 mm along the two edges of the web plus an end weld of 125 mm across the web: total $2\times140+125=405$ mm $>377$ mm ✓ (minimum length $4s=16$ mm ✓).

3. Net-section rupture with shear lag (cl. 6.3.3)

• $A_{nc}=125\times5=625$ mm² (no holes), $A_{go}=1619-625=994$ mm²
• $b_s=65+62.5-5=122.5$ mm, $L_c=140$ mm → $\beta=1.075$

$$T_{dn}=\frac{0.9\times625\times410}{1.25}+\frac{1.075\times994\times250}{1.10}=427.3\ \text{kN}>200$$

The slenderness ($r_{yy}=19.2$ mm) allows $L\le7.6$ m for the limit 400.

Answer: ISMC 125 with 4 mm fillet welds: 2 × 140 mm along the web edges and 125 mm across the end.

Assumptions (not given): factored tension 300 kN, E250 steel, M20 bolts of grade 4.6 in double shear (angles on both faces of a 10 mm gusset), $d_0=22$ mm.

1. Bolts

• $V_{dsb}=\dfrac{400}{\sqrt3\times1.25}(245+314)\times10^{-3}=103.3$ kN
• Bearing on gusset ($e=35,\ p=50$): $k_b=0.508$, $V_{dpb}=83.2$ kN → bolt value $83.2$ kN
• $n=300/83.2=3.0$ → 4 bolts (pitch 50, end 35 mm); capacity $333.0$ kN.

2. Section

$A_g\ge\dfrac{300\times1.10}{250}\times10^3=1320$ mm². Try 2 ISA 75×75×6 ($2\times866=1732$ mm²).

• $T_{dg}=393.6$ kN
• Rupture per angle (150 kN): $A_{nc}=(75-3-22)\times6=300$, $A_{go}=(75-3)\times6=432$ mm², $b_s=144$, $L_c=150$ mm, $\beta=0.844$; $T_{dn}=171.4$ kN
• Block shear per angle $=158.2$ kN
• Strength of the pair $=2\times158.2=316.3$ kN $\ge300$ ✓

Tack bolts (≤ 1000 mm) with packing between angles; $L/r\le400$ is satisfied for members up to about 9 m.

Answer: 2 ISA 75×75×6, one on each side of the gusset, with 4 M20 bolts (4.6) at 50 mm pitch.

Data: ISA 100×75×8 ($A_g=1336$ mm²), gusset 12 mm connected to the 100 mm leg, six M20 bolts (4.6) in one row. Assumed pitch $p=50$ mm, end distance 40 mm, gauge 55 mm; $d_0=22$ mm; $f_y=250$, $f_u=410$.

1. Yielding of gross section

$T_{dg}=\dfrac{A_gf_y}{\gamma_{m0}}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN

2. Net-section rupture with shear lag

• $A_{nc}=(100-\tfrac82-22)\times8=592$ mm²; $A_{go}=(75-\tfrac82)\times8=568$ mm²
• $w=75$, $b_s=w+w_1-t=75+100-8=167$ mm, $L_c=5\times50=250$ mm

$$\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{167}{250}=1.110\quad(0.7\le\beta\le1.443)$$ $$T_{dn}=\frac{0.9A_{nc}f_u}{\gamma_{m1}}+\frac{\beta A_{go}f_y}{\gamma_{m0}}=318.0\ \text{kN}$$

3. Block shear

$A_{vg}=2320$, $A_{vn}=1352$, $A_{tg}=360$, $A_{tn}=272$ mm²: $T_{db1}=384.7$ kN, $T_{db2}=312.2$ kN.

4. Bolt group (check)

$V_{dsb}=45.26$ kN; bearing ($t=8$): $V_{dpb}=66.6$ kN; six bolts: $271.6$ kN (not governing).

Answer: design tensile strength of the angle $=\min(303.6,\ 318.0,\ 312.2)=303.6$ kN (yielding of the gross section governs).

Given: tension 100 kN (taken as the design/factored load; if it is a working load multiply by 1.5), M20 bolts of product grade C, class 4.6 ($d_0=22$ mm), $f_y=250$, $f_u=410$ N/mm².

1. Bolts

• $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing on 6 mm leg ($e=35,\ p=50$): $k_b=0.508$, $V_{dpb}=49.9$ kN → bolt value $45.26$ kN
• $n=100/45.26=2.2$ → 3 bolts, pitch 50 mm, end distance 35 mm; capacity $135.8$ kN.

2. Section

$A_g\ge\dfrac{100\times1.10}{250}\times10^3=440$ mm². The leg must hold an M20 bolt with edge distances $\ge1.5d_0=33$ mm on both sides, so leg $\ge70$ mm: try ISA 75×75×6 ($A=866$ mm²), bolt gauge 40 mm from the back.

• Yielding: $T_{dg}=\dfrac{866\times250}{1.10}\times10^{-3}=196.8$ kN
• Rupture with shear lag: $A_{nc}=(75-3-22)\times6=300$ mm², $A_{go}=(75-3)\times6=432$ mm², $b_s=144$ mm, $L_c=100$ mm, $\beta=0.7$ (lower limit)

$$T_{dn}=\frac{0.9\times300\times410}{1.25}+\frac{0.7\times432\times250}{1.10}=157.3\ \text{kN}$$

• Block shear: $T_{db1}=148.8$ kN, $T_{db2}=129.5$ kN → $T_{db}=129.5$ kN

Design strength $=\min(196.8,157.3,129.5,135.8)=129.5$ kN $>100$ kN ✓.

Answer: ISA 75×75×6 connected by 3 M20 bolts (4.6) at 50 mm pitch.

Data: ISA 100×100×10 ($A_g=1903$ mm²), gusset 12 mm, connected through one 100 mm leg, M20 (4.6) bolts in a single line, threads outside the shear plane. The number of bolts is not given: assumed 4 bolts, pitch 50 mm, end distance 35 mm, gauge 55 mm from the back; $d_0=22$ mm; $f_y=250$, $f_u=410$.

1. Yielding

$T_{dg}=\dfrac{1903\times250}{1.10}\times10^{-3}=432.5$ kN

2. Rupture with shear lag

• $A_{nc}=(100-5-22)\times10=730$ mm², $A_{go}=(100-5)\times10=950$ mm²
• $b_s=100+100-10=190$ mm, $L_c=3\times50=150$ mm

$$\beta=1.4-0.076\times\frac{100}{10}\times\frac{250}{410}\times\frac{190}{150}=0.813\ \ (\text{limits }0.7\ \text{to }1.443)$$ $$T_{dn}=\frac{0.9\times730\times410}{1.25}+\frac{0.813\times950\times250}{1.10}=391.0\ \text{kN}$$

3. Block shear

$T_{db1}=343.1$ kN, $T_{db2}=286.3$ kN.

4. Bolts (threads outside the shear plane)

$V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{314}{1.25}\times10^{-3}=58.01$ kN; $V_{dpb}=83.2$ kN (10 mm angle). Four bolts: $232.0$ kN.

Answer: member design strength $=\min(432.5,\,391.0,\,286.3)=286.3$ kN (block shear governs); with the assumed 4 bolts the joint strength is $\min(286.3,\,232.0)=232.0$ kN.

Data: ISA 100×75×8 ($A_g=1336$ mm²), long (100 mm) leg bolted to a 10 mm gusset by four M16 bolts (class 8.8), pitch 40 mm, end distance 40 mm, assumed gauge 55 mm; Fe410 ($f_y=250$, $f_u=410$); $d_0=18$ mm.

1. Yielding

$T_{dg}=\dfrac{1336\times250}{1.10}\times10^{-3}=303.6$ kN

2. Net-section rupture with shear lag

• $A_{nc}=(100-4-18)\times8=624$ mm², $A_{go}=(75-4)\times8=568$ mm²
• $b_s=75+100-8=167$ mm, $L_c=3\times40=120$ mm

$$\beta=1.4-0.076\times\frac{75}{8}\times\frac{250}{410}\times\frac{167}{120}=0.795$$ $$T_{dn}=\frac{0.9\times624\times410}{1.25}+\frac{0.795\times568\times250}{1.10}=286.9\ \text{kN}$$

3. Block shear

$A_{vg}=1280$, $A_{vn}=776$, $A_{tg}=360$, $A_{tn}=288$ mm²

• $T_{db1}=253.0$ kN; $T_{db2}=214.1$ kN

4. Result

Design tensile strength of the angle $=\min(303.6,\ 286.9,\ 214.1)=214.1$ kN (block shear governs).

(Bolt check: $V_{dsb}=58.0$ kN, bearing on 8 mm leg $k_b=0.491$, $V_{dpb}=51.5$ kN; four bolts give $206.0$ kN, so the connection itself limits the joint to about $206$ kN.)

Given: ISA 90×90×12 in a truss, factored tension 100 kN, Fe410 ($f_u=410$ N/mm²), fillet welds at toe and back. Centroid of the angle $c_x\approx27$ mm from the back of the connected leg (SP 6 value $\approx27$ mm).

Weld size

Thickness of the thicker part 12 mm → minimum size 5 mm (Table 21). Maximum size at the angle edge $=t-1.5=10.5$ mm. Use $s=6$ mm (shop weld).

• $f_{wd}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²
• Strength per mm $=189.4\times0.7\times6=795.4$ N/mm

Length of weld

$$L_w=\frac{100\times10^3}{795.4}=125.7\ \text{mm}$$

Distribution so that the c.g. of the welds lies in the c.g. plane of the angle

Moments about the back weld ($L_2$): $L_1\times90=L_w\times c_x$, hence

• Toe weld: $L_1=\dfrac{125.7\times27}{90}=37.7$ mm (force $30.0$ kN)
• Back weld: $L_2=125.7-37.7=88.0$ mm (force $70.0$ kN)

Add $2s=12$ mm for end returns, and keep $L\ge4s=24$ mm.

Answer: 6 mm fillet weld; toe weld 50 mm, back weld 105 mm.

Data: 2 ISA 100×100×8 ($A_g=1539$ mm² each) on both faces of a 12 mm gusset; M16 bolts class 5.6 ($f_{ub}=500$) in one line, one shear plane through the shank and the other through the threads (double shear). Number of bolts not given: assumed 4, pitch 50 mm, end distance 40 mm, gauge 55 mm; $d_0=18$ mm; $f_y=250$, $f_u=410$.

1. Bolts

• $V_{dsb}=\dfrac{500}{\sqrt3\times1.25}(A_{nb}+A_{sb})=\dfrac{500}{\sqrt3\times1.25}(157+201)\times10^{-3}=82.7$ kN
• Bearing on gusset 12 mm ($e=40,\ p=50$): $k_b=0.676$; $V_{dpb}=106.4$ kN
• Bolt value $82.7$ kN; four bolts: $330.7$ kN

2. One angle

• $T_{dg}=\dfrac{1539\times250}{1.10}\times10^{-3}=349.8$ kN
• Rupture: $A_{nc}=(100-4-18)\times8=624$ mm², $A_{go}=(100-4)\times8=768$ mm², $b_s=100+100-8=192$ mm, $L_c=150$ mm, $\beta=0.700$ (min 0.7)

$$T_{dn}=\frac{0.9\times624\times410}{1.25}+\frac{0.700\times768\times250}{1.10}=306.4\ \text{kN}$$

• Block shear: $\min(284.5,\ 255.0)=255.0$ kN
• One angle: $\min=255.0$ kN; two angles $=510.0$ kN

3. Result

Ultimate (design) load capacity $=\min(510.0\ \text{kN (angles)},\ 330.7\ \text{kN (bolts)})=330.7$ kN, governed by shear of the bolts. (More bolts raise the capacity until the angle strength is reached.)

Reading of the figure: 4 bolts M18 (class 10.9, $f_{ub}=1000$ N/mm²) on each side of the splice, in two rows of two (gauge 110 mm, edge distances 55 mm, pitch 85 mm, end 45 mm); flats 220×18 mm; two cover plates 8 mm (total 16 mm); double shear; $d_0=20$ mm; $f_y=250$, $f_u=410$.

Plate checks

• Flat (18 mm): $T_{dg}=\dfrac{220\times18\times250}{1.10}=900.0$ kN; $T_{dn}=\dfrac{0.9(220-2\times20)\times18\times410}{1.25}=956.4$ kN
• Covers (2 × 8 = 16 mm): $T_{dg}=800.0$ kN; $T_{dn}=850.2$ kN

(i) Slip permitted at ultimate load (bearing-type)

• Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{1000}{\sqrt3\times1.25}(192+254)\times10^{-3}=206.0$ kN
• Bearing on cover plates (16 mm, thinner than the flats): $k_b=\min(\tfrac{45}{60},\tfrac{85}{60}-0.25,2.44,1)=0.750$; $V_{dpb}=\dfrac{2.5k_bdtf_u}{1.25}=177.1$ kN
• Bolt value $177.1$ kN; four bolts $=708.5$ kN
• Design load $=\min(708.5,\ 900.0,\ 956.4,\ 800.0)=708.5$ kN

(ii) Slip not permitted at ultimate load (slip-critical)

• $F_o=0.7\times1000\times192\times10^{-3}=134.4$ kN
• $V_{nsf}=\dfrac{\mu_fn_eK_hF_o}{\gamma_{mf}}=\dfrac{0.48\times2\times1.0\times134.4}{1.25}=103.2$ kN per bolt ($\mu_f=0.48$, $\gamma_{mf}=1.25$)
• Four bolts: $412.9$ kN
• Design load $=412.9$ kN (slip resistance governs).