Connections in Steel Structures

Assumptions (figure values are ambiguous): the bracket plate (10 mm) lies against the column flange (ISMB 300, $t_f=12.4$ mm) and the load $P$ acts in the plane of the bolts. The 5 bolts are M20, grade 4.6, in the pattern: four corner bolts at $(\pm60,\pm80)$ mm and one at the centre (gauges 60+60 and pitch 80+80; this makes the corner distance $r=\sqrt{60^2+80^2}=100$ mm). The bolt group centroid is 150 mm from the column face (90 mm edge + 60 mm), so the eccentricity of $P$ (250 mm from the face) is $e=250-150=100$ mm. Working load $P$ = factored load / 1.5.

     o       o      y = +80
         o          y =   0   centre
     o       o      y = -80
    x=-60   +60            P acts e = 100 mm from centroid

Bolt value (IS 800:2007, cl. 10.3)

• M20: $A_{nb}=245$ mm², $f_{ub}=400$, $f_u=410$, $d_0=22$ mm.
• Shear (single shear, thread in plane): $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing (plate 10 mm, $e=40$, $p=80$): $k_b=\min(0.606,\,0.962,\,0.976,\,1)=0.606$; $V_{dpb}=2.5k_bdtf_u/\gamma_{mb}=99.4$ kN
• Bolt value $V_{db}=45.26$ kN (shear governs).

Force on the critical (corner) bolt, per unit factored load $P_u$

• $\Sigma r^2=4\times100^2=40000$ mm²; torsion $T=P_ue=100P_u$
• Direct shear (vertical) $F_D=P_u/5=0.2P_u$
• Torsional force $F_T=\dfrac{T r}{\Sigma r^2}=\dfrac{100P_u\times100}{40000}=0.25P_u$, with components $F_{Tx}=0.25P_u\times\tfrac{80}{100}=0.20P_u$ and $F_{Ty}=0.25P_u\times\tfrac{60}{100}=0.15P_u$
• Resultant $R=\sqrt{(0.20P_u)^2+(0.2P_u+0.15P_u)^2}=0.4031P_u$

Capacity

$$0.4031P_u\le45.26\Rightarrow P_u=112.3\ \text{kN}$$

Answer: factored load $P_u\approx112.3$ kN; safe working load $P=P_u/1.5\approx74.9$ kN.

Assumptions: 8 bolts M24 (grade 4.6) in two vertical lines with 4 rows at pitch $p=100$ mm; bottom row taken as the axis of rotation (Tee stem presses the column flange at the bottom). Load acts in a plane perpendicular to the bolt plane, so bolts are in shear + tension. Factored $P=225$ kN, $e=300$ mm, $M=67.5$ kN·m.

  225 kN
   --------> |T|  o o   y=300
     300     |T|  o o   y=200
             |T|  o o   y=100
             |T|  o o   y=0  (axis)

Bolt capacities (IS 800:2007 cl. 10.3)

• M24: $A_{sb}=452$, $A_{nb}=353$ mm²; $f_{ub}=400$.
• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{353}{1.25}\times10^{-3}=65.22$ kN
• Tension: $T_{nb}=\min(0.9\times400\times353,\ 240\times452\times1.25/1.10)$; $T_{db}=T_{nb}/1.25=98.62$ kN
• Bearing: hole $d_0=26$ mm, $e=p=100$, 40: $k_b=\min(40/78,\,100/78-0.25,\,0.976,\,1)=0.513$; $V_{dpb}=2.5k_b\,d\,t\,f_u/\gamma_{mb}=100.9$ kN with $t=10$ mm (Tee flange), larger than $V_{dsb}$.

Forces

• Shear per bolt $V_{sb}=225/8=28.12$ kN
• $\Sigma y^2=2(100^2+200^2+300^2)=280000$ mm²
• $T_{max}=\dfrac{M y_{max}}{\Sigma y^2}=\dfrac{67.5\times10^6\times300}{280000}=72.32$ kN

Interaction check

$$\left(\frac{28.12}{65.22}\right)^2+\left(\frac{72.32}{98.62}\right)^2=0.724<1.0$$

$T_{max}<T_{db}$ and $V_{sb}<V_{dsb}$. Safe.

Detailing

• 8 bolts M24, grade 4.6, in 2 lines at gauge $\ge 2.5d=60$ mm (use 100 mm), pitch 100 mm (min $2.5d=60$), edge distance 40 mm ($\ge1.5d_0=39$ mm).
• The Tee flange (10 mm) is checked for prying; if required, increase thickness to 12-16 mm or provide stiffeners, and provide a bearing seat at the bottom.

Answer: 8 nos. M24 bolts of grade 4.6 in 4 rows at 100 mm pitch, utilisation 0.72.

Failure mechanisms of a bolted connection

1. Shear failure of bolt: bolt shears across one plane (single shear) or two planes (double shear). Strength $V_{dsb}=\dfrac{f_{ub}}{\sqrt3\gamma_{mb}}(n_nA_{nb}+n_sA_{sb})$.
2. Bearing failure of bolt/plate: plate elongates or tears in front of the bolt. $V_{dpb}=2.5k_bdtf_u/\gamma_{mb}$. Prevented by adequate end and pitch distances.
3. Tension failure of bolt: bolt breaks in the threaded part under direct tension, $T_{db}=0.9f_{ub}A_{nb}/\gamma_{mb}$.
4. Tearing of plate at the edge (shear-out): if end distance is too small, the plate shears out behind the bolt.
5. Tension (net-section) rupture of plate: the plate fails across the net section through the bolt holes, $T_{dn}=0.9A_nf_u/\gamma_{m1}$.
6. Yielding of gross section of plate: $T_{dg}=A_gf_y/\gamma_{m0}$.
7. Block shear failure: a block of plate tears out along shear and tension planes at the bolt group.
8. Bolt bending (for long grip): prevented by limiting grip $\le8d$ and by bearing.
9. Prying failure: extra tension in bolts due to flexible connected plates.
10. Slip in HSFG joints when friction is exceeded.

Combined stresses in eccentric connections

• Load in the plane of the bolt group (eccentricity in the plane): bolts get direct shear $P/n$ plus shear due to torsion $T r/\Sigma r^2$; the vector sum of both is the resultant shear in the critical bolt, $R=\sqrt{(F_x)^2+(F_y)^2}$, checked against $V_{db}=\min(V_{dsb},V_{dpb})$.
• Load acting perpendicular to the plane of the bolt group (bracket on column flange): bolts get direct shear $V=P/n$ and tension due to moment $T=My/\Sigma y^2$. Combined check:

$$\left(\frac{V}{V_{db}}\right)^2+\left(\frac{T}{T_{db}}\right)^2\le1.0$$

For HSFG bolts the interaction is $V_{sf}$ vs $T_f$: $\dfrac{V_{sf}}{V_{nsf}}+\dfrac{T_f}{T_{nf}}\le1$.

Assumptions: the bracket plate is shop-welded to the column flange by a C-shaped weld in the plane of the flange: two horizontal welds of 150 mm and one vertical weld of 400 mm. The load $P$ acts parallel to the flange at $e=120$ mm from the weld plane, so the weld group carries direct shear $P$ plus bending $M=Pe$. Steel E250 ($f_u=410$ N/mm²), fillet size $s=5$ mm, throat $t_t=0.7\times5=3.5$ mm.

Weld strength (IS 800:2007 cl. 10.5.7)

$$f_{wd}=\frac{f_u}{\sqrt3\gamma_{mw}}=\frac{410}{\sqrt3\times1.25}=189.37\ \text{N/mm}^2,\qquad q_{allow}=f_{wd}t_t=662.80\ \text{N/mm}$$

Weld group properties (per mm of throat)

• Total length $L=2\times150+400=700$ mm
• Neutral axis at mid-depth (symmetrical about the horizontal axis).
• $I_{xx}=\dfrac{400^3}{12}+2\times150\times200^2=17333333$ mm³ (the weld thickness itself is neglected)

Forces per unit factored load $P$ at the extreme weld (y = 200 mm)

• Shear: $q_s=\dfrac{P}{L}=0.001429P$ N/mm
• Bending: $q_b=\dfrac{M y}{I_{xx}}=\dfrac{120P\times200}{17333333}=0.001385P$ N/mm
• Resultant: $q=\sqrt{q_s^2+q_b^2}=0.001989P$ N/mm

Safe load

$$0.001989P\le662.80\Rightarrow P_u=333.2\ \text{kN}$$

Answer: design (factored) load $P_u\approx333.2$ kN, i.e. working load $P\approx222.1$ kN. (If the load were instead in the plane of the weld, torsion would be added; the bending reading is used here.)

Assumptions from the figure: main plates 10 mm and 18 mm, two cover plates 6 mm each (total 12 mm), plate width $=40+60+60+40=200$ mm; 6 bolts M20 (grade 4.6) on each side of the joint (3 across the width at 60 mm gauge, 2 along the load at 60 mm pitch), end distance 40 mm, $d_0=22$ mm. Steel E250: $f_y=250$, $f_u=410$ N/mm². The joint is in double shear; one shear plane is taken through the threads.

Strength of one bolt (IS 800:2007 cl. 10.3)

• Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{f_{ub}}{\sqrt3\gamma_{mb}}(A_{nb}+A_{sb})=\dfrac{400}{\sqrt3\times1.25}(245+314)\times10^{-3}=103.28$ kN
• Bearing on the thinnest main plate (10 mm < 12 mm of covers): $k_b=\min\left(\dfrac{40}{3\times22},\dfrac{60}{3\times22}-0.25,\dfrac{400}{410},1\right)=0.606$

$$V_{dpb}=\frac{2.5k_bdtf_u}{\gamma_{mb}}=\frac{2.5\times0.606\times20\times10\times410}{1.25}\times10^{-3}=99.39\ \text{kN}$$

• Bolt value $=\min(103.28,99.39)=99.39$ kN (bearing governs)

Strength of the joint

1. Bolts: $6\times99.39=596.4$ kN
2. Main plate 10 mm, gross section: $T_{dg}=\dfrac{200\times10\times250}{1.10}=454.5$ kN
3. Main plate net section (3 holes): $T_{dn}=\dfrac{0.9(200-3\times22)\times10\times410}{1.25}=395.6$ kN
4. Cover plates (2 × 6 = 12 mm): $T_{dg}=545.5$ kN, $T_{dn}=474.7$ kN
5. Block shear of the 10 mm plate: $T_{db}=\min(551.7,501.1)=501.1$ kN

Answer: strength of the joint $=\min(\ldots)=395.6$ kN, governed by net-section rupture of the 10 mm plate (the 18 mm plate and covers are stronger).

The load is not given, so the weld is designed to develop the full design tensile strength of the angle (ISA 100×100×10, $A_g=1903$ mm², centroid $c_x=28.4$ mm from the back), the usual approach.

Load to be carried

$$T_{dg}=\frac{A_gf_y}{\gamma_{m0}}=\frac{1903\times250}{1.10}\times10^{-3}=432.5\ \text{kN}$$

Weld size (IS 800:2007 cl. 10.5.2)

• Thicker part $=12$ mm (10 < t ≤ 20): minimum size 5 mm.
• Maximum size at the toe edge of the angle: $t-1.5=8.5$ mm. Use $s=6$ mm shop fillet weld.
• Throat $t_t=0.7\times6=4.2$ mm; $f_{wd}=\dfrac{f_u}{\sqrt3\gamma_{mw}}=\dfrac{410}{\sqrt3\times1.25}=189.4$ N/mm²
• Strength per mm: $q=189.4\times4.2=795.4$ N/mm (= 0.795 kN/mm)

Weld length

Total length $L=\dfrac{432.5\times1000}{795.4}=544$ mm.

To avoid eccentricity, the welds are placed so that their resultant passes through the centroid of the angle. Let $L_1$ be the weld at the toe (far from centroid) and $L_2$ the weld at the heel/back:

• $F_1$ (toe) $=T\dfrac{c_x}{100}=122.8$ kN → $L_1=154.4$ mm
• $F_2$ (heel) $=T\dfrac{100-c_x}{100}=309.7$ kN → $L_2=389.3$ mm

Add $2s=12$ mm for end returns (the weld must be continued round the end), so

Provide: 6 mm fillet weld, toe side length $=170$ mm and heel side length $=405$ mm (total about $544$ mm effective), with end returns of $2s$; minimum length $4s=24$ mm is satisfied. The 12 mm gusset must be long enough to take the 400 mm weld, otherwise use 8 mm size or slot/end weld.

Assumptions: the bracket plate is welded to the column flange by fillet welds on three sides: two horizontal welds of 250 mm (top and bottom) and one vertical weld of 300 mm. The factored load $P=200$ kN acts parallel to the flange at $e=100$ mm from the weld plane, so the weld carries direct shear and bending $M=Pe$. ISMB 400: $t_f=16$ mm. Shop weld, E250 ($f_u=410$).

Weld group (per mm of throat)

• $L=2\times250+300=800$ mm
• Neutral axis at mid-depth. $I_{xx}=\dfrac{300^3}{12}+2\times250\times150^2=13500000$ mm³
• $M=200\times10^3\times100=20\times10^6$ N·mm

Force per mm of weld at the extreme weld (y = 150 mm)

• Shear: $q_s=\dfrac{P}{L}=\dfrac{200\times10^3}{800}=250.0$ N/mm
• Bending: $q_b=\dfrac{My}{I_{xx}}=\dfrac{20\times10^6\times150}{13500000}=222.2$ N/mm
• Resultant: $q=\sqrt{q_s^2+q_b^2}=334.5$ N/mm

Size of weld

$$f_{wd}=\frac{410}{\sqrt3\times1.25}=189.4\ \text{N/mm}^2,\quad t_t=\frac{q}{f_{wd}}=\frac{334.5}{189.4}=1.77\ \text{mm},\quad s=\frac{t_t}{0.7}=2.52\ \text{mm}$$

The calculated size is small, so the minimum size of IS 800:2007 Table 21 governs: thicker part 16 mm (10 < t ≤ 20) gives 5 mm.

Answer: provide 5 mm fillet weld all round the three sides. Check: capacity of 5 mm weld $=662.8$ N/mm against demand $334.5$ N/mm (utilisation $0.50$).

Assumptions: the 12 mm bracket plate is welded to the ISMB 350 flange ($t_f=14.2$ mm) by a C-shaped weld: top and bottom 100 mm and vertical 160 mm. The working load of 100 kN acts parallel to the flange at $e=140$ mm from the weld plane, giving direct shear and bending in the weld. Factored load $P_u=1.5\times100=150$ kN. Shop weld, E250.

Weld group (per mm of throat)

• $L=2\times100+160=360$ mm
• $I_{xx}=\dfrac{160^3}{12}+2\times100\times80^2=1621333$ mm³
• $M=150\times10^3\times140=21.0\times10^6$ N·mm

Force per mm at the extreme weld (y = 80 mm)

• Shear: $q_s=\dfrac{150\times10^3}{360}=416.7$ N/mm
• Bending: $q_b=\dfrac{My}{I_{xx}}=\dfrac{21.0\times10^6\times80}{1621333}=1036.2$ N/mm
• Resultant: $q=\sqrt{q_s^2+q_b^2}=1116.8$ N/mm

Size of weld

$$f_{wd}=\frac{410}{\sqrt3\times1.25}=189.4\ \text{N/mm}^2,\qquad t_t=\frac{1116.8}{189.4}=5.90\ \text{mm},\qquad s=\frac{t_t}{0.7}=8.43\ \text{mm}$$

Take $s=9$ mm (capacity $0.7\times9\times189.4=1193$ N/mm $>1116.8$ N/mm; an 8 mm weld gives only 1060 N/mm).

Checks: minimum size for 14.2 mm flange = 5 mm; maximum size = plate thickness $-1.5=10.5$ mm (12 mm plate), so 9 mm is acceptable.

Answer: 9 mm fillet weld (shop) on all three sides.

Assumptions: bracket plate (12 mm) is lapped on the flange of ISMB 300 ($b_f=140$ mm, $t_f=12.4$ mm); the 220 kN factored load acts in the plane of the bolts at $e=90$ mm from the centroid of the bolt group. M20 bolts of grade 4.6 (shop bolting), $d_0=22$ mm, in two vertical lines.

Bolt value (IS 800:2007 cl. 10.3)

• Shear (single, thread in plane): $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing (12 mm plate, $e=40$, $p=60$): $k_b=\min(0.606,0.659,0.976,1)=0.606$; $V_{dpb}=119.3$ kN
• Bolt value $=45.26$ kN

Trial layout

Two vertical lines of 5 bolts: gauge $g=70$ mm (edge distance $(140-70)/2=35\ge1.5d_0=33$ mm), pitch $p=60$ mm (≥ $2.5d=50$), end distance 40 mm. Total 10 bolts.

   e = 90      o   o   y=+120
   <------     o   o   y=+60
 P  |          o   o   y=0
 220 kN        o   o   y=-60
               o   o   y=-120

Forces

• $\Sigma(x^2+y^2)=10(35^2)+2\times2(60^2+120^2)=84250$ mm²
• Moment $T=220\times90=19800$ kN·mm
• Direct shear per bolt $=220/10=22.0$ kN (vertical)
• Critical corner bolt ($x=35,y=120$): torsional components $F_x=\dfrac{Ty}{\Sigma r^2}=28.20$ kN, $F_y=\dfrac{Tx}{\Sigma r^2}=8.23$ kN
• Resultant: $R=\sqrt{28.20^2+(22.0+8.23)^2}=41.34$ kN

$R=41.34\ \text{kN}<V_{db}=45.26$ kN. Safe.

Answer: 10 bolts M20 (4.6), 2 lines of 5, pitch 60 mm, gauge 70 mm, end distance 40 mm.

Assumptions: the 200 kN is the factored load, acting in the plane of the bolts at $e=250$ mm from the centroid of the bolt group (column face taken as the bolt line). The column flange is wide enough for a gauge of 100 mm. M20 bolts, grade 4.6, in two vertical lines, plate 12 mm, E250.

Bolt value

• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing ($e=40$, $p=70$, $t=12$ mm): $k_b=0.606$, $V_{dpb}=119.3$ kN
• Bolt value $=45.26$ kN

Trial and check

The bolts are assumed at gauge 100 mm and pitch 70 mm. For $n$ bolts the critical corner bolt receives direct shear $P/n$ plus the torsion effect of $P e$ about the centroid:

For 14 bolts: $\Sigma(x^2+y^2)=309400$ mm², $T=200\times250=50000$ kN·mm

• Direct shear $=200/14=14.29$ kN
• Corner bolt ($x=50$, $y=210$): $F_x=\dfrac{Ty}{\Sigma r^2}=33.94$ kN; vertical torsion part $=8.08$ kN
• $R=\sqrt{33.94^2+(14.29+8.08)^2}=40.64$ kN $<45.26$ kN

Answer: 14 bolts M20 (4.6) in two vertical lines of 7, gauge 100 mm, pitch 70 mm, end distance 40 mm (bracket depth about 500 mm).

Assumptions: the spacing 40-80-80-80-80-40 mm gives 5 bolts in each vertical line (end 40 mm, four pitches of 80 mm), so 10 bolts M20 (grade 4.6) in all, gauge 120 mm. $P$ acts in the plane of the bolts at $e=200$ mm from the bolt centroid. Plates: flange 9.1 mm, bracket 10 mm; $f_u=410$ N/mm².

Bolt value

• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing on the thinner plate (9.1 mm): $k_b=\min(0.606,0.962,0.976,1)=0.606$; $V_{dpb}=90.4$ kN
• Bolt value $V_{db}=45.26$ kN

Force in the critical bolt per unit load $P$ (kN)

• $\Sigma(x^2+y^2)=10(60^2)+2\times2(80^2+160^2)=164000$ mm²
• Corner bolt ($x=60,\ y=160$): $F_x=\dfrac{Pey}{\Sigma r^2}=0.1951P$; $F_y=\dfrac{P}{10}+\dfrac{Pex}{\Sigma r^2}=0.1732P$
• $R=\sqrt{F_x^2+F_y^2}=0.2609P$

Safe load

$$0.2609P_u=45.26\Rightarrow P_u=173.5\ \text{kN}$$

Answer: design (factored) load $P_u\approx173.5$ kN; the safe working load $P\approx P_u/1.5=115.7$ kN.

Reading of the data: "14-16 mm diameter" is taken as 14 bolts of 16 mm diameter (M16, product grade C, class 4.6) in two vertical rows 100 mm apart, 7 per row, pitch 80 mm. The 100 kN is treated as the design (factored) load; if it is a working load multiply it by 1.5. Plate 12 mm, Fe410.

Bolt value (IS 800:2007 cl. 10.3)

• M16: $A_{nb}=157$ mm², $d_0=18$ mm (grade C hole: $d+2$).
• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=29.01$ kN
• Bearing ($e=40$, $p=80$): $k_b=\min(0.741,1.231,0.976,1)=0.741$; $V_{dpb}=2.5k_bdtf_u/\gamma_{mb}=116.6$ kN
• Bolt value $=29.01$ kN

Forces

• $\Sigma(x^2+y^2)=14(50^2)+2\times2(80^2+160^2+240^2)=393400$ mm²; $T=100\times250=25000$ kN·mm
• Direct shear per bolt $=100/14=7.14$ kN
• Critical bolt (corner, $x=50$, $y=240$): $F_x=\dfrac{Ty}{\Sigma r^2}=15.25$ kN; torsional vertical part $=3.18$ kN
• $R=\sqrt{15.25^2+(7.14+3.18)^2}=18.42$ kN

$R=18.42\ \text{kN}<29.01$ kN (utilisation 0.63).

Answer: the joint is safe. (End distance 40 mm ≥ $1.5d_0=27$ mm; pitch 80 mm lies between 40 mm and $\min(32t,300)$.)

Data: $P=100$ kN, $e=500$ mm, two rows of five bolts, gauge 100 mm, pitch 60 mm; the load lies in the plane of the bolt group (torsion). Take $x=\pm50$ mm, $y=0,\pm60,\pm120$ mm from the centroid.

   P=100 kN                  o   o   y=+120
   <------ 500 ------>       o   o   +60
                             o   o   0
                             o   o   -60
                             o   o   -120

Step 1 - direct shear

$F_d=\dfrac{P}{n}=\dfrac{100}{10}=10.0$ kN (vertical, same in every bolt)

Step 2 - torsion

$$\Sigma(x^2+y^2)=10(50^2)+2\times2(60^2+120^2)=97000\ \text{mm}^2,\qquad T=100\times500=50000\ \text{kN·mm}$$

Critical bolt: farthest corner bolt, $x=50$, $y=120$ ($r=130$ mm).

• Horizontal component: $F_x=\dfrac{Ty}{\Sigma r^2}=\dfrac{50000\times120}{97000}=61.86$ kN
• Vertical component: $\dfrac{Tx}{\Sigma r^2}=25.77$ kN

Step 3 - resultant

$$R=\sqrt{F_x^2+(F_d+F_{y,T})^2}=\sqrt{61.86^2+(10.0+25.77)^2}=71.46\ \text{kN}$$

Answer: greatest force in a bolt $\approx71.5$ kN (at the extreme corner bolt).

Plug and slot welds

A plug weld is made by filling a circular hole in one lapped plate with weld metal so that it fuses with the other plate beneath. A slot weld is the same in an elongated (slot) hole. They transfer shear between overlapping plates when the edge fillet length is not enough, and they also stop lapped plates from buckling or separating.

   top plate  =====[ O ]=====     plug (circular hole filled)
   bottom plate ===============
   top plate  ===[ ====== ]===    slot (elongated hole)

Design concepts (IS 800:2007 cl. 10.5.3)

• The weld is assumed to resist shear on the area of the hole/slot (faying surface): $\text{strength}=A_w\,f_{wd}$ with $f_{wd}=\dfrac{f_u}{\sqrt3\gamma_{mw}}$ and $A_w$ the area at the contact surface.
• The weld is not designed for tension or bending across the plate.

Requirements

1. Hole/slot size: diameter of hole not less than thickness of the part containing it $+8$ mm, nor more than the minimum diameter $+3$ mm or $2.25t$ (whichever is greater). Slot width follows the same rule; slot length not more than $10t$.
2. Spacing: centre-to-centre spacing of plug welds not less than $4\times$ hole diameter. Distance from the hole edge to the plate edge not less than $2\times$ the hole diameter.
3. Slot ends: semicircular or with corners rounded to radius not less than $t$ (or square if one end extends to the plate edge).
4. Filling: for plates up to 16 mm the hole is filled completely; thicker plates may be filled to at least $t/2$ (not less than 16 mm).
5. Fusion: weld must penetrate and fuse with the underlying plate; good welding preparation required.
6. Plug/slot welds are used mainly for lapped joints, built-up members and to stiffen thin plates, and are not used alone in primary tension members.

Given: factored $P=600$ kN at $e=300$ mm from the column face; the load is not in the plane of the bolts, so each bolt carries direct shear and tension from the moment. ISHB 300 @ 577 N/m: $b_f=250$ mm, $t_f=10.6$ mm. Bracket plate 12 mm; M24 bolts, grade 4.6, $d_0=26$ mm; E250.

Bolt capacities (IS 800:2007 cl. 10.3)

• M24: $A_{sb}=452$, $A_{nb}=353$ mm²
• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{353}{1.25}\times10^{-3}=65.22$ kN
• Tension: $T_{db}=\dfrac{\min(0.9f_{ub}A_{nb},\ f_{yb}A_{sb}\gamma_{mb}/\gamma_{m0})}{\gamma_{mb}}=98.62$ kN
• Bearing on the flange (10.6 mm, $e=40,\ p=80$): $k_b=0.513$, $V_{dpb}=107.0$ kN (not governing)

Layout

Two vertical lines of 8 bolts (16 bolts), gauge 100 mm (flange 250 mm), pitch 80 mm, end distance 40 mm ($\ge1.5d_0=39$ mm), bracket depth $=40+7\times80+40=640$ mm. The bottom bolt row is taken as the axis of rotation (conservative).

        |  o o  y=560
        |  o o  y=480
  600kN |  o o   ...
 ------>|  o o  y=80
  300   |  o o  y=0 (axis)

Forces

• Shear per bolt $V=\dfrac{600}{16}=37.5$ kN
• $M=600\times300=180\times10^3$ kN·mm; $\Sigma y^2=2\sum y_i^2=1792000$ mm²
• Maximum tension (top row): $T=\dfrac{My_{max}}{\Sigma y^2}=\dfrac{180000\times560}{1792000}=56.25$ kN

Interaction (cl. 10.3.6)

$$\left(\frac{V}{V_{db}}\right)^2+\left(\frac{T}{T_{db}}\right)^2=\left(\frac{37.5}{65.22}\right)^2+\left(\frac{56.25}{98.62}\right)^2=0.656\le1.0$$

Also $V<V_{dsb}$ and $T<T_{db}$. Safe.

Answer: 16 bolts M24 (4.6) in 2 lines of 8, pitch 80 mm, gauge 100 mm. The 10.6 mm flange should be checked for prying (provide a stiffener at the bracket level, or a thicker end plate) and the bracket should bear on a seat at the bottom.

Assumptions: the two bracket plates share the load equally, $P/2$ each. Each bracket has 4 rows of 2 bolts (pitch 60, end distance 40, gauge 120 mm) = 8 bolts M20, grade 4.6, bracket thickness at least 10 mm. The load is in the plane of the bolts at $e=300$ mm from the bolt centroid. The 30 mm off-axis position of $P$ in the section view is assumed to be taken by the column (not by the bolts).

Bolt value (IS 800:2007 cl. 10.3)

• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing ($e=40$, $p=60$, $t=10$ mm): $k_b=0.606$, $V_{dpb}=99.4$ kN
• Bolt value $=45.26$ kN

Force in the critical bolt per unit load $P_b$ on one bracket

• $\Sigma(x^2+y^2)=8(60^2)+2\times2(30^2+90^2)=64800$ mm²
• Corner bolt ($x=60,\ y=90$): $F_x=\dfrac{P_b\times300\times90}{64800}=0.4167P_b$; $F_y=\dfrac{P_b}{8}+\dfrac{P_b\times300\times60}{64800}=0.4028P_b$
• $R=\sqrt{F_x^2+F_y^2}=0.5795P_b$

Safe load

$$P_b=\frac{45.26}{0.5795}=78.1\ \text{kN per bracket}$$

Answer: $P=2P_b=156.2$ kN (design/factored load for the two brackets together; working value = this / 1.5 if needed).

Reading of the figure: five M16 bolts (class 8.6, $f_{ub}=800$ N/mm²) in a zig-zag (staggered) chain across the plate: transverse spacings 25-20-20-20-20-25 mm (so plate width $=130$ mm) and longitudinal staggering 40 mm between successive bolts. 16 mm plates, Fe410 ($f_y=250$, $f_u=410$); $d_0=18$ mm; lap joint, single shear, thread in the shear plane.

   +----------------------------+
   | o                          |  y = 25
   |    o                       |  y = 45
   |       o                    |  y = 65
   |          o                 |  y = 85
   |             o              |  y = 105
   +----------------------------+  <-- 40 mm stagger

Bolt value

• Shear: $V_{dsb}=\dfrac{800}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=58.01$ kN
• Bearing ($p=40$, $e\ge40$): $k_b=\min(0.74,\ \tfrac{40}{54}-0.25,\ 1.95,\ 1)=0.491$; $V_{dpb}=103.0$ kN
• Strength of 5 bolts $=5\times58.01=290.1$ kN

Plate strength

• Gross section: $T_{dg}=\dfrac{130\times16\times250}{1.10}=472.7$ kN
• Net section through one hole: $A_n=(130-18)\times16=1792$ mm²: $T_{dn}=\dfrac{0.9\times1792\times410}{1.25}=529.0$ kN
• Net section along the zig-zag through all five holes: $b_n=130-5\times18+4\times\dfrac{40^2}{4\times20}=120$ mm; $T_{dn}=566.8$ kN (larger, so the one-hole section governs among net sections)

Mode of failure and efficiency

Joint strength $=\min(290.1,\ 472.7,\ 529.0)=290.1$ kN. The joint fails by shearing of the bolts.

$$\eta=\frac{\text{strength of joint}}{\text{strength of solid plate}}=\frac{290.1}{472.7}\times100=61.4\%$$

Answer: failure by bolt shear; efficiency about 61%. (Gauge of 20 mm is below the code minimum $2.5d=40$ mm, so the figure values are used only for the calculation.)

Joint type: single-bolted (one row) double cover butt joint; bolts in double shear. Plates 12 mm, Fe410; M16 bolts grade 4.6, $d_0=18$ mm. Covers: two plates of 8 mm ($2\times8=16\ge12$ mm).

Strength of one bolt (IS 800:2007 cl. 10.3)

• Shear (double shear, one plane in thread): $V_{dsb}=\dfrac{400}{\sqrt3\times1.25}(157+201)\times10^{-3}=66.14$ kN
• Bearing on 12 mm main plate ($e=30\ge1.5d_0=27$ mm, $p=40$): $k_b=\min(0.556,\ 0.491,\ 0.976,\ 1)=0.491$; $V_{dpb}=\dfrac{2.5k_bdtf_u}{\gamma_{mb}}=77.3$ kN
• Bolt value $=66.14$ kN

Pitch for maximum efficiency

Efficiency is highest when plate net strength between bolt holes equals the bolt value, so that no material is wasted:

$$\frac{0.9(p-d_0)tf_u}{\gamma_{m1}}=V_b\Rightarrow p=18+\frac{66.14\times10^3}{0.9\times12\times410/1.25}=36.7\ \text{mm}$$

This is below the minimum pitch $2.5d=40$ mm, so take $p=40$ mm. Check:

• Net strength of plate in a pitch strip: $T_{dn}=\dfrac{0.9(40-18)\times12\times410}{1.25}=77.9$ kN $>V_b$
• Gross strength of plate in a pitch strip: $T_{dg}=\dfrac{40\times12\times250}{1.10}=109.1$ kN

Efficiency

$$\eta=\frac{\text{lesser of bolt value, }T_{dn}}{T_{dg}}=\frac{66.14}{109.1}\times100=60.6\%$$

Answer: M16 bolts at 40 mm pitch in one row, end distance 30 mm, with 8 mm cover plates on both sides; efficiency $\approx61\%$. (A single row cannot do better than this; efficiency can be raised by using more rows with larger-capacity bolts in zig-zag.)

Data: working load 300 kN → factored $P_u=1.5\times300=450$ kN. Flats 100×20 mm, covers 12 mm each, M16 HSFG bolts class 10.9 ($f_{ub}=1000$ N/mm²), $d_0=18$ mm, E250. "Slip permitted at design load" means the bolts are designed as bearing-type HSFG (shear and bearing, IS 800:2007 cl. 10.4). Double shear with one plane through the threads.

Bolt value

• Shear: $V_{dsb}=\dfrac{f_{ub}}{\sqrt3\gamma_{mb}}(A_{nb}+A_{sb})=\dfrac{1000}{\sqrt3\times1.25}(157+201)\times10^{-3}=165.35$ kN
• Bearing on 20 mm flat ($e=30$, $p=50$): $k_b=\min(0.556,\ 0.676,\ 2.44,\ 1)=0.556$

$$V_{dpb}=\frac{2.5k_bdtf_u}{\gamma_{mb}}=\frac{2.5\times0.556\times16\times20\times410}{1.25}\times10^{-3}=145.8\ \text{kN}$$

• Bolt value $=145.8$ kN (bearing governs)

Number of bolts

$n=\dfrac{450}{145.8}=3.08\Rightarrow$ provide 4 bolts on each side of the joint, one per row (single line at 50 mm pitch, end distance 30 mm) so the flat is weakened by one hole only.

Checks

• Flat gross yielding: $T_{dg}=\dfrac{100\times20\times250}{1.10}=454.5$ kN $\ge450$ ✓.
• Flat net rupture (1 hole): $T_{dn}=\dfrac{0.9(100-18)\times20\times410}{1.25}=484.1$ kN $\ge450$ ✓.
• Cover plates $2\times12=24>20$ mm: $T_{dg}=545.5$ kN ✓.
• Bolts: $4\times145.8=583.1$ kN $\ge450$ ✓.
• (For information, slip resistance per bolt at service $V_{nsf}=\mu n_eK_hF_o/\gamma_{mf}=95.9$ kN with $\mu=0.48$, $F_o=0.7f_{ub}A_{nb}$, $\gamma_{mf}=1.10$; total $384$ kN > 300 kN, so the joint does not slip under working load either.)

Answer: 4 M16 (10.9) HSFG bolts each side in a single line at 50 mm pitch, with 12 mm cover plates on both faces.

A riveted joint can fail in the following ways. In design each mode is checked and the weakest governs.

  P -->  [===O===O===O===]  --> P   lap joint

1. Shearing of rivets: the rivet shears across one section (single shear) or two sections (double shear). Strength $=\tau_{vf}\,\dfrac{\pi}{4}d^2$ (per shear plane).
2. Bearing of rivet or plate: the rivet crushes or the plate hole elongates under bearing pressure. Strength $=\sigma_{pf}\,d\,t$.
3. Tearing of plate between rivets (net section): the plate tears across the line of rivet holes where area is least. Strength $=\sigma_{at}(b-nd_0)t$.
4. Tearing of plate at the edge (shear-out): if the end distance is too small, the plate behind the rivet shears out. Avoided by end distance $\ge1.5d_0$.
5. Splitting / bursting of the plate edge due to small edge distance across the load (tension at the edge).
6. Rupture of the plate by block shear (tear-out of a block of plate along bolt line in shear and tension).
7. Crushing of rivets (if rivet material is softer than the plate).
8. Tension failure of rivet (head popping) when rivets are loaded in tension.
9. Shearing of cover plate or failure of the cover plates in a butt joint.

Remedies

Provide proper end distance, edge distance and pitch (min $2.5d$, max $16t$ or 200 mm), use enough rivets for the load, and use plates of adequate net section.

Efficiency of the joint $=\dfrac{\text{least of the above strengths}}{\text{strength of solid plate}}\times100$, typically 60-75% for single riveting and up to 85% for multiple riveting.

Working stress method (IS 800:1984). Hole diameters are used for the rivet areas: $d_0=d+1.5$ mm, i.e. 17.5 mm for the 16 mm rivets (A, B) and 21.5 mm for the 20 mm rivet (C).

Reading of the figure: rivets A and B are 60 mm above and below the horizontal axis and 90 mm from the load line; rivet C is on the axis, 120 mm from the load line. $P=20$ kN, vertical.

Step 1 - rivet areas and centroid of the group

• $A_A=A_B=\dfrac{\pi}{4}(17.5)^2=240.5$ mm²; $A_C=\dfrac{\pi}{4}(21.5)^2=363.1$ mm²; $\Sigma A=844.1$ mm²
• Distance of centroid from load line: $\bar x=\dfrac{2A_A(90)+A_C(120)}{\Sigma A}=102.90$ mm
• So A, B are $12.90$ mm on the load side of the centroid and C is $17.10$ mm on the far side.

Step 2 - direct shear stress (same in all rivets)

$\tau_d=\dfrac{P}{\Sigma A}=\dfrac{20000}{844.1}=23.69$ N/mm² (vertical)

Step 3 - torsion

$M=Pe=20\times10^3\times102.90=2058\times10^3$ N·mm

$$J=\Sigma A r^2=2A_A(12.90^2+60^2)+A_C(17.10^2)=1918015\ \text{mm}^4$$

Torsional shear stress in a rivet: $\tau_t=\dfrac{Mr}{J}$ (perpendicular to the radius).

Step 4 - resultant stresses

Rivet B ($x=12.90$, $y=-60$): $\tau_x=\dfrac{My}{J}=64.38$ N/mm² (magnitude), $\tau_y=\tau_d+\dfrac{Mx}{J}=37.54$ N/mm²

$$\tau_B=\sqrt{64.38^2+37.54^2}=74.53\ \text{N/mm}^2$$

Rivet C ($y=0$): $\tau_x=0$, $\tau_y=\tau_d-\dfrac{M\times17.10}{J}=5.35$ N/mm²

$$\tau_C=5.35\ \text{N/mm}^2$$

Answer: shear stress in rivet B (and A) $\approx74.5$ N/mm²; in rivet C $\approx5.3$ N/mm², both below the permissible 100 N/mm² of power-driven shop rivets.

Basis: working stress method (IS 800:1984), rivet dia 20 mm (hole 21.5 mm), $P=180$ kN (working) at $e=275$ mm. The load is shared equally by the two angles, 90 kN each. Each angle has one vertical line of 9 rivets, pitch 60 mm, end distance 40 mm, connected to the column flange. Rivets act in single shear. Thickness of angle leg 8 mm governs bearing.

Permissible stresses (IS 800:1984, Table 11.1)

Rivet values (area of hole $=363.1$ mm²)

Force in the extreme rivet (one angle)

• Moment $M=90\times275=24750$ kN·mm
• $\Sigma y^2=2(60^2+120^2+180^2+240^2)=216000$ mm²
• Horizontal force (torsion) at the extreme rivet: $F_x=\dfrac{M y_{max}}{\Sigma y^2}=\dfrac{24750\times240}{216000}=27.50$ kN
• Direct (vertical) shear: $F_d=\dfrac{90}{9}=10.00$ kN
• Resultant: $R=\sqrt{27.50^2+10.00^2}=29.26$ kN

Check

• (i) $R=29.26\ \text{kN}<36.31$ kN → safe (ratio 0.81).
• (ii) $R=29.26\ \text{kN}<32.67$ kN → safe (ratio 0.90).

Answer: provide 9 nos. 20 mm rivets at 60 mm pitch (40 mm end distance) in each of the two angles, for both hot-driven and cold-driven rivets. (18 mm rivets would be marginal, 16 mm unsafe.)