Solutions to Chapter 22 Problems

SOLUTIONS TO CHAPTER 22 PROBLEMS S.22.1

From Ex. 22.1 and noting that there are two rivets/pitch in double shear ðb
3Þ  2:5  465 ¼ 2  2  which gives

π  32  370, 4

b ¼ 12 mm:

From Eq. (22.5), the joint efficiency is η¼ S.22.2

12
3  100 ¼ 75% 12

The loading is equivalent to a shear load of 15 kN acting through the centroid of the rivet group together with a clockwise moment of 15  50 ¼ 750 kNm. The vertical shear load on each rivet is 15/9 ¼ 1.67 kN. From Ex. 22.2, the maximum shear load due to the moment will occur at rivets 3 and 9. Also qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ðrivets 1, 3, 7, 9Þ ¼ ð252 þ 252 Þ ¼ 35:4 mm r ðrivets 2, 4, 6, 8Þ ¼ 25 mm r ðrivet 5Þ ¼ 0 Then

X

r 2 ¼ 4  35:42 þ 4  252 ¼ 7500

From Eq. (22.6) Smax ¼

750  35:4 ¼ 3:54 kN 7500

Therefore, the total maximum shear force on rivets 3 and 9 is given by (see Ex. 22.2)  1=2 Smax ðtotalÞ ¼ 1:672 þ 3:542 þ 2  1:67  3:54 cos 45° ¼ 4:87 kN Then 350 ¼

4:87  103 , πd 2 =4

which gives d ¼ 4:2 mm: The plate thickness is given by 4:87  103 ¼ 600 td

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Solutions Manual from which t ¼ 1:93 mm:

S.22.3

The single shear value of one rivet is given by Single shear value ¼ 140 

π  202  10
3 ¼ 44:0 kN 4

The bearing value of one rivet is given by Bearing value ¼ 400  20  15  10
3 ¼ 120 kN Therefore, shear is the critical condition and the safe load is Safe load ¼ 4  44:0 ¼ 176:0 kN: S.22.4

The angle rivets are in double shear and the gusset plate rivets are in single shear. The double shear value of one 20 mm diameter rivet is Double shear value ¼ 2  140 

π  202  10
3 ¼ 88:0 kN 4

The bearing value of one rivet in the 12-mm-thick gusset plate within the angle is Bearing value ¼ 400  20  12  10
3 ¼ 96:0 kN Therefore, the shear load is critical and the total safe load on the angle rivets is Safe load ðangle rivetsÞ ¼ 3  88:0 ¼ 264 kN The single shear value of one gusset plate rivet is 44.0 kN. The bearing value of one rivet in the 10 mm plate is Bearing value ¼ 400  20  10  10
3 ¼ 80:0 kN Therefore, the shear load is critical and the total safe load is Total safe load ¼ 4  44 ¼ 176:0 kN: The safe load on the joint is therefore 176.0 kN, but the plate strength should be checked. Plate tensile strength ¼ ð120
20Þ  10  275  10
3 ¼ 275 kN Therefore, the safe load is 176.0 kN. S.22.5

The total length of weld is 140 mm. Therefore, the force/mm is Force=mm ¼

100  103 ¼ 714:3 N=mm 140

Solutions to Chapter 22 Problems

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The stress in the weld ¼ 100 ¼ 714.3/t, which gives t ¼ 7.1 mm. Then pffiffiffi Size of weld ¼ 2  7:1 ¼ 10:0 mm: S.22.6

I 0 for the fillet welds is given by I 0 ¼ 2  100  1002 þ

2  2003 ¼ 3:33  106 mm3 12

The bending moment on the welds is M ¼ 75  500 ¼ 37500 kNmm Load=mm due to bending ¼

37500  103  100 ¼ 1126:1 N=mm 3:33  106

Load=mm due to direct load ¼ The resultant force=mm ¼

75  103 ¼ 125 N=mm 600

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1126:12 þ 125:02 ¼ 1133:0 N=mm

pffiffiffi 2  1133:0 ¼ 16 mm: The size of weld required ¼ 100 S.22.7

Referring to Fig. S.22.7, the centroid of the weld is found by taking moments about the left hand vertical weld. i.e. 200 mm y 50 kN

200 mm

C x

150 mm

FIGURE S.22.7

x

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Solutions Manual ð2  150 þ 200Þ x ¼ 2  150  75, which gives x ¼ 45 mm The torque applied to the weld is then given by T ¼ 50ð200 þ 105Þ ¼ 15250 kNm Also J 0 ¼ Ix 0 þ Iy 0 where Ix 0 ¼ 2  150  1002 þ

2003 ¼ 3:67  106 mm3 12

 3  45 1453 ¼ 2:50  106 mm3 Iy ¼ 200  45 þ 2 þ 3 3 0

2

Then J 0 ¼ 6:17  106 mm3 Also Rmax ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   1452 þ 1002 ¼ 176:1 mm

Then the load/mm due to the torque is Load=mm ðtorqueÞ ¼

15250  103  176:1 ¼ 435:3 N=mm 6:17  106

The load/mm due to direct stress is Load=mm ðdirect stressÞ ¼

50  103 ¼ 100 N=mm 500

The resultant load/mm is then   145 1=2 Resultant ¼ 435:32 þ 1002 þ 2  435:3  100  176:1

Solutions to Chapter 22 Problems i.e. Resultant ¼ 520:7 N=mm pffiffiffi 2  520:7 ¼ 6:1 mm The required weld size is then ¼ 120 Say, a weld size of 7 mm. S.22.8

Referring to Fig. P.22.8, for the butt welds IB 0 ¼ 2  150  1902 ¼ 10:83  106 mm3 The moment taken by the butt welds is given by MB ¼

120  10:83  106  20 ¼ 130:0 kNm 200  106

Therefore, the moment resisted by the fillet welds is MF ¼ 200
130 ¼ 70 kNm For the fillet welds IF 0 ¼ 2  135  1802 þ 2 

3603 ¼ 16:52  106 mm3 12

The force in the fillet welds is then FF ¼

70  106  180 ¼ 762:7 N 16:52  106

The required thickness of the fillet weld is then 762:7 ¼ 6:9 mm 110 pffiffiffi and the weld size is ¼ 2  6:9 ¼ 9:75 mm: t¼

Say, a 10 mm fillet weld.

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