Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 27 (Q.No. 35)
35.
A circular plate 100 mm diameter is welded to another plate by means of 6 mm fillet weld. If the permissible shearing stress in the weld equals 10 kg/mm2, then the greatest twisting moment that can be resisted by the weld will be
Discussion:
3 comments Page 1 of 1.
Rahul verma said:
2 years ago
T = PD/2.
STRENGTH = P(force) = AREA * STRESS.
P = (0.7s)*πD*10.
P = (0.7*6)*(π100)*10 = 4.2X10^3π .
T = {4.2X10^3π*100}/2 kg-mm
T = 212kg-m.
STRENGTH = P(force) = AREA * STRESS.
P = (0.7s)*πD*10.
P = (0.7*6)*(π100)*10 = 4.2X10^3π .
T = {4.2X10^3π*100}/2 kg-mm
T = 212kg-m.
(2)
Noanie said:
5 years ago
T. M. =P*Dia.
=(0.707*s*l*t)*0.1(kg-m).
s=6mm, l=pi*Radius, shear stress t=10
=(0.707*6*π* 50 * 10) * 0.1 = 212.1π(Kg-m).
=(0.707*s*l*t)*0.1(kg-m).
s=6mm, l=pi*Radius, shear stress t=10
=(0.707*6*π* 50 * 10) * 0.1 = 212.1π(Kg-m).
(1)
Ranjeet sharma said:
8 years ago
Twisting moment=p*d/2.
P=k.s.l.permissible stress,
=0.7*6*d*10.
Therefore,
Twisting moment=0.7*6*d*10*50=210πkg.m,
Hence option (c) is correct.
P=k.s.l.permissible stress,
=0.7*6*d*10.
Therefore,
Twisting moment=0.7*6*d*10*50=210πkg.m,
Hence option (c) is correct.
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