Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 5)
5.
Two simply supported beams B1 and B2 have spans l and 2l respectively. Beam B1 has a cross-section of 1 x 1 units and beam B2 has a cross-section of 2 x 2 units. These beams are subjected to concentrated loads W each at the centre of their spans. The ratio of the maximum flexural stress in these beams is
4
2
Discussion:
12 comments Page 1 of 2.
Amlan ghosh said:
1 decade ago
fB1/fB2 = (MB1/i*y)/(MB2/i*y) = (wL/4ZB1)/(wL/2zB2) = Zb2/2Zb1 = 8/2 = 4.
Rahul said:
1 decade ago
M1 = Wl2/8, M2 = 4*Wl2/8 = Wl2/2,
Z1 = 1/6, Z2 = 8/6,
f1/f2 = (6*Wl2/8)/(6*Wl2/16) = 2.
Z1 = 1/6, Z2 = 8/6,
f1/f2 = (6*Wl2/8)/(6*Wl2/16) = 2.
J.P said:
1 decade ago
Stress = load/cross section area.
W/A1=W/A2.
W/1=W/4.
1=1/4.
4:1.
Ans ..(a) 4.
W/A1=W/A2.
W/1=W/4.
1=1/4.
4:1.
Ans ..(a) 4.
Saraba said:
10 years ago
M1 = wl/4; M2 = w(2l)/4 = wl/2;
Z1 = (1*1^2)/6 = 1/6;
Z2 = (2*2^2)/6 = 8/6 = 4/3;
f1 = M1/Z1 = (wl/4)/(1/6) = 3wl/2;
f2 = M2/Z2 = (wl/2)/(4/3) = 3wl/8;
f1/f2 = (3wl/2)/(3wl/8) = 4 = 4:1.....(Answer)
Z1 = (1*1^2)/6 = 1/6;
Z2 = (2*2^2)/6 = 8/6 = 4/3;
f1 = M1/Z1 = (wl/4)/(1/6) = 3wl/2;
f2 = M2/Z2 = (wl/2)/(4/3) = 3wl/8;
f1/f2 = (3wl/2)/(3wl/8) = 4 = 4:1.....(Answer)
(1)
Shivam chaudhary said:
10 years ago
To effectively compact chunks of clay the roller to be used is?
Abhay said:
10 years ago
Stress = Load/Area.
For B1 flexural stress = W/A1 = W/(1x1) = W.
For B2 flexural stress = W/A2 = W/(2x2) = W/4.
The ratio of the maximum flexural stress = W/(W/4) = 4.
For B1 flexural stress = W/A1 = W/(1x1) = W.
For B2 flexural stress = W/A2 = W/(2x2) = W/4.
The ratio of the maximum flexural stress = W/(W/4) = 4.
Rengel said:
9 years ago
Flexural stress should be equal to (3FL/2bd) right.
Load/Area is the axial stress?
Load/Area is the axial stress?
Manesh said:
9 years ago
Stress= load/ c/s of area.
There for B1 flex. Stress = load (w)/ (c/s of area) = w/1 = w.
And B2 flex. Stress = load (w)/(c/s of area) = w/(2*2) = w/4.
Then, the ratio of max flex stress= w/(w/4) = 4.
There for B1 flex. Stress = load (w)/ (c/s of area) = w/1 = w.
And B2 flex. Stress = load (w)/(c/s of area) = w/(2*2) = w/4.
Then, the ratio of max flex stress= w/(w/4) = 4.
(2)
Abiral said:
9 years ago
Stress = load/x - sectional area.
For B1 flexural stress = w/1 * 1 = w/1,
& for B2 flexural stress= w/2 * 2 = w/4,
Hence, the ratio of max. flex stress = w/(w/4) = 4.
For B1 flexural stress = w/1 * 1 = w/1,
& for B2 flexural stress= w/2 * 2 = w/4,
Hence, the ratio of max. flex stress = w/(w/4) = 4.
(2)
Habtamu said:
8 years ago
It also be 1/4 you should specify the numerator and denominator to find the ratio. The question is some what ambiguous. If it is the of the flexural stress of smallest rectangle to the largest rectangle the answers is A. If vice-versa answers will be D.
(4)
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