Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 3 (Q.No. 5)
5.
The section modulus of a rectangular light beam 25 metres long is 12.500 cm3. The beam is simply supported at its ends and carries a longitudinal axial tensile load of 10 tonnes in addition to a point load of 4 tonnes at the centre. The maximum stress in the bottom most fibre at the mid span section, is
13.33 kg/cm2 tensile
13.33 kg/cm2 compressive
26.67 kg/cm2 tensile
26.67 kg/cm2 compressive
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 1 of 2.

Aspire said:   9 years ago
How to solve this?

Ramachandraraju said:   9 years ago
Anybody give explanation about this.

Akscivilian said:   9 years ago
Question is wrong ,there is needed cross-sectional area, not given so we can't find the answer.

B.S = M/Z + P/A.
M = WL/4.

But A is not given.

Deb said:   8 years ago
Please, someone solve it clearly.

Thakur said:   8 years ago
Please explain the details.

Paul said:   8 years ago
How to find out A? Please explain in detail.

Muthu said:   6 years ago
Please explain the answer.

Pranjul said:   6 years ago
Area calculate from bending equation M = f * Z and f = force/area.

Basanth Babu said:   6 years ago
The cross sectional area of the beam is needed to solve this problem.
(2)

Vipin sainath said:   5 years ago
Bending Stress (Sigma) = M/Z x P/A.

First, we have to calculate Depth and Breadth ..From span/depth ratio we get Depth
Span/20 = 25m/20 = 1.25m or 125cm so D= 125cm, Then B/D=0.5 to 0.67 from this ratio we get breadth B,

B = Dx0.5 = 125X0.5 = 62.5 say B= 60cm, So we got B and D.
M/Z = (4000x2500/4) / 12.6 = 2,00,000 kg/cm2.

P/A = 10000/7500 =1.33 kg/cm2.
Therefore, Sigma = 200000 kg/cm2 x 1.33 kg/cm2 = 266000 Kg/cm2 or 26.6 Kg/m2
finally we got Max stress = 26.6kg/m2.
(3)


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