# Strength of Materials - Mechanical Engineering (MCQ) questions and answers for Q. 11565

Q.  A simply supported beam carries uniformly distributed load of 20 kN/m over the length of 5 m. If flexural rigidity is 30000 kN.m2, what is the maximum deflection in the beam?
- Published on 23 Sep 15

a. 5.4 mm
b. 1.08 mm
c. 6.2 mm
d. 8.6 mm

#### Discussion

• Manish singh rawat   -Posted on 08 Jul 18
It will solve with Mohr's second theorem
It it
Deflection = Ax/EI
Where A is area of bending moment diagram
X = distance of CG to point at which deflection is calculated
El = flexural rigidity
It is so easy friends
• suri   -Posted on 27 Jan 18
y = (5/384) (WL4 / EI)

= (5/ 384) (20 x 54 / 30000)

= 5.42 x 10-3 m = 5.42 mm
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• babai mukherjee   -Posted on 11 Nov 17
30000kN.mm is the valu of EI,but according your solution where is the mm is going.???
• babai mukherjee   -Posted on 11 Nov 17
30000kN.mm is the valu of EI,but according your solution where is the mm is going.???
• prince mishra   -Posted on 18 Aug 16
very good
• sandip    -Posted on 26 Jul 16
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• sandip    -Posted on 26 Jul 16
improve my knowledge
• Sravanthi   -Posted on 24 Nov 15
Given: load = 20 kN/m, length of beam = 5 m, flexural rigidity (EI) = 30000 kN.m2

Formula: Maximum deflection in the beam (y) = (5/ 384) (WL4 / EI)

Solution:

y = (5/384) (WL4 / EI)

= (5/ 384) (20 x 54 / 30000)

= 5.42 x 10-3 m = 5.42 mm

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