Page 6 - Mechanics of Structures – Chapter 1
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Mechanics of Civil Engineering Structures – Chapter 5
Example 3
A Simply Supported Beam 250 mm wide and 450 mm deep carries a shear force of 10 kN at
support. Calculate shear stress at neutral axis.
Solution
. .
Shear Stress Equation, τ =
225 mm .
Where,
225 mm S = Shear force = 10x10 N
3
A = Area above the N.A or below N.A = 225*250= 56250 mm
250mm
y = Distance of C.G of Area considered from N.A = 225/2 =112.50
mm
b = width of c/s at N.A = 250 mm
3 250 (450) 3
4
6
Ixx = Second moment of area = = = 1898.4 x 10 mm
12 12
. .
Substitute, τ = .
3
(10 10 )(56250)(112.50)
2
= = 0.133 N/mm
6
((1898.4 10 )(250)
Example 4
A simply supported beam of span 3m and cross section 150 x 250 mm carries a uniformly
distributed load of 5 kN/m over entire span. Find maximum intensity of shear stress in beam at a
section at 0.75 m from any of the support.
5kN/m
125 mm
N.A Τ max
250 mm
3 m 125 mm
150 mm
NHO JKA PSAS Page 6
