Page 7 - Mechanics of Structures – Chapter 1
P. 7
Mechanics of Civil Engineering Structures – Chapter 5
Solution
Shear Stress Equation, τ = . .
.
Where,
S = Shear at 0.75 m from diagram is as shown in sketch by similar
triangular.
5
SF at 0.75 m = 3.75 kN = 3.75 x 10 N
7.5 7.5 A = Area above the N.A or below N.A = 150*125= 18750 mm
y = Distance of C.G of Area considered from N.A = 125/2 =62.5 mm
b = width of c/s at N.A = 150 mm
0.75 0.75
3 150 (250) 3
4
6
1.5 m Ixx = Second moment of area = = = 195.31 x 10 mm
12 12
Substitute, τ N.A = . .
.
(3.75 10 )(18750)(62.50)
3
2
= = 0.15 N/mm
6
((195.31 10 )(150)
7.3.2 Shear Stress for ‘T’ section of beam
Example 5
Draw shear stress distribution diagram for cross section of beam with T section flange, 90 mm x
10 mm and web 80 mm x 10 mm, span of beam is 4m subjected to uniformly distributed load of
20 kN/m over entire simply supported beam.
90 mm
10 mm
80 mm
10 mm
NHO JKA PSAS Page 7
