Question 7.16: A cast iron beam is of T-section as shown in Fig. 7.21. The ...
A cast iron beam is of T-section as shown in Fig. 7.21. The beam is simply supported on a span of 8 m. The beam carries a uniformly distributed load of 1.5 kN/m length on the entire span. Determine the maximum tensile and maximum compressive stresses.
Given :
Length, L = 8 m
U.D.L., w = 1.5 kN/m = 1500 N/m
To find the position of the N.A., the C.G. of the section is to be calculated first. The C.G. will be lying on the y-y axis.
Let \bar{y} = Distance of the C.G. of the section from the bottom
∴ \bar{y} =\frac{A_1y_1+A_2y_2}{A_1+A_2}=\frac{(100\times 20) \times \Big(80+\frac{20}{2}\Big)+80\times 20 \times \frac{80}{2}}{(100 \times 2)+(80\times 20)}\\ \quad \quad \quad =\frac{180000+64000}{2000+1600}=\frac{244000}{3600}=67.77 mm
∴ N.A. lies at a distance of 67.77 mm from the bottom face or 100 – 67.77 = 32.23 mm from the top face.
Now moment of inertia of the section about N.A. is given by,
I=I_1+I_2
where I_1 = M.O.I. of top flange about N.A.
=\text{ M.O.I. of top flange about its C.G. + }A_1 \times \text{ (Distance of its C.G. from N.A.)}^2 \\ =\frac{100\times 20^3}{12}+(100\times 20) \times (32.23-10)^2 \\ = 66666.7 + 988345.8 = 1055012.5 mm^4 \\ I_2 = \text{ M.O.I. of web about N.A.} \\ = \text{ M.O.I. of web about its C.G. + } A_2 \times \text{ (Distance of its C.G. from N.A.)}^2 \\ = \frac{20\times 80^3}{12} + (80 × 20) × (67.77 – 40)^2 \\ = 853333.3 + 1233876.6 = 2087209.9 mm^4 \\ I = I_1 + I_2 = 1055012.5 + 2087209.9 = 3142222.4 mm^4.
For a simply supported beam, the maximum tensile stress will be at the extreme bottom fibre and maximum compressive stress will be at the extreme top fibre.
Maximum B.M. is given by,
M=\frac{w\times L^2}{8}=\frac{1500 \times 8^2}{8}=12000 Nm \\ = 12000 × 1000 = 12000000 Nmm
Now using the relation
\frac{M}{I}=\frac{\sigma}{y} \quad \text{ or } \quad \sigma=\frac{M}{I}\times y
(i) For maximum tensile stress,
y = Distance of extreme bottom fibre from N.A. = 67.77 mm
∴ \sigma=\frac{12000000 }{3142222.4}× 67.77 = \pmb{258.81 N/mm^2.}
(ii) For maximum compressive stress,
y = Distance of extreme top fibre from N.A. = 32.23 mm
∴ \sigma=\frac{M}{I}\times y=\frac{12000000}{3142222 4} \times 32.23 = \pmb{123.08 N/mm^2.}