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Chapter 7

Q. 7.2

Calculate the maximum stress* induced in a cast iron pipe of external diameter 40 mm, of internal diameter 20 mm and of length 4 metre when the pipe is supported at its ends and carries a point load of 80 N at its centre.
*The bending stress will be maximum at the section where B.M. is maximum. This is because
\frac{M}{I}=\frac{\sigma}{y} \text { or } \quad \sigma=\frac{M}{I} \times y \text {. }

Step-by-Step

Verified Solution

Given :
External dia.,    D = 40 mm
Internal dia.,     d = 20 mm
Length,               L = 4 m = 4 × 1000 = 4000 mm
Point load,         W = 80 N
In case of simply supported beam carrying a point load at the centre, the maximum bending moment is at the centre of the beam.

And maximum B.M.    =\frac{W \times L}{4}

∴    Maximum B.M.      =\frac{80 \times 4000}{4}=8 \times 10^4  Nmm

∴                                      M = 8 × 10^4  Nmm

Fig. 7.6 (b) shows the cross-section of the pipe.
Moment of inertia of hollow pipe,

I=\frac{\pi}{64}\left[D^4-d^4\right]\\ \space \\ =\frac{\pi}{64}\left[40^4-20^4\right]=\frac{\pi}{64}[2560000-160000] \\ \space \\ = 117809.7  mm^4

Now using equation (7.4),
(7.4):          \frac{M}{I}=\frac{\sigma}{y}=\frac{E}{R}

\frac{M}{I}=\frac{\sigma}{y}            …(i)

when y is maximum, stress will be maximum. But y is maximum at the top layer from the N.A.

∴                    y_{\max }=\frac{D}{2}=\frac{40}{2}=20  mm

Equation (i) can be written as

\frac{M}{I}=\frac{\sigma_{\max }}{y_{\max }}

∴                      \sigma_{\max }=\frac{M}{I} \times y_{\max }\\ \space \\ =\frac{8 \times 10^4 \times 20}{117809.7}= \pmb{1 3 . 5 8  N / mm ^2.}

7.6