Question

A test is conducted on a beam loaded by end couples. The fibres at layer CD are found to lengthen by 0.03 mm and fibres at layer AB shorten by 0.09 mm in 20 mm gauge length as shown in the given figure. Taking $E=2\times {10}^{5}N/m{m}^2$, the flexural stress at top fibres would be

• A. $900N/m{m}^{2}tensile$
• B. $1000N/m{m}^{2}tensile$
• C. $1200N/m{m}^{2}tensile$
• D. $1200N/m{m}^{2}compressive$

Answer 1

The correct option is A $900N/m{m}^{2}tensile$

The strain at layer

$CD=\frac{0.03}{20}=0.0015$

$ThestrainatlayerAB=-\frac{0.09}{20}=-0.0045$

Let the distance of fibre CD from the neutral axis is x.

$\frac{0.0015}{x}=\frac{0.0045}{100-x}$

100-x = 3 x

$\therefore$ x = 25 mm

At top fibre, the distance from neutral axis will be
= 25 + 50 = 75 mm

$Sothestrainwillbe=0.0015\times \frac{75}{25}=0.0045$

$Flexuralstress=0.0045\times 2\times {10}^5$

$=900N/m{m}^{2}tensile$