Question

A steel wire of radius 0.05 cm
having moment of inertia 49.06 x
10-7 cm4 is bent to form a circle of
10 cm radius. What is bending
moment, if Y = 2 x 1012 dynes /
cm?​

Answer 1

Bending force will be 9.8 x $10^{5}$ N/cm

Explanation:

Bending force (C) = $\frac{YI}{R}$

Y = 2 x $10^{12}$ dyne / cm

I = 49.06 x $10^{-7} cm^{4}$

R = 10 cm

Putting the values in the equation

C = $\frac{(2 * 10^{12})(49.06 * 10^{-7} )}{10}$

C = 9.8 x $10^{5}$ N/cm

Answer 2

Bending force required to bend the wire in circle will be$9.8 \times 10^{5} \mathrm{~N} / \mathrm{cm}$.

The Bending Theory

Bending theory is also known as flexure theory is defined as the axial deformation of the beam due to external load that is applied perpendicularly to a longitudinal axis which finds application in applied mechanics.

                 $\frac{\sigma}{y}=\frac{M}{I}=\frac{E}{R}$

Bending force    $(\mathrm{C})=\frac{Y I}{R}$

$\begin{aligned}&Y=2 \times 10^{12} \text { dyne } / \mathrm{cm} \\&I=49.06 \times 10^{-7} \mathrm{~cm}^{4} \\&R=10 \mathrm{~cm}\end{aligned}$

Putting the values in the equation

$\begin{aligned}&C=\frac{\left(2+10^{12}\right)\left(49.06 * 10^{-7}\right)}{10} \\&C=9.8 \times 10^{5} \mathrm{~N} / \mathrm{cm}\end{aligned}$

Thus, Bending force required to bend the wire in circle will be$9.8 \times 10^{5} \mathrm{~N} / \mathrm{cm}$.

Learn more about Bending theory here,

https://brainly.in/question/8050004?msp_poc_exp=1

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