Question

A steel girder can bear a load of 20 tons. If the thickness of girder is double, then for the same depression it can bear a load of:
1) 40 tons
2) 80 tons
3) 160 tons
4) 5 tons

Answer 1

Hint:Here there’s a formula between the depression, weight, load, thickness, length, moment of inertia and young’s modulus. There will be two expressions of the same formula. We need to equate them and then find the value of load.

Formula used:Formula for Depression of girder and its thickness.
$\delta = \dfrac{{W{L^3}}}{{4b{t^3}Y}}$;
Where,
W = Load;
L = Length;
t = Thickness;
Y = Young’s Modulus;
b = Breadth;
$\delta $= Depression;

Complete step by step solution:
Make two expressions of the above formula and equate them together.
Write the formula for the original thickness (${t_1}$)
$\delta = \dfrac{{{W_1}{L^3}}}{{4bt_1^3Y}}$;
Write the formula for the new thickness (${t_2}$)
$\delta = \dfrac{{{W_2}{L^3}}}{{4bt_2^3Y}}$;
The two thicknesses are related as:
${t_2} = 2{t_1}$;
Since both the expressions are same we can equate them together:
$\dfrac{{{W_1}{L^3}}}{{4bt_1^3Y}} = \dfrac{{{W_2}{L^3}}}{{4bt_2^3Y}}$;
Most of the variables are same so they will cancel each other out
$\dfrac{{{W_1}}}{{t_1^3}} = \dfrac{{{W_2}}}{{t_2^3}}$;
Put the value of ${t_2}$in the above equation and solve
$\dfrac{{{W_1}}}{{t_1^3}} = \dfrac{{{W_2}}}{{{2^3} \times t_1^3}}$;
Since the thickness is also the same, they will cancel out.
$\dfrac{{{W_1}}}{1} = \dfrac{{{W_2}}}{{{2^3}}}$
Take the denominator of the RHS to the LHS
${W_1} \times {2^3} = {W_2}$
Put the value of original load in the place of ${W_1}$
${W_2} = 20 \times {2^3}$
Multiply the integers
${W_2} = 160{\text{ tons}}$
Final Answer: If the thickness of girder is double, then for the same depression it can bear a load of 160 tons.

Option “3” is correct.

Note:Here most of the variables in the formula $\delta = \dfrac{{{W_1}{L^3}}}{{4bt_1^3Y}}$ are same except the thickness “t”. Make two expression of the same formula and apply the given relation for thickness i.e. ${t_2} = 2{t_1}$ and solve for the unknown value of load i.e. ${W_2}$.