Theory Of Structures MCQ Questions & Answers | Civil Engineering

$$\eqalign{ & {\text{Deflection max }}\left( {\text{A}} \right) = \frac{{{\text{P}}{{\text{L}}^3}}}{{48{\text{E}}I}} \cr & {\text{Deflection max }}\left( {\text{B}} \right) = \frac{{5{\text{w}}{{\text{L}}^4}}}{{384{\text{E}}I}};\,\,\,\left( {{\text{w}} = \frac{{\text{P}}}{{\text{L}}}} \right) \cr & = \frac{{5{\text{P}}{{\text{L}}^3}}}{{384{\text{E}}I}} \cr & {\text{Ratio}} = \frac{{{\text{P}}{{\text{L}}^3}}}{{48{\text{E}}I}} \times \frac{{384{\text{E}}I}}{{5{\text{P}}{{\text{L}}^3}}} = \frac{8}{5} \cr} $$

Principal planes are special planes within a stressed body.
They are unique because at these specific planes, something very important happens:
There are NO shear stresses acting.
This means that the only stresses present on a principal plane are normal stresses (also called direct stresses), which act perpendicular to the plane.
Therefore, the correct answer is that principal planes are subjected to normal stresses only.
The other options are incorrect because shear stress is absent on principal planes.

We have a circular bar with diameter d and length l.
It extends by an amount e under a tensile pull W.
We want to find the deflection when the same load W is applied at the mid-span of the bar, making it act like a beam.

1. Relate Elongation (e) to Tensile Stress and Strain:
When the bar is under a tensile pull W, we know:
Stress \( \sigma \) = Force (W) / Area (A), where \( A = \frac{\pi d^2}{4} \).
Strain \( \varepsilon \) = Change in Length (e) / Original Length (l).
Also, Stress \( \sigma \) = Young's Modulus \( E \) * Strain \( \varepsilon \).
Therefore, \( E = \frac{\sigma}{\varepsilon} \).

2. Find Young's Modulus (E) in terms of e, l, d, and W:
\( E = \frac{W}{\left( \frac{\pi d^2}{4} \right)} \div \frac{e}{l} \)
\( E = \frac{4Wl}{\pi d^2 e} \).

3. Deflection Formula for a Simply Supported Beam with a Point Load at Mid-Span:
The standard formula for the deflection (\( \delta \)) of a simply supported beam with a point load (W) at the mid-span is:
\( \delta = \frac{Wl^3}{48EI} \), where \( I \) is the area moment of inertia.

4. Area Moment of Inertia (I) for a Circular Cross-Section:
\( I = \frac{\pi d^4}{64} \).

5. Substitute I and E into the Deflection Formula:
\( \delta = \frac{Wl^3}{48 \times E \times \left( \frac{\pi d^4}{64} \right)} \)
\( \delta = \frac{Wl^3}{48 \times \left( \frac{4Wl}{\pi d^2 e} \right) \times \left( \frac{\pi d^4}{64} \right)} \)
\( \delta = \frac{Wl^3 \times \pi d^2 e \times 64}{48 \times 4Wl \times \pi d^4} \)
\( \delta = \frac{64Wl^3 \pi d^2 e}{192Wl \pi d^4} \)
\( \delta = \frac{e l^2}{3 d^2} \).

Therefore, the deflection is: \( \frac{el^2}{3d^2} \).
Correct answer: Option C \( \frac{el^2}{3d^2} \).

A three-hinged arch is a structural form used in civil engineering to efficiently distribute loads and resist bending moments. It consists of three hinges, which allow rotation at the supports to accommodate the arch's deformation under load. The three hinges are located at the supports and at the crown of the arch.

The crown of an arch refers to the highest point along the arch's curve, where the arch is subjected to compressive forces. Placing a hinge at the crown allows the arch to rotate and adapt to changes in loading and deformation. This rotational flexibility is crucial for the arch's stability and ability to carry loads without excessive stress.

Note: Options A, C, and D are not accurate descriptions of the typical hinge locations in a three-hinged arch. The correct configuration involves hinges at the supports and at the crown to ensure proper load distribution and structural performance.