Shear Stress and Bending Stress MCQ Quiz - Objective Question with Answer for Shear Stress and Bending Stress - Download Free PDF

Explanation:

The correct expression for the bending equation in pure bending is:

\(\rm \frac{M}{I}=\frac{f}{y}=\frac{E}{R}\)

(Where M = Bending moment, I = Moment of inertia of the cross-section about the neutral axis, f = Bending stress at a distance y from the neutral axis, y = Distance from the neutral axis, E = Modulus of elasticity, and R = Radius of curvature of the beam)

Additional Information

1. Neutral Axis: The axis within the beam where the stress is zero during bending.

2. Linear Stress Distribution: Bending stress varies linearly from the neutral axis.

3. Pure Bending Assumption: Assumes no shear forces; bending moment is constant along the length.

4. Valid for Elastic Range: The formula holds until the material behaves elastically (no plastic deformation).

5. Derivation: Based on the geometry of deformation and Hooke’s Law.

Explanation:

• Shear lag is a phenomenon that occurs in flanged sections like I-beams and box girders when shear strains modify the distribution of bending stresses.

• It results in non-uniform stress distribution across the flange, causing the edges of the flange to carry less stress than the central region.

• This uneven distribution of stress leads to warping of the section, which is a characteristic effect of shear lag.

 Additional Information

• Local buckling: This is related to instability in thin-walled sections under compressive loads, not shear strain modification.

• Web crippling: This is localized failure in the web of the beam due to high compressive loads, independent of shear lag.

• Torsional instability: This is the twisting failure of sections under torque, not related to shear strain modifications in flanges.

Concept:

Table showing relations between τmax, τN.A and τavg

Calculations:

V = 15 KN

b = 200 mm

h = 300 mm

For triangular cross-section.

\(\therefore \;{{\rm{\tau }}_{N.A}} = \frac{4}{3}{{\rm{\tau }}_{avg}} = \frac{4}{3}\left( {\frac{V}{{\frac{{b \times h}}{2}}}} \right) = \frac{4}{3} \times \left( {\frac{{15000}}{{\frac{{200 \times 300}}{2}}}} \right) =0.6666\; \approx 0.667N/m{m^2}\)

Concept:

Shear stress (τmax) at a shaft due to a torque T

τmax = \(\dfrac{16 T}{\pi D^3}\)

D = Diameter of the shaft

Explanation:

Shear Stress at the circumference is:

τ1 = \(\dfrac{16 T}{\pi D^3}\)

But, the shear stress at the center (τmin) =  0 [∵ at the centre D = 0]

Hence, the shear stress is varying from 0 at the centre to τmax to the circumference in a linear way.

Important Points

• For a hollow shaft, the shear stress also changes linearly, but a certain amount of shear force acts at the inner surface. 

Concept:

The maximum bending stress in a beam is given by:

\(σ = \frac{M}{Z}\)

Where, σ is the bending stress, M is the maximum bending moment, and Z is the section modulus.

Calculation: 

Given:

Beam length, L = 4 m; Load position, a = 1 m; Cross-section: square of side = 100 mm = 0.1 m

Maximum permissible stress, σ = 9 MN/m² = 9 × 10⁶ N/m²

For a point load P at distance a from the left support, reaction at A is:

\(R_A = P \cdot \frac{(L - a)}{L} = P \cdot \frac{3}{4}\)

Maximum moment occurs at point of load:

\(M_{\text{max}} = R_A \cdot a = \frac{3P}{4} \cdot 1 = \frac{3P}{4}~\text{Nm}\)

Section modulus Z for square cross-section of side b = 0.1 m:

\(Z = \frac{b^3}{6} = \frac{(0.1)^3}{6} = 1.6667 × 10^{-4}~\text{m}^3\)

Now, apply bending stress formula:

\(M = σ \cdot Z = 9 × 10^6 \cdot 1.6667 × 10^{-4} = 1500~\text{Nm}\)

Equating maximum moment:

\(\frac{3P}{4} = 1500 \Rightarrow P = \frac{1500 \cdot 4}{3} = 2000~\text{N} = 2~\text{kN}\)

Explanation

Maximum shear stress \(({\tau _{max}})\) in the triangular cross-section is given by:

\({\tau _{\max }} = 1.5\) \({\tau _{avg}}\) at h/2 distance

Where,

\({\tau _{avg}}\) = Average shear stress in the cross-section

Shear stress at the neutral axis of the cross-section is given by:

\({\tau _{N.A}} = 1.33\)  \({\tau _{avg}}\)

Important Points

Explanation:

Shear stress distribution in some important figures is given below.

Concept-

For any cross-section, we have the bending equation

\(\frac{\sigma }{y} = \frac{M}{I} = \frac{E}{R}\)

σ = stress at a distance y from NA, M = Bending moment at that c/s

I = MOI about the neutral axis, E = Modulus of elasticity

R = Radius of curvature

Calculation

a) For any T-section:

y₂ > y₁

As neutral axis is the centroidal axis of the cross-section.

So, the neutral axis lies near the top of the T-section.

So y₂ > y₁

\({\sigma _{bot}} = \frac{{M \times {y_2}}}{I} \ \& \;{\sigma _{top}} = \frac{{M \times {y_1}}}{I}\)

As y₂ > y₁

So σ_bot > σ_top

∴ Maximum bending stress will occur at the bottom of the section.

Concept:

Both are of equal area \(\Rightarrow {b^2} = \frac{{\pi {d^1}}}{4}\)

\(\Rightarrow b = \frac{d}{2}\sqrt \pi \)

For circular section modulus,

\({Z_c} = \frac{{\pi {d^3}}}{{32}}\)

For square section,

\(\begin{array}{l} {Z_{square}} = \frac{{{b^3}}}{6} = \frac{{\pi \sqrt \pi {d^3}}}{{48}}\\ \therefore {Z_{square}} = 1.18{Z_{circular}} \end{array}\)

So for the same cross-section area, a square section is better than the circular section in bending.

Note:-

For the same cross-sectional area, the order of sections in increasing the bending strength

Concept:

Maximum shear stress in circular beam-

\(Maximum\, shear\, stress\,(\tau_{max})=({4\over 3})\tau_{ave}={4\over 3}×{Shear force\over Area}\)

Calculation:

Given data:

Shear force (F) = 40π kN

Maximum shear stress in material (\(\tau_{max}\)) = 6 MPa or 6 N/mm²

Safe diameter of the circular beam (D) =?

The factor of safety (FOS) = 2

Factored shear force (F') = Factor of safety × Shear force (F)

Factored shear force (F') = 2 × 40π = 80π kN

Factored shear force (F') = 80π × 10³ N

\(Maximum\, shear\, stress\,(\tau_{max}) = {4\over 3}×{Shear force\over Area}\)

\(6={4\over 3}\times {80\pi \times 10^3\over {\pi\over 4}D^2}\)

\(6={4\over 3}\times {4\times 80 \times 10^3\over D^2}\)

\(D^2={4\over 3}\times {4\times 80 \times 10^3\over 6}\)

\(D^2=71.111\times 10^3\)

\(D=266.666\, mm\)

Explanation:

Maximum shear stress occurs at h/2 from base

Distance from the neutral axis \(= \frac{{\rm{h}}}{2} - \frac{{\rm{h}}}{3} = \frac{{\rm{h}}}{6}\)

Here

\({{\rm{\tau }}_{{\rm{avg}}}} = \frac{{{{\rm{V}}_{\rm{u}}}}}{{\frac{1}{2}{\rm{bh\;}}}}\)

where V_u = Maximum shear force

∴ The correct answer is h/6 from the neutral axis.

Concept:

Shear stress distribution in the triangular section:

The relation between neutral axis shear stress and average shear stress is given by:

\({{\bf{\tau }}_{{\bf{neut}}}} = \frac{4}{3}\times{{\bf{\tau }}_{{\bf{avg}}}} = \frac{4}{3}\times\frac{F}{A} = \frac{4}{3}\times\frac{F}{{\frac{1}{2}bh}}\)

\(\therefore {{\bf{\tau }}_{{\bf{neut}}}} = \frac{{8F}}{{3bh}}\)

Concept:

As per bending formula:

\(\frac{\sigma }{y} = \frac{M}{I} = \frac{E}{R}\)   

Where

M = bending moment due to load, σ = bending stress, E = Modulus of Elasticity, R = radius of Curvature, y = distance of outer fibre from the neutral axis

I is the MOI about a neutral axis and it is given as:

\(I = \frac{{b{d^3}}}{{12}}\)

Calculation:

Given:

E = 2 × 10⁵ N/mm², R = 10 m = 10 × 10³ mm, A = 120 mm × 20 mm, y = 10 mm

As we know,

\(\frac{\sigma }{y} = \frac{E}{R}\)

\(\frac{\sigma }{{10}} = \frac{{2\; ×\; {{10}^5}}}{{10\; × \;{{10}^3}}} \Rightarrow \sigma = 200\;N/{mm^2}\)

Explanation:

Direct Tensile force, T = 260 kN

Permissible stress in steel, σ_st = 130 MPa

Diameter of the tank, d = 1.3 m

It is assumed that the entire applied tensile force on the water tank has to be resisted by tensile reinforcement in the tank. Therefore, the area of steel required per meter width is calculated as:

Permissible stress in steel (σ_st) × Area of steel (A_st) per meter width = Direct tensile force (T)

130 × A_st = 260 × 1000 N

On solving, we get

A_st = 2000 mm²/m

Concept:

We have the bending equation for a beam, as

\(\frac{\text{ }\!\!\sigma\!\!\text{ }}{\text{y}}=\frac{\text{M}}{\text{I}}=\frac{\text{E}}{\text{R}}\)

Where,

σ = Bending stress, y = distance from the neutral axis, M = Bending moment of any section, I = Moment of inertia, E = Modulus of Elasticity of material, and R = Radius of curvature.

When a beam is designed such that the extreme fibers are loaded to the maximum permissible stress ρ_max by varying c/s, it will be known as beam of uniform strength.

\(\therefore \frac{\sigma }{y}=\frac{M}{I}\)

\(\text{ }\!\!\sigma\!\!\text{ }=\frac{\text{M}\times \text{y}}{\text{b}{{\text{d}}^{3}}/12}\)

\(\text{ }\!\!\sigma\!\!\text{ }=\frac{\text{M}\times {{\text{d}}_{\text{x}}}\times 12}{\text{bd}_{\text{x}}^{2}\times 2}\therefore \text{y}={{\text{d}}_{\text{x}}}/2\)

\(\text{ }\!\!\sigma\!\!\text{ }=\frac{6\text{M}}{\text{bd}_{\text{x}}^{2}}\)

\(\text{d}_{\text{x}}^{2}=\frac{6\text{M}}{\text{ }\!\!\sigma\!\!\text{ b}}\Rightarrow {{\text{d}}_{\text{x}}}=\sqrt{\frac{6\text{M}}{\text{ }\!\!\sigma\!\!\text{ b}}}\)

\(\therefore {{\text{d}}_{\text{x}}}\propto \sqrt{\text{M}}\)