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

Explanation:

Moment of Inertia → 2. Modulus of Rupture

• Moment of Inertia is the geometric property of a cross-section that indicates its resistance to bending.

• It is used in bending stress formula and helps calculate the modulus of rupture (ultimate flexural stress).

 Elongation → 1. Tensile Stress

• Elongation is the change in length of a material when subjected to tensile stress.

• Tensile stress causes a material to stretch.

 Additional InformationNeutral Axis → 4. Zero Longitudinal Stress

• The neutral axis of a beam section is where the longitudinal stress = 0 during bending.

• Fibres above this axis are in compression, below are in tension.

Top Fibre → 3. Zero Shear Stress

• At the top and bottom fibres of a beam section, shear stress is zero

• Maximum shear stress occurs at the neutral axis.

Explanation:

• In sagging, the top fibres are pushed inward, experiencing compression.

• The topmost fibre is furthest from the neutral axis, and therefore experiences the maximum compressive strain.

• The bottom fibres are stretched outward, hence they are in tension.

• The bottommost fibre has the maximum tensile strain, opposite in nature but equal in magnitude (in elastic bending) to the compressive strain at the top.

• Located between the top and bottom fibres where strain and stress are zero.

• It separates the compression zone (above) and the tension zone (below).

• The strain varies linearly across the depth of the beam section in elastic bending.

• Maximum at the top and bottom, and zero at the neutral axis.

Additional Information

• Bending refers to the deformation of a beam when an external load or moment is applied perpendicular to its longitudinal axis.

• The internal resistance developed in response to this load causes the beam to bend, resulting in internal stresses and strains within the material.

• The cross-section of the beam experiences compressive stress on one side and tensile stress on the other.

• There is a line/plane across the cross-section called the Neutral Axis, where bending stress is zero.

• The magnitude of stress increases linearly from the neutral axis to the outermost fibres.

• Beams can fail in bending when maximum stress exceeds the material strength (either tensile or compressive).

• Overloaded beams can crack, deform permanently, or collapse.

Explanation:

Assumptions in the theory of bending:

• The material of the beam is isotropic and homogeneous and follows Hook’s law.
• The stress-induced is proportional to the strain and the stress at any point does not exceed the elastic limit.
• Transverse sections of the beam that were plane before bending remains plane even after bending.
• The beam is initially straight and has a uniform cross-section.
• The modulus of elasticity is the same for the fibers of the beam under tension or compression.
• The beam is subjected to pure bending and therefore bends in an arc of a circle.
• The radius of curvature is large compared to the dimensions of the section.
• There is no resultant pull or push on the cross-section of the beam.
• The loads are applied in the plane of bending.
• The transverse section of the beam is symmetrical about a line passing through the center of gravity in the plane of bending.

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.

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.

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

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

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

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

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

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