Strength of Materials MCQ Quiz - Objective Question with Answer for Strength of Materials - Download Free PDF

Explanation:

• A ring beam is a closed circular beam subjected to a uniformly distributed load (UDL) along its span (such as from walls or roof loads).

• In a circular ring beam under UDL, the shear force is zero at the crown (mid-span).

• This is because the vertical component of load is balanced symmetrically at that point; no net transverse shear remains.

• In a ring beam under symmetric UDL, torsional moments are maximum at supports (e.g., quarter spans or support points).

• At the crown/mid-span, torsional effects from both sides cancel each other out, so torsion is zero there.

 Additional Information

• Shear Force in Circular Beams: In circular beams or ring beams under uniformly distributed load (UDL), shear force varies along the span. It is maximum near the supports and zero at the crown (mid-span) due to symmetry. This behavior is opposite to straight beams where maximum shear occurs at supports but not necessarily zero at mid-span. 
• Torsion in Ring Beams: Torsional moments develop in ring beams because the beam tends to twist due to the curvature and the eccentric nature of loading. Torsion is zero at the crown but maximum at support points or quarter-span locations. The torsional effect must be considered in design, especially when the beam supports walls or roofs.
• Bending Moment in Ring Beams:  Bending moment in a ring beam due to UDL is maximum at mid-span (crown). The curvature causes a moment that resists the vertical load, and this moment is symmetrically distributed. The design must accommodate this maximum moment with proper reinforcement.
• Importance of Ring Beams in Structures: Ring beams are crucial in circular structures like water tanks, silos, chimneys, or domes. They help to distribute radial loads uniformly, resist torsional and bending moments, and maintain the shape and stability of the structure. Proper detailing of shear, torsion, and bending reinforcement is essential for structural integrity.

Concept:

• SFD is 1° higher than the loading curve and BMD is 1° higher than the SF curve.

• If S.F. changes sign at a section then B.M at that section is either maximum or minimum.

• If BM changes sign at a point, then such a point is called a point of contraflexure or inflection. If BM changes sign curvature also changes.

• The rate of change of BM is equal to SF at that section or the slope of BM curve represents SF at that point.

• The negative slope of SF curve at a point represents the downward loading rate.

• If at a point concentrated load or reaction is present, then at that point ordinate of the SF curve will change suddenly and the slope of BM curve will also change.

• If at a point Concentrated moment or couple is present then ordinate of the BM curve will change suddenly.

• If over a span a constant SF is present only then such span is called shear span.

• If over a span BM is constant and SF = 0, then such a span is in pure bending and under pure bending the deflected shape will be an arc of a circle.

Additional Information

• A couple is a system of two equal and opposite forces whose lines of action do not coincide.

• The two forces create a rotational effect (moment) without causing any net translation of the body.

• The effect of a couple is to rotate the body around its center of mass or a fixed point.

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.

Concept:

The slenderness ratio (λ) of a column is defined as: \(\lambda=\frac{L_r}{r}\)

Where:

• 
  L_e = Effective length of column

• 
  r = Radius of gyration of the cross-section

Calculation:

L_e = 5 m = 5000 mm  (For a column hinged at both ends)

d = 160 mm

Radius of gyration \(r=\frac{d}{4}=\frac{160}{4}=40mm\)

So slenderness ratio \(\lambda=\frac{5000}{40}=125\)

Concept:

For a tapered flat bar under axial load, the extension
\Delta L is given by:

\(\Delta L=\frac{4PL}{Et(b_1+b_2)}\)

where, P = axial load, L = length, E = modulus of elasticity, t = thickness,

b₁ b₂ = widths at ends

Calculation:

Given:

P = 80 kN

\(\text{E}=2\times105\text{MPa}.\)

t = 10 mm

b₁ = 60 mm; b₂ = 40 mm

\(\Delta L=\frac{4\times80\times10^3\times600}{2\times10^5\times10(60+40)}\)

\(\Delta L=0.386mm\)

Explanation:

Ductile material fails along the principal shear plane as they are weak in shear and brittle material fails along with principal normal stress.

In Brittle materials under tension test undergoes brittle fracture i.e their failure plane is 90° to the axis of load and there is no elongation in the rod that’s why the diameter remains same before and after the load. Example: Cast Iron, concrete etc.

But in case of ductile materials, material first elongates and then fail, their failure plane is 45° to the axis of the load. After failure cup-cone failure is seen. Example Mild steel, high tensile steel etc.

Explanation:

Perfectly Plastic Material:

For this type of material, there will be only initial stress required and then the material will flow under constant stress.

The chart shows the relation between stress-strain in different materials. 

Stress-Strain Curve	Type of Material or Body	Examples
	Rigidly Perfectly Plastic Material	No material is perfectly plastic
	Ideally plastic material.	Visco-elastic (elasto-plastic) material.
	Perfectly Rigid body	No material or body is perfectly rigid.
	Nearly Rigid body	Diamond, glass, ball bearing made of hardened steel, etc
	Incompressible material	Non-dilatant material, (water) ideal fluid, etc.
	Non-linear elastic material	Natural rubbers, elastomers, and biological gels, etc

Explanation:

• Change in the temperature causes the body to expand or contract.
• Thermal stress is created when a change in size or volume is constrained due to a change in temperature.
• So an increase in temperature creates compressive stress and a decrease in temperature creates tensile stress.

Explanation:

ϵv = ϵx + ϵy + ϵz

\( {ϵ_x} = \frac{1}{E}\left[ {{σ _x} - ν \left( {{σ _y} + {σ _z}} \right)} \right] \)

\({ϵ_{\rm{y}}} = \frac{1}{{\rm{E}}}\left[ {{σ _y} - ν \left( {{σ _x} + {σ _z}} \right)} \right] \)

\({ϵ_{\rm{z}}} = \frac{1}{{\rm{E}}}\left[ {{σ _z} - ν \left( {{σ _x} + {σ _y}} \right)} \right]\)

Total strain or volumetric strain is given by 

\( {ϵ_v} = \frac{1}{E} [ {σ_x} + {σ_y} + {σ_z} ](1-2ν) \)

There will be no change in volume if volumetric strain is zero.

ϵ_v = 0 ⇒  ν = 0.5

Explanation:

Since all the faces are free to expand the stresses due to temperature rise is equal to 0.

If the cube is constrained on all six faces, the stress produced in all three directions will be the same.

∴ thermal strain in x-direction = -α(ΔT) = \(\frac{{{\sigma _x}}}{E} - \nu \frac{{{\sigma _y}}}{E} - \nu \frac{{{\sigma _z}}}{E}\)

σ_x = σ_y = σ_z = σ

\(\sigma = - \frac{{\alpha \left( {{\rm{\Delta }}T} \right)E}}{{\left( {1 - 2\nu } \right)}}\)

Concept:

Let RA and RB be the reaction at support A and B respectively.

Free body diagram of the system is:

\(R_A=\frac{Pb}{L}\;\&\;R_B=\frac{Pa}{L}\)

Calculation:

Given:

As per figure P = 10 kN, a = 1 m and b = 2 m.

\(R_A=\frac{Pb}{L}\)

\(R_A=\frac{10\times2}{3}=\frac{20}{3}\;kN\)

\(R_B=\frac{10\times1}{3}=\frac{10}{3}\;kN\)

Explanation:-

Resilience

• The total strain energy stored in a body is commonly known as resilience. Whenever the straining force is removed from the strained body, the body is capable of doing work. Hence resilience is also defined as the capacity of a strained body for doing work on the removal of the straining force.
• It is the property of materials to absorb energy and to resist shock and impact loads.
• It is measured by the amount of energy absorbed per unit volume within elastic limit this property is essential for spring materials.
• The resilience of material should be considered when it is subjected to shock loading.

Proof resilience

• The maximum strain energy, stored in a body, is known as proof of resilience. The strain energy stored in the body will be maximum when the body is stressed upto the elastic limit. Hence the proof resilience is the quantity of strain energy stored in a body when strained up to the elastic limit.
• It is defined as the maximum strain energy stored in a body.
• So, it is the quantity of strain energy stored in a body when strained up to the elastic limit (ability to store or absorb energy without permanent deformation).

Modulus of resilience

• It is defined as proof resilience per unit volume.
• It is the area under the stress-strain curve up to the elastic limit.

Toughness:

• Toughness is defined as the ability of the material to absorb energy before fracture takes place.
• This property is essential for machine components that are required to withstand impact loads.
• Tough materials have the ability to bend, twist or stretch before failure takes place.
• Toughness is measured by a quantity called modulus of toughness. Modulus of toughness is the total area under the stress-strain curve in a tension test.
• Toughness is measured by Izod and Charpy impact testing machines.
• When a material is heated it becomes ductile or simply soft and thus less stress is required to deform the material and the stress-strain curve will shift down and the area under the curve decreases thus toughness decreases.
• Toughness decreases as temperature increases.

Hardness:

• Hardness is a measure of the resistance to localized plastic deformation induced by either mechanical indentation or abrasion. 
• Hardness Testing measures a material’s strength by determining resistance to penetration.
• There are various hardness test methods, including Rockwell, Brinell, Vickers, Knoop and Shore Durometer testing.

When a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue.

The slow and continuous elongation of a material with time at constant stress and high temperature below the elastic limit is called creep.

Explanation:

Elastic recovery/strain: The strain recovered after the removal of the load is known as elastic strain.

Plastic strain: The permanent changes in dimension after the removal of load is known as plastic strain.

The load is removed when the stress was 200 MPa and the corresponding strain was 0.03

After the removal of load, the body recovered and the final strain found was 0.01.

∴ Elastic strain = 0.03 - 0.01 ⇒ 0.02 and Plastic strain = 0.01 respectively.

Concept:

Stress at any section of the bar is given by,

\(stress, \sigma =\frac{{Load ~(P)}}{{Cross-sectional ~area~(A)}}\)

Calculation:

Given:

Load in section BC, P = 30 kN (compressive), 

cross-sectional area, A = 15 m² = 15 × 10⁶ mm²

\(stress~ in ~section ~BC, \sigma =\frac{{30~\times~10^3}}{{15~\times ~10^6}}=0.002~N/mm^2\)

Explanation:

Strain energy:

When a body is subjected to gradual, sudden, or impact load, the body deforms and work is done upon it. If the elastic limit is not exceeded, this work is stored in the body. This work-done or energy stored in the body is called strain energy.

Strain energy = Work done.

Case-I:

When a rigid body is slowly dropped on another body, it is a case of gradual loading:

Work-done on the bar = Area of load-deformation diagram \(⇒\frac{1}{2}\;×\;P\;×\;δ l\)

Work stored in the bar = Area of resistance deformation diagram

\(⇒\frac{1}{2}\;×\;R\;×\;δ l\)

\(⇒\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\;\;\;[\because R=\sigma A]\)

We can write;

\(⇒\frac{1}{2}\;×\;P\;×\;δ l=\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\)

\(\sigma_{gradual}=\frac{P}{A}\)

Case-II:

Work-done on the bar = Area of load-deformation diagram ⇒ P × δl

Work stored on the bar = Area of resistance deformation diagram

\(⇒\frac{1}{2}\;×\;R\;×\;δ l\)

\(⇒\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\;\;\;[\because R=\sigma A]\)

We can write;

\(P\times\delta l=\frac{1}{2}\;×\;(\sigma\;×\;A)\;×\;δ l\)

\(\sigma_{sudden}=\frac{2P}{A}\)

\(\therefore \frac{\sigma_{sudden}}{\sigma_{gradual}}=2\)

∴ maximum stress intensity due to suddenly applied load is twice the stress intensity produced by the load of the same magnitude applied gradually.

Impact Loading:

When a load is dropped from a height before it commences to load the body, such loading is known as Impact loading.

The ratio of the stress or deflection produced due to impact loading to the stress or deflection produced due to static or gradual loading is known as the Impact factor.

\(IF=\frac{\sigma_{impact}}{\sigma_{gradual}}=\frac{\Delta_{impact}}{\Delta_{gradual}}\)

\(IF=\frac{\sigma_{sudden}}{\sigma_{gradual}}=\frac{\Delta_{sudden}}{\Delta_{gradual}}=2\)

∴ deflection due to sudden loading is twice that of gradual loading.