Simple Stress and Strain MCQ Quiz - Objective Question with Answer for Simple Stress and Strain - Download Free PDF

Explanation:

Given, P = 120 N at C

Free body diagram

Let R_A and R_B are the reaction at the fixed end A and B. 

RA + R_B = P

Sign convetion: +ve(Tension), -ve(Compression)

As the Beam AB is fixed, So

Δ_AB = 0

Δ_AC + Δ_CB = 0

\(\frac{(P\ - \ R_B)L}{AE} \ +\ \frac{- R_B\times 2L}{AE} = 0\)

After solving,

R_B = P/3 = 120/3 = 40 N

R_A = P - R_B = 120 - 40

R_A = 80 N

Explanation:

Thermal Expansion of a Steel Bar

• Thermal expansion refers to the increase in the dimensions of a material when it is subjected to a temperature rise. For a linear material like a steel bar, this expansion can be calculated using the coefficient of linear expansion.

Given Data:

• Coefficient of linear expansion (α) = 12 × 10⁻⁶/°C
• Original length of the steel bar (L₀) = 15 cm = 150 mm
• Expansion of the bar (ΔL) = 0.3 mm

Formula:

The formula for linear expansion is given by:

ΔL = α × L₀ × ΔT

Where:

• ΔL = Change in length (0.3 mm)
• α = Coefficient of linear expansion (12 × 10⁻⁶/°C)
• L₀ = Original length (150 mm)
• ΔT = Change in temperature (°C)

Calculation:

Rearranging the formula to solve for ΔT:

ΔT = ΔL / (α × L₀)

Substituting the given values:

ΔT = 0.3 / (12 × 10⁻⁶ × 150)

ΔT = 0.3 / (1.8 × 10⁻³)

ΔT = 166.67 °C

Concept:

To prevent yielding, the actual stress on the material must be less than or equal to the allowable stress (yield stress divided by the factor of safety).

Given:

• Axial load, P = 120 kN = 120000 N
• Width of bar, b = 40 mm
• Yield stress, σ_yield = 310 MPa
• Factor of Safety, FOS = 2

Step 1: Allowable Stress

\(\sigma_{\text{allowable}} = \frac{310}{2} = 155~\text{MPa}\)

Step 2: Use axial stress formula

\(\sigma = \frac{P}{A} = \frac{P}{bt}\) → \(t = \frac{P}{b \cdot \sigma_{\text{allowable}}}\)

\(t = \frac{120000}{40 \cdot 155} = \frac{120000}{6200} = 19.35 \approx 19.4~\text{mm}\)

Explanation:

When a prismatic bar (same cross-section throughout) is subjected to an axial tensile load  the internal resistance developed to counter this load is distributed uniformly across the cross-sectional area  A" id="MathJax-Element-70-Frame" role="presentation" style="position: relative;" tabindex="0"> A" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0">$A$ $A$ $A$ , assuming:

• The material is homogeneous and isotropic (same properties in all directions),

• The load is applied axially (along the length and centered),

• There is no stress concentration or eccentric loading.

 Additional Information

Types of Axial Stress:

• Tensile Stress: If the load pulls the bar apart

• Compressive Stress: If the load pushes the bar together

Explanation:

Strain Measuring Techniques for High-Temperature Environments

When selecting a strain measuring technique for high-temperature environments, it is critical to consider the durability and accuracy of the sensors under extreme conditions. Here, we analyze the given options:

1. Electrical resistance strain gauge (Option 1)

   • While widely used for various applications, electrical resistance strain gauges have limitations in high-temperature environments due to potential drift in resistance and insulation degradation.

2. Photoelastic coating (Option 2)

   • Photoelastic coatings are useful for visualizing strain distribution but are not ideal for high temperatures due to thermal sensitivity and potential for delamination.

3. Optical fiber strain sensors (Option 3)

   • Optical fiber strain sensors are highly suited for high-temperature environments due to their ability to withstand extreme temperatures and provide accurate measurements over a wide temperature range.

4. Brittle lacquer (Option 4)

   • Brittle lacquer is used for detecting strain by cracking patterns but is limited in high-temperature applications as it can degrade or perform inconsistently under thermal stress.

Conclusion

Analyzing the options, it is clear that optical fiber strain sensors (Option 3) are the most suitable for high-temperature environments due to their robustness and accuracy in such conditions, thus making them the best choice.

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:

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.

Explanation:

The stress-strain diagrams for different type of materials are given below: