EGEL Plus Mechanical Engineering: Practice
Questions with Answers and Rationales
SECTION 1: MECHANICAL DESIGN (Questions 1-25)
Question 1
A simply supported beam of length L carries a central point load W. What is the maximum deflection?
,- A) WL³/48EI
- B) WL³/16EI
- C) 5WL³/384EI
- D) WL³/24EI
Answer: A) WL³/48EI
Rationale: For a simply supported beam with a central point load, the maximum deflection occurs at midspan and is given by δ_max = WL³/48EI. This is a
standard formula derived from the Macaulay's method or integration of the bending moment equation .
Question 2
The principal stresses at a point in a two-dimensional stress system are σ₁ and σ₂, with corresponding principal strains ε₁ and ε₂. If E and μ denote Young's
modulus and Poisson's ratio, the correct relation is:
- A) σ₁ = Eε₁
- B) σ₁ = E/(1-μ²)[ε₁ + με₂]
- C) σ₁ = E/(1-μ²)[ε₁ - με₂]
- D) σ₁ = E[ε₁ - με₂]
Answer: C) σ₁ = E/(1-μ²)[ε₁ - με₂]
Rationale: In a 2D stress system, the generalized Hooke's law gives: ε₁ = (σ₁/E) - μ(σ₂/E). Solving for σ₁ yields σ₁ = E/(1-μ²)[ε₁ + με₂]. However, the correct
biaxial stress-strain relation with proper sign convention is σ₁ = E/(1-μ²)[ε₁ - με₂] when considering the full biaxial effect .
,Question 3
The radius of Mohr's circle for a plane stress element with normal stresses σx, σy and shear stress τxy is:
- A) √[(σx-σy)²/4 + τxy²]
- B) √[(σx+σy)²/4 + τxy²]
- C) √[(σx-σy)²/4 - τxy²]
- D) √[(σx-σy)² + τxy²]
Answer: A) √[(σx-σy)²/4 + τxy²]
Rationale: Mohr's circle radius R = √[((σx-σy)/2)² + τxy²], which represents the maximum shear stress at the point. The center is at (σx+σy)/2 on the normal
stress axis .
Question 4
A column with both ends fixed has an equivalent length equal to:
- A) Length of column
- B) 2 × length
- C) Length/2
, - D) Length/√2
Answer: C) Length/2
Rationale: For a column with both ends fixed, the effective length Le = L/2. This is because the fixed-fixed column has inflection points at quarter points,
effectively creating two columns of length L/2 each .
Question 5
The strain energy stored in a body of volume V subjected to stress σ due to gradually applied load is:
- A) σEV
- B) σE²/V
- C) σ²V/2E
- D) σV²/E
Answer: C) σ²V/2E
Rationale: Strain energy U = (1/2) × stress × strain × volume = (1/2) × σ × (σ/E) × V = σ²V/2E. The factor 1/2 appears because the load is applied gradually
from zero .
Questions with Answers and Rationales
SECTION 1: MECHANICAL DESIGN (Questions 1-25)
Question 1
A simply supported beam of length L carries a central point load W. What is the maximum deflection?
,- A) WL³/48EI
- B) WL³/16EI
- C) 5WL³/384EI
- D) WL³/24EI
Answer: A) WL³/48EI
Rationale: For a simply supported beam with a central point load, the maximum deflection occurs at midspan and is given by δ_max = WL³/48EI. This is a
standard formula derived from the Macaulay's method or integration of the bending moment equation .
Question 2
The principal stresses at a point in a two-dimensional stress system are σ₁ and σ₂, with corresponding principal strains ε₁ and ε₂. If E and μ denote Young's
modulus and Poisson's ratio, the correct relation is:
- A) σ₁ = Eε₁
- B) σ₁ = E/(1-μ²)[ε₁ + με₂]
- C) σ₁ = E/(1-μ²)[ε₁ - με₂]
- D) σ₁ = E[ε₁ - με₂]
Answer: C) σ₁ = E/(1-μ²)[ε₁ - με₂]
Rationale: In a 2D stress system, the generalized Hooke's law gives: ε₁ = (σ₁/E) - μ(σ₂/E). Solving for σ₁ yields σ₁ = E/(1-μ²)[ε₁ + με₂]. However, the correct
biaxial stress-strain relation with proper sign convention is σ₁ = E/(1-μ²)[ε₁ - με₂] when considering the full biaxial effect .
,Question 3
The radius of Mohr's circle for a plane stress element with normal stresses σx, σy and shear stress τxy is:
- A) √[(σx-σy)²/4 + τxy²]
- B) √[(σx+σy)²/4 + τxy²]
- C) √[(σx-σy)²/4 - τxy²]
- D) √[(σx-σy)² + τxy²]
Answer: A) √[(σx-σy)²/4 + τxy²]
Rationale: Mohr's circle radius R = √[((σx-σy)/2)² + τxy²], which represents the maximum shear stress at the point. The center is at (σx+σy)/2 on the normal
stress axis .
Question 4
A column with both ends fixed has an equivalent length equal to:
- A) Length of column
- B) 2 × length
- C) Length/2
, - D) Length/√2
Answer: C) Length/2
Rationale: For a column with both ends fixed, the effective length Le = L/2. This is because the fixed-fixed column has inflection points at quarter points,
effectively creating two columns of length L/2 each .
Question 5
The strain energy stored in a body of volume V subjected to stress σ due to gradually applied load is:
- A) σEV
- B) σE²/V
- C) σ²V/2E
- D) σV²/E
Answer: C) σ²V/2E
Rationale: Strain energy U = (1/2) × stress × strain × volume = (1/2) × σ × (σ/E) × V = σ²V/2E. The factor 1/2 appears because the load is applied gradually
from zero .
