**Quiz: Civil Engineering**

**Exam: SSC-JE**

**Topic: Strength of Material**

**Each question carries 1 mark**

**Negative marking: 1/4 mark**

**Time: 10 Minutes**

Q1. The shape of Bending Moment Diagram in a beam subjected to only Uniformly Distributed Load (UDL) is :

(a) Parabola

(b) Triangular

(c) Constant

(d) Cubic parabola

Q2. The ratio of normal stress to normal strain within elastic limits is called :

(a) Poisson’s Ratio

(b) Bulk Modulus

(c) Young’s Modulus

(d) Shear Modulus

Q3. The slenderness ratio of a column is zero when its length :

(a) is equal to its radius of gyration

(b) is supported on all sides throughout its length

(c) Effective length is equal to Actual length

(d) is very large

Q4. A material has identical properties in all directions, it is said to be

(a) homogeneous

(b) isotropic

(c) elastic

(d) orthotropic

Q5.The maximum bending moment in a simply supported beam subjected to point load occurs at :

(a) Supports

(b) Mid span

(c) Quarter span

(d) Under point load

Q6. For a given elastic material, the elastic modules E is 210 GPa and its Poisson’s Ratio is 0.27. What is the approximate value of its modules of rigidity?

(a) 105 GPa

(b) 83 GPa

(c) 159 GPa

(d) 165 GPa

Q7. If a ductile material is subjected to a unidirectional tensile force, then to avoid shear failure, the material should have its shear strength at least equal to

(a) its tensile strength

(b) Half the tensile strength

(c) Its compressive strength

(d) Twice the tensile strength

Q8. Find the tensile stress of a mild steel rod of 18mm diameter. Given: ultimate load = 9.0 ton.

(a) 0.03 Ton/sq.mm

(b) 0.02 Ton/sq.mm

(c) 0.05 Ton/sq.mm

(d) 0.06 Ton/sq.mm

Q9. A column that fails due to direct stress, is called:

(a) short column

(b) long column

(c) weak column

(d) medium column

Q10. A solid metal bar of uniform diameter D and length L is hung vertically from a ceiling. If the density of the material of the bar is P and the modulus of elasticity is E, then the total elongation of the bar due to its own weight is

(a) PL/2E

(b) PL²/2E

(c) PE/2L

(d) PE/2L²

SOLUTIONS

S1. Ans.(a)

Sol. Shape of bending moment diagram-

No load – Rectangular

Point load – Triangular

Uniformly distributed load – Parabola (2° curve)

Uniformly varying load (UVL) – Cubic parabola (3° curve)

S2. Ans.(c)

Sol.

▭(Young^’ s modulus of elasticity (E)=(Normal stress)/(Normal strain))

▭(Shear Modulus (G)=(Shear stress)/(Shear strain))

S3. Ans.(b)

Sol. if a column is supported on all sides throughout its length then the value of unsupported length of the column is zero so that the slenderness ratio is also zero.

S4. Ans.(b)

Sol. A material is said to be isotropic when it has identical properties in all direction and a material is said to be homogenous when it has same property at each cross-section.

S5. Ans.(d)

Sol. The maximum bending moment in a simply supported beam of length L subjected to a point load ‘W’ occurs at under point load and the magnitude of maximum bending moment will be WL/4.

S6. Ans.(b)

Sol. Given, Elastic modulus (E) = 210 GPa

Poison ratio (μ) = 0.27

Modulus of rigidity (G) = ?

E = 2G(1+μ)

G =E/2(1+μ)

=210/2(1+0.27)

▭(G=82.67 GPa≈83 GPa)

S7. Ans.(d)

Sol. if ductile material is subjected to a unidirectional tensile force, then to avoid shear failure material should have its shear strength is twice the tensile strength.

S8. Ans.(a)

Sol. Diameter = 18 mm

Ultimate load = 9.0 ton

Tensile stress = (9.0 ton)/(π/4×18²)

= 0.035 ton/mm²

S9. Ans.(a)

Sol. The short column fails primarily due to direct stress because in short column the buckling stresses are very small compared to direct stresses.

S10. Ans.(b)

Sol. The elongation of bar due to its own weight (w) is

∆=WL/2AE

Now W = PAL

∴ ∆ =PL²/2E