# Strength of Materials Questions and Answers – Bars of Composite Sections – 1

This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Bars of Composite Sections – 1”.

1. If a bar of sections of two different length and different diameters are in a line and P load is acting axially on them then what will be the change in length of the bar?
a) P/E x (L1 + L2)
b) P/E x (A1/L1 + A2/ L2)
c) P/E x (L1/A1 + L2/A2)
d) E/P x (L1/A1 + L2/A2)

Explanation: Change in length of section 1 = PL1/EA1
Change in length of section 2 = PL2/EA2
Total change in length of bar = PL1/EA1 + PL2/EA2.

2. How does the elastic constant varys with the elongation of body?
a) The elastic constant is directly proportional to the elongation
b) The elastic constant is directly proportional to the elongation
c) The elongation does not depends on the elastic constant
d) None of these

Explanation: Elongation of a composite bar of two sections = P/E x (L1/A1 + L2/A2)
E is inversely proportional to bar elongation.

3. A composite rod is 1000mm long, its two ends are 40 mm2 and 30mm2 in area and length are 400mm and 600mm respectively. If the rod is subjected to an axial tensile load of 1000N, what will be its total elongation(E = 200GPa)?
a) 0.130m
b) 0.197mm
c) 0.160mm
d) 0.150mm

Explanation: As elongation of a composite bar of two sections = P/E x (L1/A1 + L2/A2)
Putting L1, L2, A1 and A2 400mm2, 600mm2, 40mm2 and 30mm2 and P = 1000 and E = 200 x 103.

4. A mild steel wire 5mm in diameter and 1m ling. If the wire is subjected to an axial tensile load 10kN what will be its extension?
a) 2.55mm
b) 3.15mm
c) 2.45mm
d) 2.65mm

Explanation: As change in length = PL/AE
P = 10x 1000N, L = 1m, A = πd2/4 = 1.963 x 10-5 m2, E = 200 x 109 N/m2.

5. A composite rod is 1000mm long, its two ends are 40mm2 and 30mm2 in area and length are 300mm and 200mm respectively. The middle portion of the rod is 20mm2 in area. If the rod is subjected to an axial tensile load of 1000N, what will be its total elongation (E = 200GPa)?
a) 0.145mm
b) 0.127mm
c) 0.187mm
d) 0.196mm

Explanation: P = 1000N, Area A1 = 40mm2, A2 = 20mm2, A30 = 30mm2
Length, L1 = 300mm, L2 = 500mm, L3 = 200mm
E = 200GPa = 200x 1000 N/mm2
Total extension = P/E x (L1/A1 + L2/A2 + L3/A3).
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6. A rod of two sections of area 625mm2 and 2500mm2 of length 120cm and 60cm respectively. If the load applied is 45kN then what will be the elongation (E = 2.1x 105 N/mm2)?
a) 0.462mm
b) 0.521mm
c) 0.365mm
d) 0.514mm

Explanation: P = 45,000N, E =2.1x 105 N/mm2,
Area, A1 = 625mm2, A2 = 2500mm2,
Length, L1 = 1200mm, L2 = 600mm
Elongation = P/E x (L1/A1 + L2/A2).

7. What will be the elongation of a bar of 1250mm2 area and 90cm length when applied a force of 130kN if E = 1.05x 105 N/mm2?
a) 0.947mm
b) 0.891mm
c) 0.845mm
d) 0.745mm

Explanation: As change in length = PL/AE
P = 130x 1000N, L = 900mm, A = 1250 mm2, E = 1.05 x 105 N/m2.

8. A bar shown in diagram is subjected to load 160kN. If the stress in the middle portion Is limited to 150N/mm2, what will be the diameter of the middle portion?

a) 3.456 cm
b) 3.685 cm
c) 4.524 cm
d) 4.124 cm

Explanation: Let L2 and D2 be the dimensions of the middle portion and L1 and D2 be the end portion dimensions.
For middle portion area = load / stress
This gives area by which diameter can be calculated.

9. A steel bar of 20mm x 20mm square cross-section is subjected to an axial compressive load of 100kN. If the length of the bar is 1m and E=200GPa, then what will be the elongation of the bar?
a) 1.25mm
b) 2.70mm
c) 5.40mm
d) 4.05mm

Explanation: Elongation in bar = PL/ AE = (100x1000x1) / (0.2×0.2x200x106) = 1.25mm.

10. A solid uniform metal bar is hanging vertically from its upper end. Its elongation will be _________
a) Proportional to L and inversely proportional to D2
b) Proportional to L2 and inversely proportional to D
c) Proportional of U but independent of D
d) Proportional of L but independent of D

Explanation: Elongation = WL / 2AE = 4WL / 2πD2E α L/D2.

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