Strength of Material (Part-2) (General Engineering) (SSC JE)

Total Questions: 50

41. The stresses induced in a shaft due to bending and torsion load arc 80 Mpa and 30 Mpa respectively. The yield strength of the shaft material is 280 Mpa. Using maximum shear stress theory, the factor of safety will be [BPSC AE (GEN. ENGG.) 14.10.2022 ]

Correct Answer: (d) 2.8
Solution:

42. A column of length L has one end hinged and other fixed. The cross-section of the column is a circle of diameter D. The slenderness ratio of the column is [BPSC AE (GEN. ENGG.) 14.10.2022 ]

Correct Answer: (b) 2√2L / D
Solution:

43. Distortion energy theory as one of the theories of failure was proposed by [BPSC AE (GEN. ENGG.) 11.11.2022 ]

Correct Answer: (a) Von Mises and Henky
Solution:

Distortion energy theory–This theory is an important one because it comes closest of all to verifying experimental results. The derivation is based on the assumption that Hooke's law applies and so this theory is valid only for ductile materials, that is, materials that may yield.

• In the literature the distortion energy theory is also called the shear energy theory as well as the "Von Mises-Henky theory".

• It cannot be applied for material under hydrostatic pressure. • All theories will give same results if loading is uniaxial.

• A bar stressed to the elastic under the state of uniaxial stress as occurs in a simple tension/ compression test.

44. A beam of uniform rectangular cross-section is under a transverse load. The stress along the neutral axis of central cross- section is [BPSC AE (GEN. ENGG.) 11.11.2022 ]

Correct Answer: (c) maximum shearing stress only
Solution:

A beam of uniform rectangular cross-section is under a transverse load– The maximum shear stress developed in a beam of rectangular cross-section–
• It may be noted that the shear stress in a beam is uniformly distributed over the cross-section but varies from zero at outer fiber to a maximum at the neutral axis.

45. The factor of safety of an infinite slope in a sand deposit is found to be 1.732. The angle of shearing resistance of the sand is 30 degree. The average slope of the sand deposit is given by [BPSC AE (Civil) 30.03.2019 ]

Correct Answer: (c) tan⁻¹(0.333)
Solution:

46. The absolute maximum bending moment in a simply supported beam of span 20 m due to moving u.d.l. of 4 kN/m spanning over 5 m is [BPSC AE (Civil) 30.03.2019 ]

Correct Answer: (d) 87.5 kNm at the midpoint
Solution:

47. In a two-hinged arch, an increase in temperature induces [BPSC AE (Civil) 30.03.2019 ]

Correct Answer: (c) the maximum bending moment at the crown
Solution:

Maximum bending at the crown if the temperature increase is greater at the crown of the arch rib than at the hinges, there will be a maximum bending moment induced at the crown. This is because the arch rib expands more at the crown than at the hinges and this results in a maximum bending moment at the crown.

48. A symmetrical two-hinged parabolic arch when subjected to a uniformly distributed load on the entire horizontal span, is subjected to [BPSC AE (Civil) 30.03.2019 ]

Correct Answer: (b) normal thrust alone
Solution:

Parabolic arch subjected to UDC (w/per unit length) our entire span
H = wL²/8h
Two hinged parabolic arch subjected to uniformly distributed load on the half span.
• A two hinged parabolic arch carries a UDL of w per unit run on entire span then it is subjected to normal thrust alone.
Radial shear and bending moment at every point on arch is found to be zero.

49. A fixed beam of span L is carrying a point load P at its mid-span. If the moment of inertia of the middle half-length is two times that of the remaining length, then the fixed end moments will be [BPSC AE (Civil) 30.03.2019 ]

Correct Answer: (b) 5PL/48
Solution:

The slope of fixed and will be zero using moment area theorem.

50. The influence line for horizontal thrust in twohinged parabolic arch is [BPSC AE (Civil) 30.03.2019 ]

Correct Answer: (a) parabolic
Solution:

The ILD for thrust in parabolic arch horizontal thrust due to a concentrated load P at a distance x from X

Then ILD for horizontal thrust is parabolic.