Mechanical and structural analysis miscellaneous
- The frame below shows three beam elements OA, OB and OC, with identical length L and flexural rigidity EI, subject to an external moment M applied at the rigid joint O. The correct set of bending moments [MOX MOB, MOC] that develop at O in the three beam elements OA, OB and OC respectively, is
,img src="http://images.interviewmania.com/wp-content/uploads/2019/10/63-2.png">
EI/L is constant for all three members
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KOA = 3EI l KOC = 4EI l
KOB = 0DFOA = KOA KOA + KOC = 3 = 3 EI cancels 3 + 4 7 l DFOC = KOC KOC + KOA = 4 = 4 4 + 3 7
DFOB = 0Correct Option: D
KOA = 3EI l KOC = 4EI l
KOB = 0DFOA = KOA KOA + KOC = 3 = 3 EI cancels 3 + 4 7 l DFOC = KOC KOC + KOA = 4 = 4 4 + 3 7
DFOB = 0
- Consider the following two statements related to reinforced concrete design, and identify whether they are TRUE or FALSE:
I. Curtailment of bars in the flexural tension zone in beams reduces the shear strength at the cut-off locations.
II. When a rectangular column section is subject to biaxially eccentric compression, the neutral axis will be parallel to the resultant axis of bending.
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Statement I is TRUE, and Statement II is FALSE
Correct Option: B
Statement I is TRUE, and Statement II is FALSE
- For the loading given in the figure below, two statements (I and II) are made).
I. Member AB carries shear force and bending moment
II. Member BC carries axial load and shear force.
Which of the following is true?
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Member AB carries shear force and bending moment. Member BC carries only axial load and no shear force
Correct Option: A
Member AB carries shear force and bending moment. Member BC carries only axial load and no shear force
- The shear modulus (G), modulus of elasticity (E) and the Poisson's ratio (v) of a material are related as,
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E =2G (1 + v)
∴ G = E 2(1 + v) Correct Option: A
E =2G (1 + v)
∴ G = E 2(1 + v)
- Identify, from the following, the correct value of the bending moment MA (in kNm) units) at the fixed end A in the statically determinate beam shown below (with internal hinges at B and D), when a uniformly distributed load of 10 kN/m is placed on all spans.
(Hint: Sketching the influence line for MA or applying the Principle of Virtual Displacements makes the solution easy).
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(Influence line Diagram, ILD for MA)
Ordinate at B, 2/y = 1
⇒ y = 2
Ordinate at D = 2
BM at A = loading intensity X area under load= 10. 1 × 4 × 2 + 10 - 1 × 4 × 2 = 0 2 2 Correct Option: C
(Influence line Diagram, ILD for MA)
Ordinate at B, 2/y = 1
⇒ y = 2
Ordinate at D = 2
BM at A = loading intensity X area under load= 10. 1 × 4 × 2 + 10 - 1 × 4 × 2 = 0 2 2