66. If a small concrete cube is submerged deep in still water in such a way that the pressure exerted on all faces of the cube is p, then the maximum shear stress developed inside the cube is

1. 1. 0

   
   
   

   2. p/2

   
   
   

   3. p

   
   
   

   4. 2p

   
   
   

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   NA

   Correct Option: A

   NA

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67. As per IS 800:2007, the cross-section in which the extreme fiber can reach the yield stress, but cannot develop the plastic moment of resistance due to failure by local buckling is classified as
    (a)
    (b)
    (c)
    (d)

1. 1. plastic section

   
   
   

   2. compact section

   
   
   

   3. semi-compact section

   
   
   

   4. slender section

   
   
   

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   semi-compact section

   Correct Option: C

   semi-compact section

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68. The creep strains are

1. 1. caused due to dead loads only

   
   
   

   2. caused due to live loads only

   
   
   

   3. caused due to cyclic loads only

   
   
   

   4. independent of loads

   
   
   

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   caused due to dead loads only

   Correct Option: A

   caused due to dead loads only

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69. A simply supported beam AB of span, L = 24 m is subjected to two wheel loads acting at a distance, d = 5 m apart as shown in the figure below. Each wheel transmits is a load, P = 3 kN and may occupy any position along the beam. If the beam is an I-section having section modulus, S = 16.2 cm3, the maximum bending stress (in GPa) due to the wheel loads is_______________
    

1. 1. 1759

   
   
   

   2. 175

   
   
   

   3. 1759.9

   
   
   

   4. 1759.2

   
   
   

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   	σ	=	m
   	y		I

   
   

   ⇒ σ =	m	. y =	m
   	I		z

   
   

   =	28.5 × 106
   	16.2 × 103

   
   (m = 3 × 9.5 = 28.5)
   = 1759.2 GPa

   Correct Option: D

   
   

   	σ	=	m
   	y		I

   
   

   ⇒ σ =	m	. y =	m
   	I		z

   
   

   =	28.5 × 106
   	16.2 × 103

   
   (m = 3 × 9.5 = 28.5)
   = 1759.2 GPa

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70. A horizontal beam ABC is loaded as shown in the figure below. The distance of the point of contraflexure from end A (in m) is ____
    

1. 1. 0.25

   
   
   

   2. 0.75

   
   
   

   3. 0.24

   
   
   

   4. 0.45

   
   
   

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   Take R_C at redundant
   By Marshall Reciprocal Theorem
   Deflection at C due to load B, ∆_BC
   

   ∆BC =	10 ×(0.75)3	+	10 ×(0.75)3	× 0.25
   	3EI		2EI

   
   

   =	2.11	↓
   	EI

   
   Deflection at C due to redundant R_c, ∆_cc
   

   ∆cc =	Rc ×(0.75)3	=	0.141Rc
   	3EI		EI

   
   ∴ ∆_c = 0
   

   	2.11	-	0.141Rc	= 0
   	EI		EI

   
   

   Rc =	2.11	= 15 KN
   	0.141

   
   
   M_x = 10x – 15 × (x – 0.25) = 0
   ⇒ = 10x – 15x + 3.75 = 0
   ⇒ x = 0.75m
   ∴ Distanced contraflexure from C = 0.75m
   ⇒ Distanced contraflexure from A = 1 – 0.75 = 0.25m

   Correct Option: A

   
   Take R_C at redundant
   By Marshall Reciprocal Theorem
   Deflection at C due to load B, ∆_BC
   

   ∆BC =	10 ×(0.75)3	+	10 ×(0.75)3	× 0.25
   	3EI		2EI

   
   

   =	2.11	↓
   	EI

   
   Deflection at C due to redundant R_c, ∆_cc
   

   ∆cc =	Rc ×(0.75)3	=	0.141Rc
   	3EI		EI

   
   ∴ ∆_c = 0
   

   	2.11	-	0.141Rc	= 0
   	EI		EI

   
   

   Rc =	2.11	= 15 KN
   	0.141

   
   
   M_x = 10x – 15 × (x – 0.25) = 0
   ⇒ = 10x – 15x + 3.75 = 0
   ⇒ x = 0.75m
   ∴ Distanced contraflexure from C = 0.75m
   ⇒ Distanced contraflexure from A = 1 – 0.75 = 0.25m

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