101. Force in the member AB of the truss is

1. 1. P/√2

   
   
   

   2. P/√3

   
   
   

   3. P/2

   
   
   

   4. P

   
   
   

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1. View Hint View Answer Discuss in Forum

   (Method of joints)
   
   ∑ F_x = 0
   F_BD cos θ + F_AB = 0
   ∑ F_y = 0
   

   FBD sin θ +	P	= 0
   	2

   
   
   

   	- P	+	1	+ FAB = 0
   	√2		√2

   
   

   ∴ FAB =	P
   	2

   Correct Option: C

   (Method of joints)
   
   ∑ F_x = 0
   F_BD cos θ + F_AB = 0
   ∑ F_y = 0
   

   FBD sin θ +	P	= 0
   	2

   
   
   

   	- P	+	1	+ FAB = 0
   	√2		√2

   
   

   ∴ FAB =	P
   	2

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102. The vertical reaction at support H is

1. 1. 15 kN upward

   
   
   

   2. 9.84 kN upward

   
   
   

   3. 15 kN downward

   
   
   

   4. 9.84 kN downward

   
   
   

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1. View Hint View Answer Discuss in Forum

   Consider section E H.
   
   ∑M_G = 0
   (RH × 4) + 10.31 – (9.48 × 2) + (10 × 1) – (5 × 4 × 2) = 0
   R_H = 9.84 kN

   Correct Option: B

   Consider section E H.
   
   ∑M_G = 0
   (RH × 4) + 10.31 – (9.48 × 2) + (10 × 1) – (5 × 4 × 2) = 0
   R_H = 9.84 kN

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103. The magnitude of the shear force immediate to the left and immediate to the right of section B are, respectively

1. 1. 0 and 20 kN

   
   
   

   2. 10 kN and 10 kN

   
   
   

   3. 20 kN and 0

   
   
   

   4. 9.84 kN and 10.16 kN

   
   
   

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1. View Hint View Answer Discuss in Forum

   SF at right of B = 20 kN
   SF at left of B = 0

   Correct Option: A

   SF at right of B = 20 kN
   SF at left of B = 0

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104. In a system, two connected rigid bars AC and BC are of identical length L with pin supports at A and B. The bars are interconnected at C by a friction less hinge. The rotation of the hinge is restrained by a rotational spring of stiffness, k. The system initially assumes a straight line configuration, ACB. Assuming both the bars as weightless, the rotation at supports, A and B, due to a transverse load, P applied at C is:

1. 1. PL/4k

   
   
   

   2. PL/2k

   
   
   

   3. P/4k

   
   
   

   4. Pk/4L

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   External work = (1/2)P.L θ
   Strain energy in spring
   =1/2 × k.(2θ)(2θ) = 2 K θ²
   External work = strain energy
   ∴ (1/2)P.L θ = 2 K θ²
   ∴ θ = PL/4k

   Correct Option: A

   
   External work = (1/2)P.L θ
   Strain energy in spring
   =1/2 × k.(2θ)(2θ) = 2 K θ²
   External work = strain energy
   ∴ (1/2)P.L θ = 2 K θ²
   ∴ θ = PL/4k

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105. A fixed end beam is subjected to a load, W at 1/ 3rd span from the left support as shown in the figure. The collapse load of the beam is
     

1. 1. 16.5 M_p /L

   
   
   

   2. 15.5 M_p /L

   
   
   

   3. 15.0 M_p /L

   
   
   

   4. 16.0 M_p /L

   
   
   

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1. View Hint View Answer Discuss in Forum

   Plastic hinges = 3
   
   

   α =	θ
   	2

   
   ⇒ θ = 2α
   

   θ =	∆
   	L/3

   
   

   ⇒ ∆ = θ	L
   	3

   
   

   = 2α	L
   	3

   
   2M_Pθ + 2M_Pθ + 2M_Pα + M_Pα = W∆
   

   Correct Option: C

   Plastic hinges = 3
   
   

   α =	θ
   	2

   
   ⇒ θ = 2α
   

   θ =	∆
   	L/3

   
   

   ⇒ ∆ = θ	L
   	3

   
   

   = 2α	L
   	3

   
   2M_Pθ + 2M_Pθ + 2M_Pα + M_Pα = W∆
   

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