31. The tension and shear force (both in kN) in each bolt of the joint, as shown below, respectively are
    

1. 1. 30.33 and 20.00

   
   
   

   2. 30.33 and 25.00

   
   
   

   3. 33.33 and 20.00

   
   
   

   4. 33.33 and 25.00

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Horizontal component = P_u cos θ
   Tension in each bolt,
   

   	Pu cos θ
   	6

   
   

   =	250 × 4	= 33.3 kN
   	5 × 6

   
   Shear in each bolt = Vertical component
   

   =	3 × 250	= 25 kN
   	5 × 6

   Correct Option: D

   
   Horizontal component = P_u cos θ
   Tension in each bolt,
   

   	Pu cos θ
   	6

   
   

   =	250 × 4	= 33.3 kN
   	5 × 6

   
   Shear in each bolt = Vertical component
   

   =	3 × 250	= 25 kN
   	5 × 6

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32. A steel section is subjected to a combination of shear and bending actions. The applied shear force is V and the shear capacity of the section is V_s. For such a section, high shear force (as per IS:800-2007) is defined as

1. 1. V > 0.6 V_s

   
   
   

   2. V > 0.7 V_s

   
   
   

   3. V > 0.8 V_s

   
   
   

   4. V > 0.9 V_s

   
   
   

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1. View Hint View Answer Discuss in Forum

   As per IS 800:2007
   V > 0.6 V_s

   Correct Option: A

   As per IS 800:2007
   V > 0.6 V_s

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33. Two steel columns P (length L and yield strength f_y = 250 MPa) and Q (length 2L and yield strength f_y = 500 MPa) have the same cross-sections and end-conditions. The ratio of buckling load of column P to that of column Q is

1. 1. 0.5

   
   
   

   2. 1.0

   
   
   

   3. 2.0

   
   
   

   4. 4.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Buckling load,
   

   P =	π2EI
   	(leff)2

   
   

   ∴ PP	π2EI	PQ =	π2EI
   	L2		(2L)2

   
   

   ∴	PP	= 4
   	PQ

   Correct Option: D

   Buckling load,
   

   P =	π2EI
   	(leff)2

   
   

   ∴ PP	π2EI	PQ =	π2EI
   	L2		(2L)2

   
   

   ∴	PP	= 4
   	PQ

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34. Two plates are connected by fillet welds of size 10 mm and subjected to tension, as shown in the figure. The thickness of each plate is 12 mm. The yield stress and the ultimate tensile stress of steel are 250 MPa and 410 MPa, respectively. The welding is done in the workshop (γ_mw = 1.25). As per the Limit State Method of IS 800: 2007, the minimum length (rounded off to the nearest higher multiple of 5 mm) of each weld to transmit a force P equal to 270 kN is
    

1. 1. 100 mm

   
   
   

   2. 105 mm

   
   
   

   3. 110 mm

   
   
   

   4. 115 mm

   
   
   

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1. View Hint View Answer Discuss in Forum

   P =		fy		.tess.Less.
   		rmax

   
   t_ess = 0.7 × 10 = 7 mm
   

   ⇒ 210 × 103 =	250	× 7 × Less
   	1.25

   
   ∴ L_ess = 192.82 ≈ 190
   Provide extra widening
   ∴ resultant L_ess = 190 + 20 = 210 mm
   ∴ Each weld = 105 mm

   Correct Option: B

   P =		fy		.tess.Less.
   		rmax

   
   t_ess = 0.7 × 10 = 7 mm
   

   ⇒ 210 × 103 =	250	× 7 × Less
   	1.25

   
   ∴ L_ess = 192.82 ≈ 190
   Provide extra widening
   ∴ resultant L_ess = 190 + 20 = 210 mm
   ∴ Each weld = 105 mm

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35. As per IS 456: 2000, in the Limit State Design of a flexural member, the strain in reinforcing bars under tension at ultimate state should not be less than

1. 1. 	fy
      	Es

   
   
   

   2. 	fy	+ 0.002
      	Es

   
   
   

   3. 	fy
      	1.15Es

   
   
   

   4. 	fy	+ 0.002
      	1.15Es

   
   
   

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1. View Hint View Answer Discuss in Forum

   	fy	+ 0.002
   	1.15Es

   Correct Option: D

   	fy	+ 0.002
   	1.15Es

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