131. A rectangular region in a solid is in a state of plane strain. The (x, y) coordinates of the corners of the undeformed rectangle are given by P(0, 0), Q(4, 3), S(0, 3). The'rectangle is subjected to uniform strain ε_xx = 0.001, ε_yy = 0.002, γ_xy = 0.003. The deformed length of the elongated xy diagonal, upto three decimal places, is _______ units.

1. 1. 5.015 units

   
   
   

   2. 15 units

   
   
   

   3. 5 units

   
   
   

   4. None of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Length of diagonal PR = √4² + 3² = 5
   Strain at an angle 45°
   ε = ε_xcos² θ + ε_ysin² θ + γ_xysinθcos θ
   = 0.001 × cos² 45 + 0.002 sin² 45 + 0.003 sin 45 cos 45
   = 0.003
   

   ε =	Δl	= 0.003
   	l

   
   Δl = = 5 × 0.003 = 0.015 units
   Deformed length = 5 + 0.015 = 5.015 units

   Correct Option: A

   
   Length of diagonal PR = √4² + 3² = 5
   Strain at an angle 45°
   ε = ε_xcos² θ + ε_ysin² θ + γ_xysinθcos θ
   = 0.001 × cos² 45 + 0.002 sin² 45 + 0.003 sin 45 cos 45
   = 0.003
   

   ε =	Δl	= 0.003
   	l

   
   Δl = = 5 × 0.003 = 0.015 units
   Deformed length = 5 + 0.015 = 5.015 units

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132. An initially stress-free massless elastic beam of length L and circular cross-section with diameter d(d<< L) is held fixed between two walls as shown. The beam material has Young's modulus E and coefficient of thermal expansion α.
     
     If the beam is slowly and uniformly heated, the temperature rise required to cause the beam to buckle is proportional to
     

1. 1. d
      

   
   
   

   2. d²
      

   
   
   

   3. d³
      

   
   
   

   4. d⁴

   
   
   

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1. View Hint View Answer Discuss in Forum

   By Buckling load, P =	4π2EI
   	l2

   
   Force due to change in temperature,
   

   σ =	P	= lα∆T
   	A

   
   P = Alα∆T
   

   	4π2EI	= A l α∆T
   	l2

   
   

   	I	∝ ∆T
   	A

   
   

   ∆T ∝	d4
   	d2

   
   ∆T ∝ d²
   

   Correct Option: B

   By Buckling load, P =	4π2EI
   	l2

   
   Force due to change in temperature,
   

   σ =	P	= lα∆T
   	A

   
   P = Alα∆T
   

   	4π2EI	= A l α∆T
   	l2

   
   

   	I	∝ ∆T
   	A

   
   

   ∆T ∝	d4
   	d2

   
   ∆T ∝ d²
   

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133. Consider a steel (Young's modulus E = 200 GPa) column hinged on both sides. Its heights is 1.0 m and cross-section is 10 mm x 20 mm. The lowest Euler critical buckling load (in N) is ___.

1. 1. 3289 N

   
   
   

   2. 3289.8681 N

   
   
   

   3. 32.898 N

   
   
   

   4. 3298.8186 N

   
   
   

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1. View Hint View Answer Discuss in Forum

   Euler’s critical load =	π2EI
   	l2

   
   

   ⇒	π2 × 200 × 109 × 0.2 × 0.13	= 3289.8681 N
   	12

   Correct Option: B

   Euler’s critical load =	π2EI
   	l2

   
   

   ⇒	π2 × 200 × 109 × 0.2 × 0.13	= 3289.8681 N
   	12

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134. For a long slender column of uniform cross section, the ratio of critical buckling load for the case with both ends clamped to the case with both ends hinged is
     

1. 1. 1
      

   
   
   

   2. 2
      

   
   
   

   3. 4
      

   
   
   

   4. 8

   
   
   

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1. View Hint View Answer Discuss in Forum

   Critical Buckling load for column fixed at both ends
   

   =	4π2EI
   	L2

   
   

   Critical Bucking load for a column hinged at both ends =	π2EI
   	L2

   
   

   Hence ,	(Pcr)1	= 4
   	(Pcr)2

   Correct Option: C

   Critical Buckling load for column fixed at both ends
   

   =	4π2EI
   	L2

   
   

   Critical Bucking load for a column hinged at both ends =	π2EI
   	L2

   
   

   Hence ,	(Pcr)1	= 4
   	(Pcr)2

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135. The rod PQ of length L and with flexural rigidity EI is hinged at both ends. For what minimum force F is it expected to buckle?
     
     

1. 1. 	π2EI
      	L2

      
      

   
   
   

   2. 	√2π2EI
      	L2

      
      

   
   
   

   3. 	π2EI
      	√2L2

      
      

   
   
   

   4. 	π2EI
      	2L2

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since both ends hinged, therefore, L_e = L
   

   Buckling load, W =	π2EI
   	L2

   
   
   Also W = F cos 45°
   

   ∴ F =	π2EI	/ cos 45° =	√2π2EI
   	L2		L2

   Correct Option: C

   Since both ends hinged, therefore, L_e = L
   

   Buckling load, W =	π2EI
   	L2

   
   
   Also W = F cos 45°
   

   ∴ F =	π2EI	/ cos 45° =	√2π2EI
   	L2		L2

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