Miscellaneous MCQ’s on Strength of Materials along with Answers and Solutions (Part-I)

 

Miscellaneous MCQ’s on Strength of Materials along with Answers and Solutions  (Part-I)

 

Q.01c) A member ABCD is subjected to point loads as shown in fig. find total change in length of the bar.    

Fig. 1c

Objective Questions:-

1)      What will b magnitude of force exerted in portion BC,

a)      10kN tensile

b)     30kN compressive

c)      20kN tensile

d)     None of these

2)      Elongation in portion AB of the bar,

a)      0.033mm

b)     0.180mm

c)      0.050mm

d)     0.097mm

3)      what will be the contraction in portion BC of the bar,

a)      0.033mm

b)     0.180mm

c)      0.050mm

d)     0.097mm

4)      Elongation in portion CD of the bar,

a)      0.033mm

b)     0.180mm

c)      0.050mm

d)     0.097mm

5)      Total Elongation of the bar ABCD,

a)      0.033mm

b)     0.180mm

c)      0.050mm

d)     0.097mm

Q.No.1 c)

Solution:- Given data; E = 200 x 103 N/mm2,                                       A1 = 300 mm2, L1 = 200 mm,                                       A2 = 500 mm2, L2 = 600 mm,                                       A3 = 400 mm2, L3 = 200 mm. Fig 1, 2 & 3  Carries 10 kN, 30kN & 20 kN Load respectively. As, dL = P L / A E               (+ve (extension), -ve (contraction) )   Here, dL = dL1 + dL2 + dL3               = (P1 L1 / A1 E) – (P2 L2 / A2 E) + (P3 L3 / A3 E)               = (10 x 1000 x 200/ 300 x 200 x 1000) – -          ( 30 x 1000 x 600 / 500 x 200 x 1000) +            + ( 20 x 1000 x 200 / 400 x 200 x 1000)               = 0.033 – 0.18 + 0.05         dL = 0.097 mm ( contraction)

02 Marks                     02 marks     02    arks

 

Q.02 A metallic bar of 300 mm x 100 mm x 40 mm is subjected to a force of 5 kN (tensile), 6 kN (tensile) and 4 kN (tensile) along x, y & z directions respectively. Determine the change in the volume of the block.  Take E = 2 x 105 N/mm2 and poisson’s ratio = 0.25.

Objective Questions:-

1)      What will be the normal stress in X-direction,

a)      1.25 N/mm2

b)      0.50 N/mm2

c)      0.133N/mm2

d)     None of these

 

2)      What will be the normal stress in Y-direction,

a)      1.25 N/mm2

b)     0.50 N/mm2

c)      0.133N/mm2

d)     1.75 N/mm2

3)      What will be the normal stress in Z-direction,

a)      1.25 N/mm2

b)      0.50 N/mm2

c)      0.133 N/mm2

d)     0.633 N/mm2

4)      Volumetric strain in the member,

a)      4.6575x10^-6 mm3/mm3

b)      4.6575x10^-5 mm3/mm3

c)      3.6575x10^-6 mm3/mm3

d)     5.5895x10^-4 mm3/mm3

5)      Change in volume under stresses induced will be,

a)      4.6575 mm3

b)      3.6575 mm3

c)      5.8950 mm3

d)     None of these

 

Q.No.2

01 Marks     ½ mark     ½ mark       ½ mark      ½ mark     02 marks

 

Q.03 A simply supported beam is subjected to a combination of loads as shown in figure. Sketch the shear force and bending moment diagrams and find the position and magnitude of maximum bending moment.

Objective Questions:-

1)      Support Reaction at A i.e, Ra,

a)      8 kN

b)     4 kN

c)      12 kN

d)     None of these

 

2)      Support Reaction at B i.e, Rb,

a)      8 kN

b)      4 kN

c)      12 kN

d)     None of these

3)      Shear Force at C i.e, SF @C,

a)      8 kN

b)     4 kN

c)      12 kN

d)     None of these

4)      Shear Force at D i.e, SF @D,

a)      8 kN

b)      4 kN

c)      12 kN

d)     None of these

5)      Bending Moment at C i.e, BM@C,

a)      16 kNm

b)      20 kNm

c)      24 kNm

d)     12 kNm

6)      Bending Moment at D i.e, BM @D,

a)      16 kNm

b)     20 kNm

c)      24 kNm

d)     12 kNm

7)      Maximum Bending Moment in Beam i.e, Mmax,

a)      16 kNm

b)      20 kNm

c)      24 kNm

d)     None of these

8)      Position of Point of Contra-shear i.e, Position of Mmax,

a)      5m from Support A

b)      2m from Support A

c)      2m from point D,

d)     Both a & c

Q.No. 3

Solution: To draw the shear force diagram and bending moment diagram we need RA and RB. Fig. Q.3  Shear force and bending moment By taking moment of all the forces about point A. We get RB × 10 – 8 × 9 – 2 × 4 × 5 – 4 × 2 = 0 RB = 12 kN From condition of static equilibrium ΣFy = 0 RA + RB – 4 – 8 – 8 = 0 RA = 20 – 12 = 8 kN To draw shear force diagram we need shear force at all salient points: For AC; FA = + RA = 8 kN For CD, FC = + 8 – 4 = 4 kN FD = 4 kN For DE, Fx = 8 – 4 – 2 (x – 3) = 10 – 2x At x = 3 m; FD = 10 – 6 = 4 kN At x = 7 m; FE = 10 – 2 × 7 = –4 kN The position for zero SF can be obtained by 10 – 2x = 0 x = 5 m For EF; Fx = 8 – 4 – 8 = – 4 kN For FB; Fx = 8 – 4 – 8 – 8 = – 12 kN To draw BMD, we need BM at all salient points. For region AC, Mx = + 8x At x = 0; MA = 0 x = 2; MC = 8 × 2 = 16 kN m For region CD; Mx = + 8x – 4 (x – 2) At x = 2 m; MC = 8 × 2 – 4 (2 – 2) = 16 kN m At x = 3 m; MD = 8 × 3 – 4 (3 – 2) = 20 kN m For region DE, Mx = + 8x – 4 (x – 2) – 2(x – 3)2 / 2 = 10x – x2 – 1 At x = 3 m; MD = 8 × 3 – 4 (3 – 2) – 2(32 – 3)2 / 2 = 20 kN m At x = 7 m; ME = 10 × 7 – (7)2 – 1 = 20 kN m At x = 5 m; MG = 10 × 5 – (5)2 – 1 = 24 kN m For region EF, Mx = 8x – 4 (x – 2) – 2 × 4 (x – 5) = 48 – 4x At x = 9 m, MF = 120 – 12 × 9 = 12 KN m At x = 10 m; MB = 120 – 12 × 10 = 0 The shear force diagram and bending moment diagram can now be drawn by using the various values of shear force and bending moment. For bending moment diagram the bending moment is proportional to x, so it depends, linearly on x and the lines drawn are straight lines. The maximum bending moment exists at the point where the shear force is zero, and also dM/dx = 0 in the region of DE d/dx (10 x – x2 – 1) = 0 10 – 2x = 0 X = 5 m Mmax = 10 × 5 – (5)2 – 1 = 24 kN m Thus, the maximum bending is 24 kN m at a distance of 5 m from end A.

02 Marks               02 marks                   01 marks              02 marks              01 marks             02 marks                                 02 marks