Structural Analysis Questions and Answers - Influence Lines for Beams

Directions to solve : While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
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Data for Questions 1 to 5 : AC= 1m, CB =3 m
C is just an arbitrary point. A is pin support and B is a roller type support.
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1. What will be the equation of ILD of shear force at point C for AC part?
a) .25X – 1.25
b) .25X – 2.25
c) .25X – .25
d) .25X + .25

Answer: c
Explanation: Just assume force at any point between AC and conserve moment about point A.

2. if we have to apply a concentrated load in the above shown beam, such that shear at C becomes max. , where should we apply that load?
a) At A
b) At B
c) At C
d) Midway between A and C

Answer: c
Explanation: If we draw ILD according to the above given equations, we will see that peak of ILD comes at point C.

3. If a concentrated load of 50KN is applied at point C, then what will be the shear developed at point C?
a) 17.5 KN
b) 27.5 KN
c) 37.5 KN
d) 47.5 KN

Answer: c
Explanation: Position of ILD at point C is 0.75 (peak). So, shear developed will be 0.75 multiplied by 50KN.

4. What will be the shear developed at point C if a uniform load of 10KN/m is applied between point B and C?
a) 10.25 KN
b) 11.25 KN
c) 12.25 KN
d) 13.25 KN

Answer: b
Explanation: In case of uniform load, area of ILD curve multiplied by uniform load gives the shear.

5. If both, a load of 50KN at point C and a uniform load of 10KN/m between CB acts, then what will be the shear generated at point C?
a) 48.75
b) 50.75
c) 46.75
d) 52.75

Answer: a
Explanation: Net shear generated will be the sum of individually generated shear which has been already calculated earlier

Data for Questions 6 to 8 : AB= 2m, BC= 3m, CD= 3m
B is pin support, D is roller and C is just an arbitrary point.
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6. What will be the ILD equation for ILD of shear at point B?
a) 1.33 – 0.116625X
b) 2.33 – 0.16625X
c) 3.33 – 0.16625X
d) 1.33 – 0.16625X

Answer: d
Explanation: Apply unit load at any point at a distance X and conserve moment about point D.

7. What will be the ILD equation for ILD of shear at point C for AB part of beam?
a) -0.33 + 0.165X
b) -0.33 + 0.265X
c) -0.43 + 0.165X
d) -0.33 + 0.365X

Answer: a
Explanation: Apply unit load between point A and B and conserve moment about point B.

8. What will be the ILD equation for ILD of shear at point D?
a) -.43 + 0.16625X
b) -.33 + 0.16625X
c) -.53 + 0.16625X
d) -.33 + 0.216625X

Answer: b
Explanation: Apply load at any point and conserve moment about point B.

9. If we require to construct ILD of vertical support at a pin joint, then according to Muller-Breslau principle, by which type of support should it be replaced?
a) Roller guide
b) Pin roller
c) Fixed support
d) Hinge

Answer: a
Explanation: We need to remove the force for which we need to construct ILD and roller guide would remove the vertical reaction.

10. A is a pin support, B is a hinge, C and D are roller type support.
AB = BC = 1m, CD = 2m
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For ILD of shear at a point just left to C, what will be the equation for it on BC part of beam?
a) X
b) -X
c) 2X
d) 1

Answer: d
Explanation: On applying Muller-Breslau principle, we will see that the part right to the point can’t move as point C is right next to it. So, ILD will have to remain parallel to x axis.