Structural Analysis Multiple Choice Questions on “Influence Lines for Beams”.
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
1. If on ILD analysis peak force comes out to be 2 KN, then what will be the peak force if loading is 2KN?
a) 1 KN
b) 2 KN
c) 3 KN
d) 4 KN
Answer: d
Clarification: Peak force will be load multiplied by earlier peak load i.e. 2*2.
Following figure is for Q2-Q7.
AC= 1m, CB =3 m
C is just an arbitrary point. A is pin support and B is a roller type support.
2. What will be the equation of ILD of shear force at point C for CB part?
a) 0.75 – 0.375X
b) 0.75 – 0.475X
c) 0.85 – 0.375X
d) 0.75 – 0.1375X
Answer: a
Clarification: Just assume force at any point between BC and conserve moment about point B.
3. 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
Clarification: Just assume force at any point between AC and conserve moment about point A.
4. 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
Clarification: If we draw ILD according to the above given equations, we will see that peak of ILD comes at point C.
5. 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
Clarification: Position of ILD at point C is 0.75 (peak). So, shear developed will be 0.75 multiplied by 50KN.
6. 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
Clarification: In case of uniform load, area of ILD curve multiplied by uniform load gives the shear.
7. 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
Clarification: Net shear generated will be the sum of individually generated shear which has been already calculated earlier.
Following figure is for Q8-Q10.
AB= 2m, BC= 3m, CD= 3m
B is pin support, D is roller and C is just an arbitrary point.
8. 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
Clarification: Apply unit load at any point at a distance X and conserve moment about point D.
9. 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
Clarification: Apply unit load between point A and B and conserve moment about point B.
10. 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
Clarification: Apply load at any point and conserve moment about point B.