B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the reaction force at support C?
Answer: c
Explanation: Balance moment about point B.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the shape of SFD in this case?
Answer: c
Explanation: Since loads are not uniform, SFD will be linear and support at point C and B will lead to discontinuity.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What is the shape of BMD for this diagram?
Answer: b
Explanation: It will be 0 till point B, and then will increase till C and then again decrease till D (every time linearly), thus making it triangular
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the peak value of SFD?
Answer: c
Explanation: It will decrease to 20 at point B and then increase by 80 at point C and then will remain constant. So, peak value will be 80-20 i.e. 60KN.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. Where would peak value of BMD lie?
Answer: c
Explanation: It will be 0 till point B, and then will increase till C and then again decrease till D (every time linearly), thus making it triangular.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. Which type of joint would replace point A in its conjugate beam?
Answer: d
Explanation: Since point A is a free end, a fixed joint would replace it in the conjugate beam.
How many fixed joint will be there in conjugate beam?
Answer: b
Explanation: Since there are 2 fixed ends in initial beam, there will be two fixed joints.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the shear developed at hinge B in conjugate beam?
Answer: a
Explanation: Balance moment about point C in the beam BC.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the shear developed at hinge C in conjugate beam?
Answer: c
Explanation: Balance moment about point B in the beam BC.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the modulus of slope at point A?
Answer: c
Explanation: Just balance shear forces in the conjugate beam AB.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the modulus of slope at point D?
Answer: a
Explanation: Just balance shear forces in the conjugate beam CD after balancing moment.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the modulus of deflection of point A?
Answer: b
Explanation: Balance moment about point B in beam AB.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
Q. What will be the modulus of deflection of point D?
Answer: c
Explanation: Balance moment about point C in beam CD.
B is a hinge support and C is roller support. A and D are free ends. A load of 60 KN acts in downward direction at point D. Sign conventions are as usual.
AB = CD = 1m and BC = 3m
All force options are in kN.
All moment options are in KNM.
All deformation options are in M.
E and I are given.
There won’t be any hinge in the conjugate beam.
State whether the above statement is true or false.
Answer: b
Explanation: Initial beam comprises of one internal pin and one internal roller, so conjugate beam would contain two hinges.
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