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All questions of Influence Lines for Civil Engineering (CE) Exam
A and C is a pin support while B is roller support.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
Q. 3 moment method can be used in:-- a)single span beams
- b)only double span beams
- c)double and single span beams
- d)all types of beams
Correct answer is option 'D'. Can you explain this answer?
A and C is a pin support while B is roller support.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
Q. 3 moment method can be used in:-
a)
single span beams
b)
only double span beams
c)
double and single span beams
d)
all types of beams
|
Nandita Datta answered |
Answer: d
Explanation: Although it is not valid for single span beams, it can be used for them using some particular techniques.
Explanation: Although it is not valid for single span beams, it can be used for them using some particular techniques.
The bed of an alluvial channel along the flow will always be- a)flat
- b)way
- c)duned and rippled
- d)and of the above is possible
Correct answer is option 'D'. Can you explain this answer?
The bed of an alluvial channel along the flow will always be
a)
flat
b)
way
c)
duned and rippled
d)
and of the above is possible
|
Avinash Mehta answered |
The correct answer is 'D', all of the above is possible. The bed of an alluvial channel along the flow can take on different forms depending on the specific conditions of the channel and the flow. For example, in a stable channel with a relatively low sediment load, the bed may be relatively flat. However, in a channel with a high sediment load or with high velocity flow, the bed may be rippled or even duned. Additionally, in meandering channels, the bed may be way. The bed morphology of an alluvial channel is a constantly changing and influenced by many factors such as sediment load, discharge, channel slope, and channel width.
The maximum bending moment at the left quarter point of a simple beam due to crossing of UDL of length shorter than the span in the direction left to right, would occur after the load had just crossed the section by- a)One-fourth of its length
- b)Half of its length
- c)Three-fourth of its length
- d)Its full length
Correct answer is option 'C'. Can you explain this answer?
The maximum bending moment at the left quarter point of a simple beam due to crossing of UDL of length shorter than the span in the direction left to right, would occur after the load had just crossed the section by
a)
One-fourth of its length
b)
Half of its length
c)
Three-fourth of its length
d)
Its full length
|
Jyoti Choudhury answered |
The load should be positioned such that section divides the span and load in the same ratio.
What will be the value of Rda?- a)1
- b)0
- c)0.481
- d)0.681
Correct answer is option 'C'. Can you explain this answer?
What will be the value of Rda?
a)
1
b)
0
c)
0.481
d)
0.681
|
|
Kirti Sharma answered |
Answer: c
Explanation: It will be the ration of Δda and Δaa. Δaa is 1944/EI and Δca is 936/EI. Join all these point for ILD.
Explanation: It will be the ration of Δda and Δaa. Δaa is 1944/EI and Δca is 936/EI. Join all these point for ILD.
The most important shape parameter in sediment analysis is- a)sphericity
- b)Shape factor
- c)roundness
- d)form factor
Correct answer is option 'A'. Can you explain this answer?
The most important shape parameter in sediment analysis is
a)
sphericity
b)
Shape factor
c)
roundness
d)
form factor
|
Aditya Deshmukh answered |
Yes, that is correct. Sphericity is the most important shape parameter in sediment analysis. It is a measure of the degree to which the shape of a particle approximates a perfect sphere. Sphericity is typically expressed as a dimensionless ratio, with a value of 1 indicating a perfect sphere and values less than 1 indicating a particle that is more elongated or irregular in shape.
Sphericity is an important parameter in sediment analysis because it can affect the behavior of particles in a number of ways. For example, more spherical particles tend to roll more easily and may be more resistant to abrasion than more elongated particles. Sphericity can also affect the way that particles settle out of a suspension and the way that they pack together.
So, the correct answer is:
a) Sphericity
Three wheel loads 10t, 26t and 24t spaced 2m apart roll on a girder from left to right with the 10t load leading. The girder has a span of 20 meter. For the condition of maximum bending moment at a section 8 meter from the left end.- a)The 10t load should be placed at the section.
- b)The 26t load should be placed at the section.
- c)The 24t load should be placed at the section.
- d)Either the 26t load or the 24t load should be placed at the section.
Correct answer is option 'B'. Can you explain this answer?
Three wheel loads 10t, 26t and 24t spaced 2m apart roll on a girder from left to right with the 10t load leading. The girder has a span of 20 meter. For the condition of maximum bending moment at a section 8 meter from the left end.
a)
The 10t load should be placed at the section.
b)
The 26t load should be placed at the section.
c)
The 24t load should be placed at the section.
d)
Either the 26t load or the 24t load should be placed at the section.
|
|
Ishani Basu answered |
Maximum bending moment at a section occurs when a particular load is on the section which changes the ratio 
as the load passes over the section .
where R1, → resultant of load on left sid e of section Resultant of all loads (R)
= 10 + 26 + 24 = 60t
When 10t load crosses section C.
When 26t load crosses the section C,
It means that maximum bending moment is obtained when 26t load is on the section.
as the load passes over the section .
where R1, → resultant of load on left sid e of section Resultant of all loads (R)
= 10 + 26 + 24 = 60t
When 10t load crosses section C.
When 26t load crosses the section C,
It means that maximum bending moment is obtained when 26t load is on the section.
When a series of wheel loads crosses a simply supported girder, the maximum bending moment under any given wheel load occurs when
- a)the centre of span is midway between the centre of gravity of the load system and the wheel load under consideration
- b)the centre of gravity of the load system is midway between the centre of span and wheel load under consideration
- c)the wheel load under consideration is midway between the centre of gravity of the load system
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
When a series of wheel loads crosses a simply supported girder, the maximum bending moment under any given wheel load occurs when
a)
the centre of span is midway between the centre of gravity of the load system and the wheel load under consideration
b)
the centre of gravity of the load system is midway between the centre of span and wheel load under consideration
c)
the wheel load under consideration is midway between the centre of gravity of the load system
d)
None of the above
|
Arnab Saini answered |
The centre of span is midway between the centre of gravity of the load system and the wheel load under consideration
The absolute maximum Bending Moment in a simply supported beam of span 20 m due to a moving UDL of 4 t/m spanning over 5 m is- a)87.5 t-m at the support
- b)87.5 t-m near the midpoint
- c)3.5 t-m at the midpoint
- d)87.5 t-m at the midpoint
Correct answer is option 'D'. Can you explain this answer?
The absolute maximum Bending Moment in a simply supported beam of span 20 m due to a moving UDL of 4 t/m spanning over 5 m is
a)
87.5 t-m at the support
b)
87.5 t-m near the midpoint
c)
3.5 t-m at the midpoint
d)
87.5 t-m at the midpoint
|
|
Ishani Basu answered |
Absolute maximum bending moment will occur at the centre when the load is spread equally on either side of the centre.
Consider the following statements:
1. Influence Line Diagram (ILD) for SF at the fixed end of a cantilever and SFD due to unit load at the free end are same.
2. ILD for BM at the fixed end of a cantilever and BMD due to unit load at the free end are same.
Which of these statements is/are correct?- a)Only 1
- b)Only 2
- c)Both 1 and 2
- d)Neither 1 nor 2
Correct answer is option 'A'. Can you explain this answer?
Consider the following statements:
1. Influence Line Diagram (ILD) for SF at the fixed end of a cantilever and SFD due to unit load at the free end are same.
2. ILD for BM at the fixed end of a cantilever and BMD due to unit load at the free end are same.
Which of these statements is/are correct?
1. Influence Line Diagram (ILD) for SF at the fixed end of a cantilever and SFD due to unit load at the free end are same.
2. ILD for BM at the fixed end of a cantilever and BMD due to unit load at the free end are same.
Which of these statements is/are correct?
a)
Only 1
b)
Only 2
c)
Both 1 and 2
d)
Neither 1 nor 2
|
|
Arya Menon answered |
ILD for BM at fixed end will have maximum ordinate when the unit load is at free end. While the BMD due to unit load at free end will have zero ordinate at free end and maximum ordinate at fixed end.
A uniformly distributed line load of 60 kN per metre run of length 5 meters on a girder of span 16 metres. What is the maximum positive shear force at a section 6 metres from the left end.- a)140.625 kN
- b)65.625 kN
- c)90.625 kN
- d)45.625 kN
Correct answer is option 'A'. Can you explain this answer?
A uniformly distributed line load of 60 kN per metre run of length 5 meters on a girder of span 16 metres. What is the maximum positive shear force at a section 6 metres from the left end.
a)
140.625 kN
b)
65.625 kN
c)
90.625 kN
d)
45.625 kN
|
|
Puja Sharma answered |
We must first draw the influence line diagram for the SF at the section D,
For maximum positive SF at D, the loading should be applied as shown in the figure.
Maximum positive = load x area of ILD SF at D intensity covered by the load
For maximum positive SF at D, the loading should be applied as shown in the figure.
Maximum positive = load x area of ILD SF at D intensity covered by the load
Ray = reaction at support A in y direction.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Q. What will be the value of Ray in the above figure?- a)Fad/Faa
- b)–Fad/Faa
- c)Fda/Faa
- d)–Fda/Faa
Correct answer is option 'B'. Can you explain this answer?
Ray = reaction at support A in y direction.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Q. What will be the value of Ray in the above figure?
a)
Fad/Faa
b)
–Fad/Faa
c)
Fda/Faa
d)
–Fda/Faa
|
|
Pallabi Chavan answered |
Answer: b
Explanation: If we remove the support at point A and replace it by force Faa. Due to sign convention, its direction will be initially considered upward.
Explanation: If we remove the support at point A and replace it by force Faa. Due to sign convention, its direction will be initially considered upward.
Following figure is for questions 5 and 5.
AB is of length 1m.Q. What will be the equation for vertical reaction at point A (RAY)?- a)RAY = 1-X
- b)RAY = 2-X
- c)RAY = 3-X
- d)RAY = 4-X
Correct answer is option 'A'. Can you explain this answer?
Following figure is for questions 5 and 5.
AB is of length 1m.
AB is of length 1m.
Q. What will be the equation for vertical reaction at point A (RAY)?
a)
RAY = 1-X
b)
RAY = 2-X
c)
RAY = 3-X
d)
RAY = 4-X
|
|
Poulomi Patel answered |
Answer: a
Explanation: Place unit load at any variable distance X and calculate RAY by conserving moment about point B.
Explanation: Place unit load at any variable distance X and calculate RAY by conserving moment about point B.
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
Q. Which support should replace fixed support at point A for above condition?- a)pin
- b)roller
- c)hinge
- d)slider mechanism
Correct answer is option 'D'. Can you explain this answer?
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
Q. Which support should replace fixed support at point A for above condition?
a)
pin
b)
roller
c)
hinge
d)
slider mechanism
|
|
Kavya Mehta answered |
Answer: d
Explanation: Only slider will give both moment and vertical reaction.
Explanation: Only slider will give both moment and vertical reaction.
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.What is the DSI for this structure?- a)1
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
What is the DSI for this structure?
a)
1
b)
2
c)
3
d)
4
|
Sankar Dasgupta answered |
Answer: a
Explanation: Total 4 unknowns are there and 3 equations are there.
Explanation: Total 4 unknowns are there and 3 equations are there.
What will be the value of Rca?- a)1
- b)0
- c)0.652
- d)0.852
Correct answer is option 'D'. Can you explain this answer?
What will be the value of Rca?
a)
1
b)
0
c)
0.652
d)
0.852
|
|
Ishani Basu answered |
Answer: d
Explanation: It will be the ration of Δca and Δaa. Δaa is 1944/EI and Δca is 1656/EI
Explanation: It will be the ration of Δca and Δaa. Δaa is 1944/EI and Δca is 1656/EI
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
Q. In BMD and SFD :-- a)Points remain fixed, position of load changes
- b)Points change, position of loads remain fixed
- c)Both of them changes
- d)Neither of them changes
Correct answer is option 'B'. Can you explain this answer?
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
Q. In BMD and SFD :-
a)
Points remain fixed, position of load changes
b)
Points change, position of loads remain fixed
c)
Both of them changes
d)
Neither of them changes
|
|
Sahil Chawla answered |
Answer: b
Explanation: In BMD and SFD, we analyze the structure by fixing loads initially.
Explanation: In BMD and SFD, we analyze the structure by fixing loads initially.
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.Q. For drawing ILD, what value of test load is assumed?- a)1 unit
- b)Arbitrary
- c)Depends upon structure
- d)0
Correct answer is option 'A'. Can you explain this answer?
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
Q. For drawing ILD, what value of test load is assumed?
a)
1 unit
b)
Arbitrary
c)
Depends upon structure
d)
0
|
|
Pallabi Tiwari answered |
Answer: a
Explanation: 1 unit load is assumed as calculations are easy then. Actual loads are multiplies with the results obtained to calculate further.
Explanation: 1 unit load is assumed as calculations are easy then. Actual loads are multiplies with the results obtained to calculate further.
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.
Q. 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
Correct answer is option 'C'. Can you explain this answer?
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.
AC= 1m, CB =3 m
C is just an arbitrary point. A is pin support and B is a roller type support.
Q. 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
|
|
Bhavya Ahuja answered |
Answer: c
Explanation: Just assume force at any point between AC and conserve moment about point A.
Explanation: Just assume force at any point between AC and conserve moment about point A.
The influence line for horizontal thrust of a two- hinged parabolic arch of span i’ and rise 'h' will be shown in- a)
- b)
- c)
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The influence line for horizontal thrust of a two- hinged parabolic arch of span i’ and rise 'h' will be shown in
a)
b)
c)
d)
None of these
|
|
Gitanjali Menon answered |
Figure in (a) is the I.L.D. of horizontal, thrust for three hinged arch
Following figure is for questions 7 and 8.
AB = BC = 1mQ. What will be the equation for vertical reaction at point B (RBY)?- a)RBY = 4X
- b)RBY = 2X
- c)RBY = 3X
- d)RBY = X
Correct answer is option 'D'. Can you explain this answer?
Following figure is for questions 7 and 8.
AB = BC = 1m
AB = BC = 1m
Q. What will be the equation for vertical reaction at point B (RBY)?
a)
RBY = 4X
b)
RBY = 2X
c)
RBY = 3X
d)
RBY = X
|
Anshul Chakraborty answered |
Answer: d
Explanation: Take load at any variable distance X and conserve load about point A.
Explanation: Take load at any variable distance X and conserve load about point A.
What will be the value of Rba?- a)1
- b)0
- c)2
- d)3
Correct answer is option 'B'. Can you explain this answer?
What will be the value of Rba?
a)
1
b)
0
c)
2
d)
3
|
|
Tanishq Nair answered |
Answer: b
Explanation: Δba will be zero as it is roller supported.
Explanation: Δba will be zero as it is roller supported.
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.
Q. 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
Correct answer is option 'A'. Can you explain this answer?
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.
AB= 2m, BC= 3m, CD= 3m
B is pin support, D is roller and C is just an arbitrary point.
Q. 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
|
|
Gauri Sarkar answered |
Answer: a
Explanation: Apply unit load between point A and B and conserve moment about point B.
Explanation: Apply unit load between point A and B and conserve moment about point B.
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.Q. ILD of statically determinate beams consists of curve as well as straight lines.
State whether the above statement is true or false.- a)True
- b)False
Correct answer is option 'B'. Can you explain this answer?
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
Q. ILD of statically determinate beams consists of curve as well as straight lines.
State whether the above statement is true or false.
State whether the above statement is true or false.
a)
True
b)
False
|
|
Anshul Chakraborty answered |
Answer: b
Explanation: ILD of statically determinate beams consist of only straight lines.
Explanation: ILD of statically determinate beams consist of only straight lines.
A and C is a pin support while B is roller support.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.Q. What is the change in slope at point B?- a)AR . XR/LR
- b)AL . XL/IL
- c)0
- d)AR . XR/E . LR
Correct answer is option 'C'. Can you explain this answer?
A and C is a pin support while B is roller support.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
Q. What is the change in slope at point B?
a)
AR . XR/LR
b)
AL . XL/IL
c)
0
d)
AR . XR/E . LR
|
|
Ishani Basu answered |
Answer: c
Explanation: Since point B is roller support, there won’t be any change at of slope at point B.
Explanation: Since point B is roller support, there won’t be any change at of slope at point B.
What will be the slope of ILD at point A?- a)1
- b)-1
- c)0
- d)can’t say
Correct answer is option 'C'. Can you explain this answer?
What will be the slope of ILD at point A?
a)
1
b)
-1
c)
0
d)
can’t say
|
|
Kirti Sharma answered |
Answer: c
Explanation: Since point A is fixed joint, slope will always be zero there.
Explanation: Since point A is fixed joint, slope will always be zero there.
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.Q. How many redundant forces are there in the above structure?- a)0
- b)1
- c)2
- d)3
Correct answer is option 'B'. Can you explain this answer?
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
Q. How many redundant forces are there in the above structure?
a)
0
b)
1
c)
2
d)
3
|
|
Aditi Sarkar answered |
Answer: b
Explanation: 1 redundant force is there as it is equal to DSI.
We have taken the horizontal force at point A as redundant force.
Explanation: 1 redundant force is there as it is equal to DSI.
We have taken the horizontal force at point A as redundant force.
Following figure is for questions 7 and 8.
AB = BC = 1mQ. Where will be the maximum point of ILD lie?- a)A
- b)B
- c)C
- d)Between A and B
Correct answer is option 'C'. Can you explain this answer?
Following figure is for questions 7 and 8.
AB = BC = 1m
AB = BC = 1m
Q. Where will be the maximum point of ILD lie?
a)
A
b)
B
c)
C
d)
Between A and B
|
|
Anirudh Kulkarni answered |
Answer: c
Explanation: On drawing curve based on the above equation, we will find that maximum point will lie at C.
Explanation: On drawing curve based on the above equation, we will find that maximum point will lie at C.
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.Q. 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
Correct answer is option 'C'. Can you explain this answer?
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.
AC= 1m, CB =3 m
C is just an arbitrary point. A is pin support and B is a roller type support.
Q. 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
|
|
Akanksha Mehta answered |
Answer: c
Explanation: Position of ILD at point C is 0.75 (peak). So, shear developed will be 0.75 multiplied by 50KN.
Explanation: Position of ILD at point C is 0.75 (peak). So, shear developed will be 0.75 multiplied by 50KN.
A Warren truss is supported as shown in the given figure. Which one of the following diagrams represents the influence line for the force in member U3U4?
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A Warren truss is supported as shown in the given figure. Which one of the following diagrams represents the influence line for the force in member U3U4?
a)
b)
c)
d)
|
|
Sahil Mehra answered |
Introduce a cut through U3U4, U3L4and L3L4. The member forces of L3L4 and U3L4 pass through support at L4 . So w hen load is at L0, L1 L2, L3 and L4. the total moment of all the force except force in U3U4 on the right part of cut section is zero. So from moment equilibrium the force in U3U4 will remain zero when load is between. L0 and L4.
When load is at L5, the force in member U3U4
When load is at L5, the force in member U3U4
Influence line diagram for a truss member is shown in the above figure. Positive values indicate tension. Dead load of the truss is 20 kN/m and the live load is 10 kN/m. Live load is longer than the span. Maximum tensile force in the member is:- a)600 kN
- b)400 kN
- c)300 kN
- d)200 kN
Correct answer is option 'B'. Can you explain this answer?
Influence line diagram for a truss member is shown in the above figure. Positive values indicate tension. Dead load of the truss is 20 kN/m and the live load is 10 kN/m. Live load is longer than the span. Maximum tensile force in the member is:
a)
600 kN
b)
400 kN
c)
300 kN
d)
200 kN
|
|
Subhankar Khanna answered |
Tensile force due to D.L.
= (20 - 10) x 20 = 200 kN
Maximum Tensile force due to L.L.
= 20 x 10 = 200 kN
Thus maximum tensile force
= 200 + 200 = 400 kN
= (20 - 10) x 20 = 200 kN
Maximum Tensile force due to L.L.
= 20 x 10 = 200 kN
Thus maximum tensile force
= 200 + 200 = 400 kN
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.Q. 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
Correct answer is option 'D'. Can you explain this answer?
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.
AB= 2m, BC= 3m, CD= 3m
B is pin support, D is roller and C is just an arbitrary point.
Q. 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
|
|
Akanksha Mehta answered |
Answer: d
Explanation: Apply unit load at any point at a distance X and conserve moment about point D
Explanation: Apply unit load at any point at a distance X and conserve moment about point D
A and C is a pin support while B is roller support.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.Q. Which of the following is method for solving indeterminate structures?- a)1 moment equation
- b)2 moment equation
- c)3 moment equation
- d)4 moment equation
Correct answer is option 'C'. Can you explain this answer?
A and C is a pin support while B is roller support.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
Q. Which of the following is method for solving indeterminate structures?
a)
1 moment equation
b)
2 moment equation
c)
3 moment equation
d)
4 moment equation
|
|
Kirti Sharma answered |
Answer: c
Explanation: We will be discussing this method in detail subsequently.
Explanation: We will be discussing this method in detail subsequently.
Ray = reaction at support A in y direction.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.Q. Which of the following is equal to Ray?- a)Faa/Fda
- b)–Faa/Fda
- c)Fda/Faa
- d)–Fda/Faa
Correct answer is option 'C'. Can you explain this answer?
Ray = reaction at support A in y direction.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Q. Which of the following is equal to Ray?
a)
Faa/Fda
b)
–Faa/Fda
c)
Fda/Faa
d)
–Fda/Faa
|
|
Pallabi Bajaj answered |
Answer: c
Explanation: Just replacing the value of Fad in the answer of Q1 will give us the result. This relation will be the basic equation for ILD of indeterminate structures.
Explanation: Just replacing the value of Fad in the answer of Q1 will give us the result. This relation will be the basic equation for ILD of indeterminate structures.
What will be the shape of overall ILD?- a)straight line
- b)parabola
- c)hyperbola
- d)arbitrary curve
Correct answer is option 'D'. Can you explain this answer?
What will be the shape of overall ILD?
a)
straight line
b)
parabola
c)
hyperbola
d)
arbitrary curve
|
|
Aditi Chakraborty answered |
Answer: d
Explanation: ILD will be basically(approximately) a line passing through all above points but it will be a type of curve as slope near A will be zero and then slope will change.
Explanation: ILD will be basically(approximately) a line passing through all above points but it will be a type of curve as slope near A will be zero and then slope will change.
For the continuous beam shown in figure, the influence line diagram for support reaction at D is best represented as
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
For the continuous beam shown in figure, the influence line diagram for support reaction at D is best represented as
a)
b)
c)
d)
|
|
Athira Pillai answered |
The ILD for support reaction at D can be obtained by:giving unit displacement in the direction of reaction. The deflected shape of beam will represent ILD as in figure (c).
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.Q. What will be the value of Δaa?- a)544/EI
- b)944/EI
- c)1544/EI
- d)1944/EI
Correct answer is option 'D'. Can you explain this answer?
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
Q. What will be the value of Δaa?
a)
544/EI
b)
944/EI
c)
1544/EI
d)
1944/EI
|
|
Aditi Chakraborty answered |
Answer: d
Explanation: Applying a redundant load of unity on point A and applying conjugate beam method will give the answer easily.
Explanation: Applying a redundant load of unity on point A and applying conjugate beam method will give the answer easily.
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.Q. 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
Correct answer is option 'C'. Can you explain this answer?
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.
AC= 1m, CB =3 m
C is just an arbitrary point. A is pin support and B is a roller type support.
Q. 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
|
|
Lakshmi Datta answered |
Answer: c
Explanation: If we draw ILD according to the above given equations, we will see that peak of ILD comes at point C
Explanation: If we draw ILD according to the above given equations, we will see that peak of ILD comes at point C
A simply supported beam is shown in the figure-I. The influence line for the beam is drawn in figure-II, Which one of the following statements is correctly associated with the figure-II?
- a)influence line for shear force at B
- b)Influence line for bending moment at B
- c)Influence line for vertical reaction at A
- d)Influence line for vertical reaction at B
Correct answer is option 'C'. Can you explain this answer?
A simply supported beam is shown in the figure-I. The influence line for the beam is drawn in figure-II, Which one of the following statements is correctly associated with the figure-II?
a)
influence line for shear force at B
b)
Influence line for bending moment at B
c)
Influence line for vertical reaction at A
d)
Influence line for vertical reaction at B
|
|
Madhurima Banerjee answered |
Since the unit value of I.L.D. is 1,0 at A. So this is for reaction at support A
The Muller-Breslau principle can be used to
1. Determine the shape of the influence line
2. Indicate the parts of the structure to be loaded to obtain the maximum effect
3. Calculate the ordinates of the influence linesWhich of the these statements is/are correct? - a)only 1
- b)both 1 and 2
- c)both 2 and 3
- d)1, 2 and 3
Correct answer is option 'D'. Can you explain this answer?
The Muller-Breslau principle can be used to
1. Determine the shape of the influence line
2. Indicate the parts of the structure to be loaded to obtain the maximum effect
3. Calculate the ordinates of the influence lines
1. Determine the shape of the influence line
2. Indicate the parts of the structure to be loaded to obtain the maximum effect
3. Calculate the ordinates of the influence lines
Which of the these statements is/are correct?
a)
only 1
b)
both 1 and 2
c)
both 2 and 3
d)
1, 2 and 3
|
Tarun Shah answered |
The Muller-Breslau principle is a method used for calculating influence lines for determining the effects of moving loads on a structure. It is based on the principle that the displacement of a structure due to a unit load at any point is proportional to the area of the influence line.
The principle can be used to:
1. Determine the shape of the influence line: The shape of the influence line can be determined by considering the structure as a continuous beam and calculating the deflection at any point due to a unit load moving along the beam. The influence line is then plotted by taking the ratio of the deflection to the length of the beam.
2. Indicate the parts of the structure to be loaded to obtain the maximum effect: The Muller-Breslau principle can also be used to determine the position of the load on the structure that will produce the maximum effect. This is done by finding the point along the influence line where the product of the load and the area under the influence line is maximum.
3. Calculate the ordinates of the influence lines: The Muller-Breslau principle can also be used to calculate the ordinates of the influence lines at any point along the beam. This is done by dividing the area under the influence line up to that point by the length of the beam.
Therefore, all three statements are correct and the correct answer is option 'D'.
The principle can be used to:
1. Determine the shape of the influence line: The shape of the influence line can be determined by considering the structure as a continuous beam and calculating the deflection at any point due to a unit load moving along the beam. The influence line is then plotted by taking the ratio of the deflection to the length of the beam.
2. Indicate the parts of the structure to be loaded to obtain the maximum effect: The Muller-Breslau principle can also be used to determine the position of the load on the structure that will produce the maximum effect. This is done by finding the point along the influence line where the product of the load and the area under the influence line is maximum.
3. Calculate the ordinates of the influence lines: The Muller-Breslau principle can also be used to calculate the ordinates of the influence lines at any point along the beam. This is done by dividing the area under the influence line up to that point by the length of the beam.
Therefore, all three statements are correct and the correct answer is option 'D'.
A and C is a pin support while B is roller support.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.Q. 3 moment method is valid in:-- a)single span beams
- b)only double span beams
- c)double and single span beams
- d)only multiple span beams
Correct answer is option 'D'. Can you explain this answer?
A and C is a pin support while B is roller support.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
XL and XR are centroid of uniform loading on beams AB and BC respectively.
IL and IR are given.
LR and LL are length of beams.
AR and AL are areas of beams.
Q. 3 moment method is valid in:-
a)
single span beams
b)
only double span beams
c)
double and single span beams
d)
only multiple span beams
|
|
Sreemoyee Chauhan answered |
Answer: d
Explanation: This method is valid in only multiple span beam i.e. including double span.
Explanation: This method is valid in only multiple span beam i.e. including double span.
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.Q. 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
Correct answer is option 'B'. Can you explain this answer?
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.
AB= 2m, BC= 3m, CD= 3m
B is pin support, D is roller and C is just an arbitrary point.
Q. 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
|
|
Pallabi Tiwari answered |
Answer: b
Explanation: Apply load at any point and conserve moment about point B.
Explanation: Apply load at any point and conserve moment about point B.
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.Q. What will be the value of Ray for ILD?- a)0
- b)1
- c)2
- d)-1
Correct answer is option 'B'. Can you explain this answer?
A is a fixed support and B is roller.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
AB = 18m.
C is a point at a distance of 6m from point A.
D is a point at a distance of 6m from point B.
Assume E and I to be given.
Q. What will be the value of Ray for ILD?
a)
0
b)
1
c)
2
d)
-1
|
|
Sankar Dasgupta answered |
Answer: b
Explanation: Since redundant force is applied at point A, Fda/Faa will be the result unity.
Explanation: Since redundant force is applied at point A, Fda/Faa will be the result unity.
Following figure is for questions 7 and 8.
AB = BC = 1mQ. Maximum point of ILD always lies at the point at which load is applied.
State whether the above statement is true or false.- a)True
- b)False
Correct answer is option 'B'. Can you explain this answer?
Following figure is for questions 7 and 8.
AB = BC = 1m
AB = BC = 1m
Q. Maximum point of ILD always lies at the point at which load is applied.
State whether the above statement is true or false.
State whether the above statement is true or false.
a)
True
b)
False
|
|
Anshul Chakraborty answered |
Answer: b
Explanation: In the above example above sentence is proved wrong.
Explanation: In the above example above sentence is proved wrong.
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.Q. 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
Correct answer is option 'A'. Can you explain this answer?
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.
AC= 1m, CB =3 m
C is just an arbitrary point. A is pin support and B is a roller type support.
Q. 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
|
|
Lakshmi Datta answered |
Answer: a
Explanation: Net shear generated will be the sum of individually generated shear which has been already calculated earlier.
Explanation: Net shear generated will be the sum of individually generated shear which has been already calculated earlier.
Which one of the following equations represents influence line of fixed end moment at B of the fixed beam AB of length l with origin at A?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which one of the following equations represents influence line of fixed end moment at B of the fixed beam AB of length l with origin at A?
a)
b)
c)
d)
|
|
Partho Choudhary answered |
For the influence line of fixed end moment at B, release the moment at B and give unit rotation, the deformed shape will represent the influence line.
due to unit load at x distance from A.
due to unit load at x distance from A.
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
Q. 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
Correct answer is option 'D'. Can you explain this answer?
While writing influence line equations, left most point is always considered as origin and following sign convention is followed.
Q. 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
|
|
Lakshmi Datta answered |
Answer: d
Explanation: Peak force will be load multiplied by earlier peak load i.e. 2*2.
Explanation: Peak force will be load multiplied by earlier peak load i.e. 2*2.
Select the correct influence line diagram for shear force at x of the following beam
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Select the correct influence line diagram for shear force at x of the following beam
a)
b)
c)
d)
|
|
Anisha Chakraborty answered |
Maximum shear force is at either of the support due to a point load
A uniformly distributed load whose length is shorter than the span, crosses a simply supported beam from left to right. The maximum bending moment at a section occurs when- a)the load remains only on the left of the section
- b)the load is only on the right of the section
- c)the section divides the load in the same ratio as it divides the span
- d)half of the load remains to the left of the section and other half on the right of section
Correct answer is option 'C'. Can you explain this answer?
A uniformly distributed load whose length is shorter than the span, crosses a simply supported beam from left to right. The maximum bending moment at a section occurs when
a)
the load remains only on the left of the section
b)
the load is only on the right of the section
c)
the section divides the load in the same ratio as it divides the span
d)
half of the load remains to the left of the section and other half on the right of section
|
Srestha Tiwari answered |
Explanation:
When a uniformly distributed load crosses a simply supported beam from left to right, the maximum bending moment at a section occurs when the section divides the load in the same ratio as it divides the span. Let's understand this concept in detail.
Uniformly Distributed Load:
A uniformly distributed load is a load that is spread evenly over a given length or area. In the case of a simply supported beam, a uniformly distributed load is a load that is distributed evenly over the entire length of the beam.
Maximum Bending Moment:
Bending moment is the moment that is developed in a beam due to the external loads applied to it. The maximum bending moment in a simply supported beam occurs at the mid-span of the beam.
Section Divides the Load:
When a section is taken in a simply supported beam, it divides the load into two parts: the load on the left side of the section and the load on the right side of the section. The ratio of these two loads is determined by the distance of the section from the left support and the distance of the section from the right support.
When the section is at the mid-span of the beam, the load on the left side of the section is equal to the load on the right side of the section. Therefore, the section divides the load in the same ratio as it divides the span.
The maximum bending moment in a simply supported beam occurs when the load on one side of the section is greater than the load on the other side of the section. This is because the bending moment is directly proportional to the load and the distance from the load to the section.
Conclusion:
In conclusion, the maximum bending moment at a section in a simply supported beam occurs when the section divides the load in the same ratio as it divides the span. Therefore, option 'C' is the correct answer.
When a uniformly distributed load crosses a simply supported beam from left to right, the maximum bending moment at a section occurs when the section divides the load in the same ratio as it divides the span. Let's understand this concept in detail.
Uniformly Distributed Load:
A uniformly distributed load is a load that is spread evenly over a given length or area. In the case of a simply supported beam, a uniformly distributed load is a load that is distributed evenly over the entire length of the beam.
Maximum Bending Moment:
Bending moment is the moment that is developed in a beam due to the external loads applied to it. The maximum bending moment in a simply supported beam occurs at the mid-span of the beam.
Section Divides the Load:
When a section is taken in a simply supported beam, it divides the load into two parts: the load on the left side of the section and the load on the right side of the section. The ratio of these two loads is determined by the distance of the section from the left support and the distance of the section from the right support.
When the section is at the mid-span of the beam, the load on the left side of the section is equal to the load on the right side of the section. Therefore, the section divides the load in the same ratio as it divides the span.
The maximum bending moment in a simply supported beam occurs when the load on one side of the section is greater than the load on the other side of the section. This is because the bending moment is directly proportional to the load and the distance from the load to the section.
Conclusion:
In conclusion, the maximum bending moment at a section in a simply supported beam occurs when the section divides the load in the same ratio as it divides the span. Therefore, option 'C' is the correct answer.
Ray = reaction at support A in y direction.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.Q. What will be the difference between magnitudes of Fad and Fda?- a)Faa
- b)0
- c)Fad
- d)Fda
Correct answer is option 'B'. Can you explain this answer?
Ray = reaction at support A in y direction.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Q. What will be the difference between magnitudes of Fad and Fda?
a)
Faa
b)
0
c)
Fad
d)
Fda
|
|
Sreemoyee Chauhan answered |
Answer: b
Explanation: Due to validity of Maxwell’s reciprocal theorem, magnitudes of Fad and Fda will be equal.
Explanation: Due to validity of Maxwell’s reciprocal theorem, magnitudes of Fad and Fda will be equal.
Ray = reaction at support A in y direction.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.Q. What is the relation between Fad and Fda?- a)Fad = -Fda
- b)Fad = Fda
- c)Fad = -2Fda
- d)Fad = 2Fda
Correct answer is option 'A'. Can you explain this answer?
Ray = reaction at support A in y direction.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Sign convention is as usual.
Beam is linear elastic.
Point A is a pin joint and point B and C are roller support. D is just an arbitrary point.
Unit load is applied at point D.
Q. What is the relation between Fad and Fda?
a)
Fad = -Fda
b)
Fad = Fda
c)
Fad = -2Fda
d)
Fad = 2Fda
|
|
Aditi Sarkar answered |
Answer: a
Explanation: Magnitudes will be same as stated above. Now, sign will be opposite as slope at B won’t change in deflection diagram and there won’t be and vertical deflection at point C.
Explanation: Magnitudes will be same as stated above. Now, sign will be opposite as slope at B won’t change in deflection diagram and there won’t be and vertical deflection at point C.
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