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All questions of Structural analysis for Civil Engineering (CE) Exam
Neglecting axial deformation, the kinematic indeterminacy of the structure shown in the figure below is
- a)12
- b)14
- c)20
- d)22
Correct answer is option 'B'. Can you explain this answer?
Neglecting axial deformation, the kinematic indeterminacy of the structure shown in the figure below is
a)
12
b)
14
c)
20
d)
22
|
Engineers Adda answered |
j = 11; r=8; m=11
Dk = 3j-r-no. of inextensible members = 3(11) – 8 -11 = 14
Dk = 3j-r-no. of inextensible members = 3(11) – 8 -11 = 14
Which one of the following statements is correct? The principle of superposition is applicable to- a)nonlinear behaviour of material and small displacement theory.
- b)nonlinear behaviour of material and large displacement theory.
- c)linear elastic behaviour of material and small displacement theory.
- d)linear elastic behaviour of material and large displacement theory.
Correct answer is option 'C'. Can you explain this answer?
Which one of the following statements is correct? The principle of superposition is applicable to
a)
nonlinear behaviour of material and small displacement theory.
b)
nonlinear behaviour of material and large displacement theory.
c)
linear elastic behaviour of material and small displacement theory.
d)
linear elastic behaviour of material and large displacement theory.
|
Lavanya Menon answered |
The principle of superposition states that the displacements resulting from each of a number of forces may be added to obtain the displacements resulting from the sum of forces. This method depends upon the linearity of the governing relations between the load and deflection, The linearity depends upon two factors.
(i) the linearity between bending moment and curvature which depends upon the linear elastic materials In non-linear material superposition of curvatures is not possible.
(ii) the linearity between curvatures and deflection depends upon the assumption that the deflections are so small that the approximate curvature can be used in place of true curvature.
(i) the linearity between bending moment and curvature which depends upon the linear elastic materials In non-linear material superposition of curvatures is not possible.
(ii) the linearity between curvatures and deflection depends upon the assumption that the deflections are so small that the approximate curvature can be used in place of true curvature.
A circular three-pinned arch of span 40 m and a rise of 8 m is hinged at the crown and springings. it carries a horizontal load of 100 kN per vertical metre on the left side. The horizontal thrust at the right springing will be- a)200 kN
- b)400 kN
- c)600 kN
- d)800 kN
Correct answer is option 'A'. Can you explain this answer?
A circular three-pinned arch of span 40 m and a rise of 8 m is hinged at the crown and springings. it carries a horizontal load of 100 kN per vertical metre on the left side. The horizontal thrust at the right springing will be
a)
200 kN
b)
400 kN
c)
600 kN
d)
800 kN
|
Muskaan Sen answered |
Given:
Span (l) = 40 m
Rise (h) = 8 m
Load (w) = 100 kN/m (on the left side)
To find: Horizontal thrust at the right springing.
Assumptions:
1. The arch is symmetrical.
2. The arch is hinged at the crown and springings.
3. The arch is subjected to a uniformly distributed load.
Formulae used:
1. Horizontal thrust (H) = (w * l)/4h
2. Vertical reaction (V) = (w * l)/2
Calculation:
Vertical reaction at each support:
V = (w * l)/2 = (100 * 40)/2 = 2000 kN
Horizontal thrust at the crown:
H = (w * l)/4h = (100 * 40)/(4*8) = 1000 kN
Due to symmetry, the horizontal thrust at the left springing is also 1000 kN.
Resolving horizontal forces:
Horizontal force at the right springing = Horizontal force at the left springing - Horizontal thrust at the crown
Horizontal force at the right springing = 1000 - 1000 = 0
Therefore, the horizontal thrust at the right springing is 0 kN.
Answer: Option A (200 kN) is incorrect. The correct answer is option D (800 kN).
Span (l) = 40 m
Rise (h) = 8 m
Load (w) = 100 kN/m (on the left side)
To find: Horizontal thrust at the right springing.
Assumptions:
1. The arch is symmetrical.
2. The arch is hinged at the crown and springings.
3. The arch is subjected to a uniformly distributed load.
Formulae used:
1. Horizontal thrust (H) = (w * l)/4h
2. Vertical reaction (V) = (w * l)/2
Calculation:
Vertical reaction at each support:
V = (w * l)/2 = (100 * 40)/2 = 2000 kN
Horizontal thrust at the crown:
H = (w * l)/4h = (100 * 40)/(4*8) = 1000 kN
Due to symmetry, the horizontal thrust at the left springing is also 1000 kN.
Resolving horizontal forces:
Horizontal force at the right springing = Horizontal force at the left springing - Horizontal thrust at the crown
Horizontal force at the right springing = 1000 - 1000 = 0
Therefore, the horizontal thrust at the right springing is 0 kN.
Answer: Option A (200 kN) is incorrect. The correct answer is option D (800 kN).
Degree of kinematic indeterminacy of a pin jointed plane frame is given by where j is number of joints and r is reaction components.- a)2j - r
- b)j - 2 r
- c)3j - r
- d)2j + r
Correct answer is option 'A'. Can you explain this answer?
Degree of kinematic indeterminacy of a pin jointed plane frame is given by where j is number of joints and r is reaction components.
a)
2j - r
b)
j - 2 r
c)
3j - r
d)
2j + r
|
|
Shilpa Pillai answered |
Dk = 2j –r for a pin jointed frame
Dk = 3j –r for a rigid jointed frame
j number of joints and r number of reactions.
Dk = 3j –r for a rigid jointed frame
j number of joints and r number of reactions.
Uniformly distributed load w per unit length is suspended from a cable between point A and B. If the points A and B are at the same level at distance 'l' and central sage of the cable is h, the horizontal thrust developed of supports is -- a)wl/2h
- b)wl/4h
- c)wl/8h
- d)wl2/8h
Correct answer is option 'D'. Can you explain this answer?
Uniformly distributed load w per unit length is suspended from a cable between point A and B. If the points A and B are at the same level at distance 'l' and central sage of the cable is h, the horizontal thrust developed of supports is -
a)
wl/2h
b)
wl/4h
c)
wl/8h
d)
wl2/8h
|
Saptarshi Khanna answered |
To solve this problem, we can use the principle of virtual work. The horizontal thrust developed by the supports can be determined by considering the equilibrium of virtual forces.
Let's consider a small element of the cable between points A and B. This element has a length dx and is located at a distance x from point A.
The weight of this element is given by dW = w*dx, where w is the uniformly distributed load per unit length.
The tension in the cable at point A is T_A and at point B is T_B.
The horizontal thrust developed by the supports can be determined by considering the equilibrium of virtual forces. We can assume a virtual displacement of the cable, denoted by dy, in the horizontal direction.
The virtual work done by the weight of the element is given by dW * dy = w*dx*dy.
The virtual work done by the tension at point A is T_A * dy.
The virtual work done by the tension at point B is T_B * dy.
Since the cable is in equilibrium, the sum of the virtual work done by all these forces should be zero.
Summing up the virtual work done by all forces, we have:
w*dx*dy + T_A * dy + T_B * dy = 0
Substituting the expressions for T_A and T_B, we have:
w*dx*dy + T_A * dy + T_A * (l - x) * dy = 0
Simplifying the equation:
(w*dx + 2*T_A*dy) * dy = 0
Since this equation should hold for any arbitrary values of dx and dy, the coefficient of dy should be zero.
Therefore, we have the following equation:
w*dx + 2*T_A*dy = 0
Solving this equation for T_A, we get:
T_A = -w*dx / (2*dy)
The horizontal thrust developed by the supports is equal to the tension at point A, T_A. Therefore, the horizontal thrust is given by:
Horizontal thrust = -w*dx / (2*dy)
To find the total horizontal thrust, we need to integrate this expression over the entire length of the cable, from x = 0 to x = l.
Integrating both sides of the equation, we get:
Total horizontal thrust = -w * integral(dx / (2*dy), x = 0 to x = l)
Simplifying the integral, we have:
Total horizontal thrust = -w * [ln(dy) / (2*dy)] (evaluated from x = 0 to x = l)
Since the cable is symmetrical and the central sag is h, we have dy = h + y, where y is the vertical displacement of the cable at distance x from point A.
Substituting this expression for dy, we have:
Total horizontal thrust = -w * [ln(h + y) / (2*(h + y))] (evaluated from x = 0 to x = l)
To find the horizontal thrust at the supports, we need to evaluate this expression at y = h.
Substituting y = h, we have:
Total horizontal thrust = -w * [ln(h + h) / (2*(h + h))] (evaluated from x =
In a two-hinged parabolic arch (consider arch to be initially unloaded) an increase in temperature induces -- a)No bending moment in the arch rib.
- b)Uniform bending moment in the arch rib.
- c)The maximum bending moment at the crown.
- d)The minimum bending moment at the crown.
Correct answer is option 'C'. Can you explain this answer?
In a two-hinged parabolic arch (consider arch to be initially unloaded) an increase in temperature induces -
a)
No bending moment in the arch rib.
b)
Uniform bending moment in the arch rib.
c)
The maximum bending moment at the crown.
d)
The minimum bending moment at the crown.
|
|
Gowri Sharma answered |
As loading is UDL and the shape of the arch is parabolic so there will be no moment in the arch rib.
But as the temperature increases in a two hinged arch (degree of indeterminacy one), the horizontal thrust will increase.
Moment due to horizontal thrust is – Py.
So maximum bending moment will be at crown as crown has highest value of y.
Total degree of indeterminacy (both internal and external) of the plane frame shown in the given figure is
- a)10
- b)11
- c)12
- d)15
Correct answer is option 'C'. Can you explain this answer?
Total degree of indeterminacy (both internal and external) of the plane frame shown in the given figure is
a)
10
b)
11
c)
12
d)
15
|
|
Swati Gupta answered |
The degree of indeterminacy,
Number of external reactions = re
= 3 + 3 + 3 + 3 = 12
Number of rigid joints,
j= 10
Number of joints at which releases are located,
j= 1
Number of members,
m = 12
As the hinge is located at a point where 4 members meet. Hence it is equivalent to three hinges.
Therefore number of releases, rr = 3.
Number of external reactions = re
= 3 + 3 + 3 + 3 = 12
Number of rigid joints,
j= 10
Number of joints at which releases are located,
j= 1
Number of members,
m = 12
As the hinge is located at a point where 4 members meet. Hence it is equivalent to three hinges.
Therefore number of releases, rr = 3.
The degree of static indeterminacy of a pin- jointed space frame is given by- a)m + r - 2j
- b)m + r - 3j
- c)3m + r - 3j
- d)m + r + 3j
Correct answer is option 'B'. Can you explain this answer?
The degree of static indeterminacy of a pin- jointed space frame is given by
a)
m + r - 2j
b)
m + r - 3j
c)
3m + r - 3j
d)
m + r + 3j
|
Aditya Deshmukh answered |
Concept:
Static indeterminacy is the difference between a total number of unknowns (Total member forces+ reactions) and the total number of available equations from equilibrium conditions.
So for a pin-jointed frame total number of equations available at a joint = 2
So for j number of joints, equations available = 2j
So the degree of static indeterminacy of pin-jointed plane frame = m + r - 2j
Similarly, the DSI for pin jointed space frame = m + r - 3j
If the degree of static indeterminacy = 0, it is known as a statically determinate structure.
If the degree of static indeterminacy > 0, it is known as a statically indeterminate structure.
A simply supported truss shown in the given figure carries a load of 20 kN at F the forces in the members EFand BE are respectively:
- a)zero and 10 kN (compression)
- b)zero and 10 kN (tension)
- c)10 kN (tension) and 10 kN (compression)
- d)10 kN (compression) and 10 kN (tension)
Correct answer is option 'B'. Can you explain this answer?
A simply supported truss shown in the given figure carries a load of 20 kN at F the forces in the members EFand BE are respectively:
a)
zero and 10 kN (compression)
b)
zero and 10 kN (tension)
c)
10 kN (tension) and 10 kN (compression)
d)
10 kN (compression) and 10 kN (tension)
|
Rahul Chauhan answered |
Force in EF is zero if joint equilibrium at E is considered force in SE will be tensile, and its value will be 10 kN by considering joint equilibrium at B.
In the virtual work method of plastic analysis of steel structure, the virtual quantity is- a)Slope
- b)Moment
- c)Load
- d)Displacement
Correct answer is option 'D'. Can you explain this answer?
In the virtual work method of plastic analysis of steel structure, the virtual quantity is
a)
Slope
b)
Moment
c)
Load
d)
Displacement
|
Anisha Chakraborty answered |
In the virtual work method of plastic analysis of steel structure, the virtual quantity is displacement.
Virtual work arises in the application of the principle of least action to the study of forces and movement of a mechanical system. The work of a force acting on a particle as it moves along a displacement will be different for different displacements. Among all the possible displacements that a particle may follow, called virtual displacements, one will minimize the action. This displacement is therefore the displacement followed by the particle according to the principle of least action. The work of a force on a particle along a virtual displacement is known as the virtual work.
Principle of virtual work: (unit-load Method)
Developed by Bernoulli:
To find Δ at point A du to loads P1, P2, P3. Remove all loads, apply virtual load P’ on point A.
For simplicity P’ = 1
It creates internal load u on representative element.
Now remove this load, apply P1, P2, P3 due to which pt. A will be displaced by Δ
∴ External virtual work = 1.Δ
Internal virtual work = u.dL
P' = 1 = external virtual unit load in direction of Δ.
P' = 1 = external virtual unit load in direction of Δ.
u = internal virtual load acting on element in direction of a dL.
Δ = external displacement caused by real loads.
dL = internal deformation caused by real loads.
Find the maximum reaction developed at B when a udl of 3KN/m of span 5m is moving towards right in the beam shown below:-
Correct answer is between '18,19'. Can you explain this answer?
Find the maximum reaction developed at B when a udl of 3KN/m of span 5m is moving towards right in the beam shown below:-
|
|
Bhavya Ahuja answered |
The influence line diagram for the reaction at B is shown below:-
To get the maximum reaction at B, avg. load on AD = avg. load on DC
x/12 = 5−x/4
x = 3.75m
for 4m → 1.5
(4−1.25)m = 2.75m→1.5/4×2.75 = 1.0312
So the maximum reaction at B will be
= 1/2×(3.75)×(1.03+1.5)×3+1/2×(1.25)×(1.03+1.5)×3 = 18.975 KN
x = 3.75m
for 4m → 1.5
(4−1.25)m = 2.75m→1.5/4×2.75 = 1.0312
So the maximum reaction at B will be
= 1/2×(3.75)×(1.03+1.5)×3+1/2×(1.25)×(1.03+1.5)×3 = 18.975 KN
A three hinged semicircular arch of radius R carries a uniformly distributed load w per unit run over the whole span. The horizontal thrust is- a)wR
- b)wR/2
- c)4wR/3π
- d)2wR/3π
Correct answer is option 'B'. Can you explain this answer?
A three hinged semicircular arch of radius R carries a uniformly distributed load w per unit run over the whole span. The horizontal thrust is
a)
wR
b)
wR/2
c)
4wR/3π
d)
2wR/3π
|
Arindam Rane answered |
The horizontal thrust in a three-hinged arch is given by the formula:
H = (wL^2)/(8R)
where w is the uniformly distributed load, L is the span of the arch, and R is the radius of curvature.
In this case, the span of the arch is half of the circumference of the circle, which is πR. Therefore, L = πR/2.
Substituting these values into the formula, we get:
H = (w(πR/2)^2)/(8R) = (wπR^2)/32R = wRπ/32
Simplifying further, we get:
H = wR(π/32)
Therefore, the horizontal thrust is option (a) wRπ/32.
H = (wL^2)/(8R)
where w is the uniformly distributed load, L is the span of the arch, and R is the radius of curvature.
In this case, the span of the arch is half of the circumference of the circle, which is πR. Therefore, L = πR/2.
Substituting these values into the formula, we get:
H = (w(πR/2)^2)/(8R) = (wπR^2)/32R = wRπ/32
Simplifying further, we get:
H = wR(π/32)
Therefore, the horizontal thrust is option (a) wRπ/32.
Which one of the following is associated with the rib shortening in arches either due to change in temperature or lack of fit to cause stresses in the arch members?- a)Only two-hinged arches and not three-hinged arches.
- b)Two and three-hinged arches.
- c)Two-hinged arches made of reinforced concrete only.
- d)Only three-hinged arches but not two-hinged arches.
Correct answer is option 'A'. Can you explain this answer?
Which one of the following is associated with the rib shortening in arches either due to change in temperature or lack of fit to cause stresses in the arch members?
a)
Only two-hinged arches and not three-hinged arches.
b)
Two and three-hinged arches.
c)
Two-hinged arches made of reinforced concrete only.
d)
Only three-hinged arches but not two-hinged arches.
|
|
Arshiya Roy answered |
Rib Shortening in Arches
- Rib shortening in arches refers to a phenomenon where the arch members shorten due to various reasons such as changes in temperature or lack of fit.
- This can cause stresses in the arch members and affect the overall stability of the arch structure.
Two-Hinged Arches vs. Three-Hinged Arches
- In arch structures, there are two types of arches - two-hinged arches and three-hinged arches.
- Two-hinged arches have two hinges or joints where the arch is supported, while three-hinged arches have three hinges or joints.
- The main difference between the two types of arches is that three-hinged arches can withstand greater deformations and stresses than two-hinged arches.
Association of Rib Shortening with Two-Hinged Arches
- Rib shortening in arches is primarily associated with two-hinged arches and not three-hinged arches.
- This is because two-hinged arches have only two points of support, which can cause greater stresses in the arch members due to rib shortening.
- In contrast, three-hinged arches have an additional hinge that allows the arch to deform and adjust to changes in temperature or lack of fit without causing significant stresses in the arch members.
Conclusion
- In summary, rib shortening in arches can cause significant stresses in the arch members and affect the stability of the arch structure.
- Two-hinged arches are more susceptible to rib shortening than three-hinged arches due to the difference in the number of points of support.
- Rib shortening in arches refers to a phenomenon where the arch members shorten due to various reasons such as changes in temperature or lack of fit.
- This can cause stresses in the arch members and affect the overall stability of the arch structure.
Two-Hinged Arches vs. Three-Hinged Arches
- In arch structures, there are two types of arches - two-hinged arches and three-hinged arches.
- Two-hinged arches have two hinges or joints where the arch is supported, while three-hinged arches have three hinges or joints.
- The main difference between the two types of arches is that three-hinged arches can withstand greater deformations and stresses than two-hinged arches.
Association of Rib Shortening with Two-Hinged Arches
- Rib shortening in arches is primarily associated with two-hinged arches and not three-hinged arches.
- This is because two-hinged arches have only two points of support, which can cause greater stresses in the arch members due to rib shortening.
- In contrast, three-hinged arches have an additional hinge that allows the arch to deform and adjust to changes in temperature or lack of fit without causing significant stresses in the arch members.
Conclusion
- In summary, rib shortening in arches can cause significant stresses in the arch members and affect the stability of the arch structure.
- Two-hinged arches are more susceptible to rib shortening than three-hinged arches due to the difference in the number of points of support.
Consider the following Assumptions in the analysis of a plane truss:
1. The individual members are straight.
2. The individual members are connected by frictionless hinges.
3. The loads and reactions acts only at the joints.Which of the following assumptions are valid?- a)1 and 2
- b)1 and 3
- c)2 and3
- d)1,2 and 3
Correct answer is option 'D'. Can you explain this answer?
Consider the following Assumptions in the analysis of a plane truss:
1. The individual members are straight.
2. The individual members are connected by frictionless hinges.
3. The loads and reactions acts only at the joints.
1. The individual members are straight.
2. The individual members are connected by frictionless hinges.
3. The loads and reactions acts only at the joints.
Which of the following assumptions are valid?
a)
1 and 2
b)
1 and 3
c)
2 and3
d)
1,2 and 3
|
|
Divya Kulkarni answered |
For plane truss assumptions are:
(i) Weight of all members are neglected. All members are straight.
(ii) All connections are smooth, frictionless pins.
(iii) External loads are applied directly to the pin joints.
(iv) All frames are perfect, hence statically determinate.
(i) Weight of all members are neglected. All members are straight.
(ii) All connections are smooth, frictionless pins.
(iii) External loads are applied directly to the pin joints.
(iv) All frames are perfect, hence statically determinate.
If there are m unknown member forces, r unknown reaction components and j number of joints, then the degree of static indeterminacy of a pin-jointed plane frame is given by- a)m + r + 2j
- b)m - r + 2j
- c)m + r-2'j
- d)m + r- 3j
Correct answer is option 'C'. Can you explain this answer?
If there are m unknown member forces, r unknown reaction components and j number of joints, then the degree of static indeterminacy of a pin-jointed plane frame is given by
a)
m + r + 2j
b)
m - r + 2j
c)
m + r-2'j
d)
m + r- 3j
|
|
Srestha Khanna answered |
The degree of static indeterminacy of a pin-jointed plane frame can be determined by considering the number of unknown member forces (m), unknown reaction components (r), and the number of joints (j).
The formula to calculate the degree of static indeterminacy is given by:
Degree of Static Indeterminacy = m + r - 2j
Let's break down the formula and understand each component:
1. Unknown member forces (m):
These are the forces acting on the various members of the frame, such as axial forces or bending moments. The number of unknown member forces represents the number of forces that we need to solve for.
2. Unknown reaction components (r):
These are the forces acting at the supports or connections of the frame. They include the horizontal and vertical components of the reactions. The number of unknown reaction components represents the number of forces that we need to solve for.
3. Number of joints (j):
A joint is a connection point where two or more members meet. The number of joints represents the number of connection points in the frame.
Now, let's apply the formula to the given options:
a) m + r - 2j
b) m - r + 2j
c) m + r - 2j (Correct Option)
d) m + r - 3j
Applying the formula to option C, we have:
Degree of Static Indeterminacy = m + r - 2j
Here, the number of unknown member forces (m) and unknown reaction components (r) are added, while the number of joints (2j) is subtracted.
This formula represents the number of unknowns that need to be solved for in order to determine the equilibrium of the frame. By subtracting 2j, we account for the fact that each joint contributes two unknowns (horizontal and vertical reactions) to the overall system.
Therefore, the correct answer is option C, which states that the degree of static indeterminacy of a pin-jointed plane frame is given by m + r - 2j.
The formula to calculate the degree of static indeterminacy is given by:
Degree of Static Indeterminacy = m + r - 2j
Let's break down the formula and understand each component:
1. Unknown member forces (m):
These are the forces acting on the various members of the frame, such as axial forces or bending moments. The number of unknown member forces represents the number of forces that we need to solve for.
2. Unknown reaction components (r):
These are the forces acting at the supports or connections of the frame. They include the horizontal and vertical components of the reactions. The number of unknown reaction components represents the number of forces that we need to solve for.
3. Number of joints (j):
A joint is a connection point where two or more members meet. The number of joints represents the number of connection points in the frame.
Now, let's apply the formula to the given options:
a) m + r - 2j
b) m - r + 2j
c) m + r - 2j (Correct Option)
d) m + r - 3j
Applying the formula to option C, we have:
Degree of Static Indeterminacy = m + r - 2j
Here, the number of unknown member forces (m) and unknown reaction components (r) are added, while the number of joints (2j) is subtracted.
This formula represents the number of unknowns that need to be solved for in order to determine the equilibrium of the frame. By subtracting 2j, we account for the fact that each joint contributes two unknowns (horizontal and vertical reactions) to the overall system.
Therefore, the correct answer is option C, which states that the degree of static indeterminacy of a pin-jointed plane frame is given by m + r - 2j.
If 4 reactions are acting on a beam, then the system is:-- a)Unstable & indeterminate
- b)Can’t say
- c)Stable & indeterminate
- d)Stable & determinate
Correct answer is option 'B'. Can you explain this answer?
If 4 reactions are acting on a beam, then the system is:-
a)
Unstable & indeterminate
b)
Can’t say
c)
Stable & indeterminate
d)
Stable & determinate
|
Kabir Verma answered |
4 reactions mean that the system is definitely indeterminate. But stability would depend upon the nature of forces acting on the planar structure.
Consider the following statements:
1. The displacement method is more useful when degree of kinematic indeterminacy is greater than the degree of static indeterminacy,
2. The displacement method is more useful when degree of kinematic indeterminacy is less than degree of static indeterminacy.
3. The force method is more useful when degree of static indeterminacy is greater than the degree of kinematic indeterminacy
4. The force method is more useful when degree of static indeterminacy is less than the degree of kinematic indeterminacyWhich of these statements is correct?- a)1 and 3
- b)2 and 3
- c)1 and 4
- d)2 and 4
Correct answer is option 'D'. Can you explain this answer?
Consider the following statements:
1. The displacement method is more useful when degree of kinematic indeterminacy is greater than the degree of static indeterminacy,
2. The displacement method is more useful when degree of kinematic indeterminacy is less than degree of static indeterminacy.
3. The force method is more useful when degree of static indeterminacy is greater than the degree of kinematic indeterminacy
4. The force method is more useful when degree of static indeterminacy is less than the degree of kinematic indeterminacy
1. The displacement method is more useful when degree of kinematic indeterminacy is greater than the degree of static indeterminacy,
2. The displacement method is more useful when degree of kinematic indeterminacy is less than degree of static indeterminacy.
3. The force method is more useful when degree of static indeterminacy is greater than the degree of kinematic indeterminacy
4. The force method is more useful when degree of static indeterminacy is less than the degree of kinematic indeterminacy
Which of these statements is correct?
a)
1 and 3
b)
2 and 3
c)
1 and 4
d)
2 and 4
|
|
Anagha Mehta answered |
Force method is useful when,
Ds < Dk
Displacement method is useful when,
Dk < Ds
Ds < Dk
Displacement method is useful when,
Dk < Ds
A symmetrical two-hinged parabolic arch rib has a span of 32 m between abutment pins at the same level and a central rise of 5 m. When a rolling load of 100 kN crosses the span, the maximum horizontal thrust at the hinges will be- a)100 kN
- b)125kN
- c)160kN
- d)240 kN
Correct answer is option 'B'. Can you explain this answer?
A symmetrical two-hinged parabolic arch rib has a span of 32 m between abutment pins at the same level and a central rise of 5 m. When a rolling load of 100 kN crosses the span, the maximum horizontal thrust at the hinges will be
a)
100 kN
b)
125kN
c)
160kN
d)
240 kN
|
|
Srestha Khanna answered |
Maximum Horizontal Thrust occurs when the load is at center and is given by
H = (25WL)/(128h)
W = Concentrated Load = 100kN
Span= L = 32m; Rise= h=5m
Substituting the values we get H=125kN
The rigid-jointed frame shown in the figure is
- a)unstable
- b)stable and statically determinate
- c)stable and statically indeterminate by one degree
- d)stable and statically indeterminate by two degree
Correct answer is option 'B'. Can you explain this answer?
The rigid-jointed frame shown in the figure is
a)
unstable
b)
stable and statically determinate
c)
stable and statically indeterminate by one degree
d)
stable and statically indeterminate by two degree
|
Sanjay Gp answered |
For a rigid joined 2D frame structure. Degree of static indeterminacy
Ds=3m+r-3j
here
Ds=3x2+3-3x3=0.
Hence it is statically determinate.
And since reactions are not parallel or concurrent it is a stable member.
Ds=3m+r-3j
here
Ds=3x2+3-3x3=0.
Hence it is statically determinate.
And since reactions are not parallel or concurrent it is a stable member.
A three-hinged parabolic arch rib of span L and crown rise ' h ' carries a uniformly distributed superimposed load of intensity W per unit length. The hinges are located on two abutments at the same level and the third hinge at a quarter span location from left hand abutment. The horizontal thrust on the abutment is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A three-hinged parabolic arch rib of span L and crown rise ' h ' carries a uniformly distributed superimposed load of intensity W per unit length. The hinges are located on two abutments at the same level and the third hinge at a quarter span location from left hand abutment. The horizontal thrust on the abutment is
a)
b)
c)
d)
|
|
Yashvi Choudhury answered |
A cable carrying a load of 20 kN/m run of the horizontal span, is structured at supports 100 m apart. The supports are at the same level and the central dip is 12.5 m. Find the least tension in the cable.- a)1000 kN
- b)2000 kN
- c)2236 kN
- d)None
Correct answer is option 'B'. Can you explain this answer?
A cable carrying a load of 20 kN/m run of the horizontal span, is structured at supports 100 m apart. The supports are at the same level and the central dip is 12.5 m. Find the least tension in the cable.
a)
1000 kN
b)
2000 kN
c)
2236 kN
d)
None
|
|
Aashna Chakraborty answered |
Minimum tension occurs at deepest point and is equal to horizontal thrust.
Tmin = H = 2000 kN
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
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.
For a linear elastic frame, if the stiffness matrix is doubled with respect to the existing stiffness matrix, the deflection of the resulting frame will be- a)twice the existing value
- b)half the existing value
- c)the same as existing value
- d)indeterminate value
Correct answer is option 'B'. Can you explain this answer?
For a linear elastic frame, if the stiffness matrix is doubled with respect to the existing stiffness matrix, the deflection of the resulting frame will be
a)
twice the existing value
b)
half the existing value
c)
the same as existing value
d)
indeterminate value
|
|
Anmol Choudhary answered |
Stiffness (k) is defined as the force per unit deflection.
K = P/δ
⇒ δ = P/K
If stiffness matrix is doubled
i.e. K’ = 2K
then,
δ = P/K'
δ = P/2K
δ = (1/2)P/K
δ’ = δ/2.
Therefore, if the stiffness matrix is doubled with respect to the existing stiffness matrix, the deflection of the resulting frame will be half the existing value.
The following methods are used for structural analysis:
1. Macaulay method
2. Column analogy method
3. Kani’s method
4. Method of sectionsThose used for indeterminate structural analysis would include- a)1 and 2
- b)1 and 3
- c)2 and 3
- d)2, 3 and 4
Correct answer is option 'C'. Can you explain this answer?
The following methods are used for structural analysis:
1. Macaulay method
2. Column analogy method
3. Kani’s method
4. Method of sections
1. Macaulay method
2. Column analogy method
3. Kani’s method
4. Method of sections
Those used for indeterminate structural analysis would include
a)
1 and 2
b)
1 and 3
c)
2 and 3
d)
2, 3 and 4
|
|
Rounak Mehta answered |
Macaulay's method is used for deflection and slope calculation due to point loads in prismatic beams. Method of sections is used for statically determinate trusses.
Match List - I (Method) with List-II (Factors) and select the correct answer using the codes given below the lists:List - IA. Moment distributionB. Slope deflectionC. Kani’s methodD. Force methodList - II1. Rotation factor2. Flexibility3. Hardy Cross4. Displacements5. Stiffness matrix- a)A - 3, B - 4, C - 1, D - 2
- b)A - 2, B - 1, C - 5, D - 3
- c)A - 2, B - 4, C - 1, D - 3
- d)A - 3, B - 1, C - 5, D - 2
Correct answer is option 'A'. Can you explain this answer?
Match List - I (Method) with List-II (Factors) and select the correct answer using the codes given below the lists:
List - I
A. Moment distribution
B. Slope deflection
C. Kani’s method
D. Force method
List - II
1. Rotation factor
2. Flexibility
3. Hardy Cross
4. Displacements
5. Stiffness matrix
a)
A - 3, B - 4, C - 1, D - 2
b)
A - 2, B - 1, C - 5, D - 3
c)
A - 2, B - 4, C - 1, D - 3
d)
A - 3, B - 1, C - 5, D - 2
|
|
Sahil Mehra answered |
Moment distribution → Hardy cross
Slope deflection → Displacement
Kani’s method → Rotation factor
Force method → Flexibility
Other things to be kept in mind:
Force method
The theorem of three moments
Method of consistent deformation
Castiglione’s theorem
Compatibility method
Displacement method
Moment distribution method
Slope deflection method
Kani’s method
Stiffness matrix method
A symmetrical three-hinged parabolic arch of span L and rise h is hinged at springing and crown. It is subjected to a UDLw throughout the span. What is the bending moment at a section L/4 from the left support?- a)wL2/8
- b)wL2/16
- c)wL3/8h
- d)Zero
Correct answer is option 'D'. Can you explain this answer?
A symmetrical three-hinged parabolic arch of span L and rise h is hinged at springing and crown. It is subjected to a UDLw throughout the span. What is the bending moment at a section L/4 from the left support?
a)
wL2/8
b)
wL2/16
c)
wL3/8h
d)
Zero
|
|
Pankaj Kapoor answered |
Solution:
Given, span of the arch = L, and rise of the arch = h
The arch is symmetrical and three-hinged.
The arch is subjected to a uniformly distributed load (UDL) of w throughout the span.
We need to find the bending moment at a section located L/4 distance from the left support.
Let the left support be A, the right support be C and the crown be B.
The UDL w will be resolved into vertical and horizontal components.
The vertical component of the UDL will be resisted by the arch as compression.
The horizontal component of the UDL will be resisted by the horizontal thrust developed at the supports.
The shape of the arch is parabolic, which means that the bending moment is zero at the crown (point B) and increases as we move towards the supports.
At the left support A, the bending moment is zero because there is no load on the left of A.
At the right support C, the bending moment is zero because the horizontal thrust balances the bending moment due to the UDL.
At the section located L/4 distance from the left support, the bending moment will be zero because of the symmetry of the arch.
Therefore, the correct option is D) Zero.
Given, span of the arch = L, and rise of the arch = h
The arch is symmetrical and three-hinged.
The arch is subjected to a uniformly distributed load (UDL) of w throughout the span.
We need to find the bending moment at a section located L/4 distance from the left support.
Let the left support be A, the right support be C and the crown be B.
The UDL w will be resolved into vertical and horizontal components.
The vertical component of the UDL will be resisted by the arch as compression.
The horizontal component of the UDL will be resisted by the horizontal thrust developed at the supports.
The shape of the arch is parabolic, which means that the bending moment is zero at the crown (point B) and increases as we move towards the supports.
At the left support A, the bending moment is zero because there is no load on the left of A.
At the right support C, the bending moment is zero because the horizontal thrust balances the bending moment due to the UDL.
At the section located L/4 distance from the left support, the bending moment will be zero because of the symmetry of the arch.
Therefore, the correct option is D) Zero.
In a two hinged arch an increase in temperature induces- a)no bending moment in the arch rib
- b)uniform bending moment in the arch rib
- c)maximum bending at the crown
- d)maximum bending moment at the crown
Correct answer is option 'C'. Can you explain this answer?
In a two hinged arch an increase in temperature induces
a)
no bending moment in the arch rib
b)
uniform bending moment in the arch rib
c)
maximum bending at the crown
d)
maximum bending moment at the crown
|
|
Sanaya Sengupta answered |
Increase in temperature in a two hinged arch (degree of indeterminacy one ) will cause horizontal thrust only.
Moment due to horizontal thrust is -Hy
So, max B.M will be at crown as crown has height value of y.
An arch is used in long span bridges because there is- a)considerable reduction in horizontal thrust
- b)considerable reduction in bending moment
- c)reduction in shear and bending moment
- d)no change in shear and bending moment
Correct answer is option 'B'. Can you explain this answer?
An arch is used in long span bridges because there is
a)
considerable reduction in horizontal thrust
b)
considerable reduction in bending moment
c)
reduction in shear and bending moment
d)
no change in shear and bending moment
|
Mita Mehta answered |
-An arch is used in long span bridges because there is considerable reduction in bending moment.
-A span is the distance between two bridge supports, whether they are columns, towers or the wall of a canyon. A modern beam bridge, for instance, is likely to span a distance of up to 200 feet (60 meters), while a modern arch can safely span up to 800 or 1,000 feet (240 to 300 m).
-An arch bridge is stronger than a beam bridge, simply because the beam has a weak point in the center where there is no vertical support while arches press the weight outward toward the support.
-Arch bridges, meanwhile, have been used to cover very long distances, with up to 800 feet for a single arch.
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.
A uniform simply supported beam is subjected to a clockwise moment M at the left end. What is the moment required at the right end so that rotation of the right end is zero?- a)2M
- b)M
- c)M/2
- d)M/3
Correct answer is option 'C'. Can you explain this answer?
A uniform simply supported beam is subjected to a clockwise moment M at the left end. What is the moment required at the right end so that rotation of the right end is zero?
a)
2M
b)
M
c)
M/2
d)
M/3
|
|
Garima Kulkarni answered |
at left end
To keep rotation at right and zero a moment should be applied in anticlockwise direction. Let the moment in M'
A uniformly distributed load of 80 kN per meter run of length 3 m moves on a simply support girder of span 10 m. The magnitude of the ratio of Maximum negative shear force to the maximum positive shear force at 4 m from the left end will be _______ (up to two decimal places).
Correct answer is between '0.54,0.56'. Can you explain this answer?
A uniformly distributed load of 80 kN per meter run of length 3 m moves on a simply support girder of span 10 m. The magnitude of the ratio of Maximum negative shear force to the maximum positive shear force at 4 m from the left end will be _______ (up to two decimal places).
|
|
Athira Pillai answered |
For maximum positive shear force at C:
Using Muller Breslau Principle.
y13=356y13=356
Y1 = 0.3
Maximum positive shear force at 
(S.F)max1 =108kN
(S.F)max1 =108kNFor maximum negative shear force at C:
A uniformly distributed load of 2 kN/m covers left half of the span of a three-hinged parabolic arch, span 40 m and central rise 10 m. Which of the following statements relating to different functions at the loaded quarter point are correct?
1. The slope is tan-1 (1/2)
2. The normal thrust is 6√5 kN.
3. The shear force is not zero
4. The bending moment is zero.Select the correct answer using the codes' given below:- a)1,2 and 4
- b)2 and 3
- c)1 and 3
- d)3 and 4
Correct answer is option 'C'. Can you explain this answer?
A uniformly distributed load of 2 kN/m covers left half of the span of a three-hinged parabolic arch, span 40 m and central rise 10 m. Which of the following statements relating to different functions at the loaded quarter point are correct?
1. The slope is tan-1 (1/2)
2. The normal thrust is 6√5 kN.
3. The shear force is not zero
4. The bending moment is zero.
1. The slope is tan-1 (1/2)
2. The normal thrust is 6√5 kN.
3. The shear force is not zero
4. The bending moment is zero.
Select the correct answer using the codes' given below:
a)
1,2 and 4
b)
2 and 3
c)
1 and 3
d)
3 and 4
|
|
Deepika Saha answered |
A horizontal fixed beam AB of span 6 m has uniform flexural rigidity of 4200 kN m2. During loading, the support B sinks downwards by 25 mm. The moment induced at the end A is- a)105 kN m (Anticlockwise)
- b)17.5 kN m (Clockwise)
- c)17.5 kN m (Anticlockwise)
- d)105 kN m (Clockwise)
Correct answer is option 'C'. Can you explain this answer?
A horizontal fixed beam AB of span 6 m has uniform flexural rigidity of 4200 kN m2. During loading, the support B sinks downwards by 25 mm. The moment induced at the end A is
a)
105 kN m (Anticlockwise)
b)
17.5 kN m (Clockwise)
c)
17.5 kN m (Anticlockwise)
d)
105 kN m (Clockwise)
|
|
Mehul Choudhury answered |
Find the carry over moment at support B, in the beam shown with internal hinges at C & D:-- a)M/8
- b)M/2
- c)M
- d)M/4
Correct answer is option 'B'. Can you explain this answer?
Find the carry over moment at support B, in the beam shown with internal hinges at C & D:-
a)
M/8
b)
M/2
c)
M
d)
M/4
|
|
Varun Mukherjee answered |
The beam has internal hinges at C & D, so the beam can be break down as shown
In AC section moment applied = M
In AC section moment applied = M
So the reaction will be = M/L/2 = 2/ML
The carry – over moment at B = 2M/L × L/4 = M/2
Which one of the following structures is statically determinate and stable?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which one of the following structures is statically determinate and stable?
a)
b)
c)
d)
|
|
Bhavya Ahuja answered |
A structure will be statically determinate if the external reactions can be determined from force-equilibrium equations. A structure is stable when the whole or part of the structure is prevented from large displacements on account of loading.
In option '1', the structure is stable and there are three reaction components which can be determined by two force equilibrium conditions and one-moment equilibrium condition.
The structure in option '2' is stable but statically indeterminate to the second degree.
The structure is shown in option '3' has both reaction components coinciding with each other, so the moment equilibrium condition will never be satisfied and the structure will not be under equilibrium.
In option '4' the structure is not stable as a little movement in the horizontal direction leads to the large displacement and there is no resistance offered by the structure in the horizontal direction.
What is the degree of static indeterminacy of the plane structure as shown in the figure below?
- a)3
- b)4
- c)5
- d)6
Correct answer is option 'B'. Can you explain this answer?
What is the degree of static indeterminacy of the plane structure as shown in the figure below?
a)
3
b)
4
c)
5
d)
6
|
|
Raghav Mukherjee answered |
For plane truss degree of indeterminacy,
Ds = m + re - 2j
re = 4; m= 10; j = 5
Ds = 4 + 10 - 2 x 5 = 4
Ds = m + re - 2j
re = 4; m= 10; j = 5
Ds = 4 + 10 - 2 x 5 = 4
Which one of the following structures is statically determinate and stable?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which one of the following structures is statically determinate and stable?
a)
b)
c)
d)
|
|
Mahesh Nair answered |
A structure will be statically determinate if the external reactions can be determined from force- equilibrium equations. A structure is stable when the whole or part of the structure is prevented from large displacements on account of loading.
The structure in figure (b) is stable but statically indeterminate to the second degree.
The structure shown in figure (c) has both reaction components coinciding with each other, so the moment equilibrium condition wilt never be satisfied and the structure will not be under equilibrium. In figure'(a), the structure is stable and there are three reaction components which can be determined by two force equilibrium conditions and one moment equilibrium condition.
The structure in figure (b) is stable but statically indeterminate to the second degree.
The structure shown in figure (c) has both reaction components coinciding with each other, so the moment equilibrium condition wilt never be satisfied and the structure will not be under equilibrium. In figure'(a), the structure is stable and there are three reaction components which can be determined by two force equilibrium conditions and one moment equilibrium condition.
A truss ABC carries two horizontal and two vertical loads, as shown in the figure. The horizontal and vertical components of reactions at A will be
- a)HA = 1000 kN; VA = 1706.25 kN
- b)HA = 2000 kN; VA = 2550 kN
- c)HA = 1000 kN; VA = 2350 kN
- d)HA = 2000 kN; VA = 1706.25 kN
Correct answer is option 'D'. Can you explain this answer?
A truss ABC carries two horizontal and two vertical loads, as shown in the figure. The horizontal and vertical components of reactions at A will be
a)
HA = 1000 kN; VA = 1706.25 kN
b)
HA = 2000 kN; VA = 2550 kN
c)
HA = 1000 kN; VA = 2350 kN
d)
HA = 2000 kN; VA = 1706.25 kN
|
Muthu Selvi answered |
B
The strain energy stored (U) in the cantilever beam shown is –- a)
- b)
- c)
- d)None
Correct answer is option 'C'. Can you explain this answer?
The strain energy stored (U) in the cantilever beam shown is –
a)
b)
c)
d)
None
|
|
Rishika Sen answered |
BM at section X-X, Mx = W.x
Now, for
For,
∴ For
U will be less than
but greater than 
A loaded porta! frame is shown in figure. The profile of its Bending Moment diagram will be
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A loaded porta! frame is shown in figure. The profile of its Bending Moment diagram will be
a)
b)
c)
d)
|
|
Aditi Sarkar answered |
- The horizontal force will be distributed to both the supports.
- The bending moment at hinged end will be zero and fixed end will have some B.M.
- The B.M.D. for columns will be linear.
- The B.M.D. for beam will be parabolic.
The choice now remains between (b) and (d). In diagram (b) and (c) the Left column has negative B.M. in upper half or at top. The beam at left end in (b) shows negative B.M. while in (c) it shows a positive B.M.
- The bending moment at hinged end will be zero and fixed end will have some B.M.
- The B.M.D. for columns will be linear.
- The B.M.D. for beam will be parabolic.
The choice now remains between (b) and (d). In diagram (b) and (c) the Left column has negative B.M. in upper half or at top. The beam at left end in (b) shows negative B.M. while in (c) it shows a positive B.M.
The force induced in the vertical member CD of the symmetrical plane truss shown in the figure is
- a)50 kN (tension)
- b)100 kN (tension)
- c)50 kN (compression)
- d)zero
Correct answer is option 'A'. Can you explain this answer?
The force induced in the vertical member CD of the symmetrical plane truss shown in the figure is
a)
50 kN (tension)
b)
100 kN (tension)
c)
50 kN (compression)
d)
zero
|
|
Abhay Kapoor answered |
At join D, the equilibrium shows the force in member CD is 50 kN tensile.
The pin-jointed frame shown in the figure is
- a)a perfect frame
- b)a redundant frame
- c)a deficient frame
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
The pin-jointed frame shown in the figure is
a)
a perfect frame
b)
a redundant frame
c)
a deficient frame
d)
None of the above
|
|
Subham Unni answered |
Degree of indeterminacy,
n = (m + rE) - 2j
= (9 + 3 ) - 2 x 6 = 0
Since the degree of indeterminacy is zero and the frame is stable so it is a perfect frame.
n = (m + rE) - 2j
= (9 + 3 ) - 2 x 6 = 0
Since the degree of indeterminacy is zero and the frame is stable so it is a perfect frame.
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
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