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All questions of Solid Mechanics for Civil Engineering (CE) Exam
The maximum positive bending moment in a fixed beam of span 8m and subjected to a central point load of 120 kN is (in kN-m)- a)240
- b)120
- c)80
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The maximum positive bending moment in a fixed beam of span 8m and subjected to a central point load of 120 kN is (in kN-m)
a)
240
b)
120
c)
80
d)
None of these
|
Tanvi Shah answered |
Bending moment at centre of the span
= (W×L)/4 ie (120 kn × 8 m )/4
= 240 kn-m.
Since the load acting in the middle of span. Hence the load will be equally distributed at both endsTherefore max bending moment in fixed beam
= 240 kn-m / 2= 120 kn-m .
A simple supported beam with rectangular cross section is subjected to a concentrated load at mid span. If the width and depth of beam are doubled, the deflection at mid span will be reduced to:-- a)6.25%
- b)12.5%
- c)50%
- d)25%
Correct answer is option 'A'. Can you explain this answer?
A simple supported beam with rectangular cross section is subjected to a concentrated load at mid span. If the width and depth of beam are doubled, the deflection at mid span will be reduced to:-
a)
6.25%
b)
12.5%
c)
50%
d)
25%
|
Rhea Reddy answered |
Max. deflection at mid span(δ)=5WL4/384EI
Hence,deflection is inversely proportional to the moment of inertia.
For rectangular section,I=bd3/12
Therefore,if width and depth both are doubled,deflection will get reduced to 6.25%.
For a given material, the modulus of rigidity is 120 GPa and Poisson’s ratio is 0.4. The value of modulus of elasticity in GPa is- a)120
- b)276
- c)312
- d)336
Correct answer is option 'D'. Can you explain this answer?
For a given material, the modulus of rigidity is 120 GPa and Poisson’s ratio is 0.4. The value of modulus of elasticity in GPa is
a)
120
b)
276
c)
312
d)
336
|
Jay Sharma answered |
E = 2G (1 + μ) = 2 × 120 (1 + 0.4) = 336 GPa
The stress in a body due to suddenly applied load compared to when it is applied gradually is _______.- a)Same
- b)Half
- c)Two times
- d)Four times
Correct answer is option 'C'. Can you explain this answer?
The stress in a body due to suddenly applied load compared to when it is applied gradually is _______.
a)
Same
b)
Half
c)
Two times
d)
Four times
|
Bank Exams India answered |
Gradually applied load is given as σ = (F/A) ------ (F is the gradually applied load)
here, work done is given as (F δL) / 2 and strain energy stored = (σ2 / 2E) AL
Work done is equal to the strain energy stored.
(F δL) / 2 = (σ2 / 2E) AL
Therefore, σ = (F/A) ----------- (1)
Suddenly applied load is given as σ = (2F/A), here work done = (F δL)
(F δL) = (σ2 / 2E) AL
Therefore, σ = (2F/A) ------------- (2)
From (1) and (2), it can be concluded that
Hence, suddenly applied load is twice the gradually applied load.
here, work done is given as (F δL) / 2 and strain energy stored = (σ2 / 2E) AL
Work done is equal to the strain energy stored.
(F δL) / 2 = (σ2 / 2E) AL
Therefore, σ = (F/A) ----------- (1)
Suddenly applied load is given as σ = (2F/A), here work done = (F δL)
(F δL) = (σ2 / 2E) AL
Therefore, σ = (2F/A) ------------- (2)
From (1) and (2), it can be concluded that
Hence, suddenly applied load is twice the gradually applied load.
If the deflection at the free end of a cantilever beam of span 1 m, subjected UDL over entire span is 15 mm, then the slope at the free end is ________- a)0.02 radians
- b)0.03 radians
- c)0.04 radians
- d)0.06 radians
Correct answer is option 'A'. Can you explain this answer?
If the deflection at the free end of a cantilever beam of span 1 m, subjected UDL over entire span is 15 mm, then the slope at the free end is ________
a)
0.02 radians
b)
0.03 radians
c)
0.04 radians
d)
0.06 radians
|
Anirudh Kulkarni answered |
Deflection at the free end of a cantilever beam subjected to udl, 
Slope at free end subjected to udl, 
The radius of Mohr’s circle for two equal and unlike principal stresses of magnitude “p” is:- a)p
- b)p/2
- c)Zero
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
The radius of Mohr’s circle for two equal and unlike principal stresses of magnitude “p” is:
a)
p
b)
p/2
c)
Zero
d)
None of the above
|
|
Nikhil Majumdar answered |
Radius of Mohr’s circle is given by:
At principal plane, σx = σ1, σy = σ2
τxy = 0
σ1 = p , σ2 = -p
r = p
Consider the following statements:1. On a Principal plane, only normal stress acts2. On a Principal plane, both normal and shear stresses acts3. On a Principal plane, only shear stress acts4. Isotropic state of stress is independent of frame of referenceWhich of the above statements is/are correct?- a)1 and 4
- b)2 only
- c)2 and 4 only
- d)2 and 3
Correct answer is option 'A'. Can you explain this answer?
Consider the following statements:
1. On a Principal plane, only normal stress acts
2. On a Principal plane, both normal and shear stresses acts
3. On a Principal plane, only shear stress acts
4. Isotropic state of stress is independent of frame of reference
Which of the above statements is/are correct?
a)
1 and 4
b)
2 only
c)
2 and 4 only
d)
2 and 3
|
|
Prasad Desai answered |
Principal planes are the planes at which only normal stress is acting and shear stress is zero at that plane,
Option 2 and 3 are wrong. So, correct option is 1 and 4
A simply supported beam is of rectangular section. It carries a uniformly distributed load over the whole span. The deflection at the centre is “y”. If the depth of the beam is doubled, the deflection at centre will be:- a)2 y
- b)4 y
- c)y/8
- d)y/2
Correct answer is option 'C'. Can you explain this answer?
A simply supported beam is of rectangular section. It carries a uniformly distributed load over the whole span. The deflection at the centre is “y”. If the depth of the beam is doubled, the deflection at centre will be:
a)
2 y
b)
4 y
c)
y/8
d)
y/2
|
|
Anuj Verma answered |
Deflection of beam is inversely proportional to area moment of inertia of the beam.
Moment of inertia for rectangular section, 
Let the final deflection is yfinal when the depth is doubled.
The equivalent length of a column of length L having one end fixed and the other end free is- a)2L
- b)L
- c)L/2
- d)L/√2
Correct answer is option 'A'. Can you explain this answer?
The equivalent length of a column of length L having one end fixed and the other end free is
a)
2L
b)
L
c)
L/2
d)
L/√2
|
|
Sarthak Kulkarni answered |
Effective length of columns under different end conditions are:
1. For both ends hinged: Le = L
2. One end fixed and another end free: Le = 2L
3. Both ends fixed = 
4. One end fixed and other is hinges: 
A steel bar of 5mm is heated from 5°C to 40°C and is free to expand. The bar will induce- a)No stress
- b)Tensile stress
- c)Shear stress
- d)Compressive stress
Correct answer is option 'A'. Can you explain this answer?
A steel bar of 5mm is heated from 5°C to 40°C and is free to expand. The bar will induce
a)
No stress
b)
Tensile stress
c)
Shear stress
d)
Compressive stress
|
|
Aarav Kulkarni answered |
When a material undergoes a change in temperature, it either elongates or contracts depending upon whether temperature is increased or decreased of the material. If the elongation or contraction is not restricted, i. e. free then the material does not experience any stress despite the fact that it undergoes a strain.
The shear stress on a beam section is maximum:-- a)On the extreme bottom surface fibres
- b)At the neutral axis of the section
- c)At the free edges
- d)On the extreme top surface fibres
Correct answer is option 'B'. Can you explain this answer?
The shear stress on a beam section is maximum:-
a)
On the extreme bottom surface fibres
b)
At the neutral axis of the section
c)
At the free edges
d)
On the extreme top surface fibres
|
Lavanya Menon answered |
Shown below is a rectangular beam in pure bending. As shown above, shear stresses vary quadratically with the distance y1 from the neutral axis. The maximum shear stress occurs at the neutral axis and is zero at both the top and bottom surface of the beam. Shear flow has the units of force per unit distance.
Two bars of different materials are of the same size and are subjected to same tensile forces. If the bars have unit elongations in the ratio of 3 : 8, then the ratio of moduli of elasticity of the two materials is:- a)7 : 4
- b)3 : 8
- c)8 : 3
- d)64 : 9
Correct answer is option 'C'. Can you explain this answer?
Two bars of different materials are of the same size and are subjected to same tensile forces. If the bars have unit elongations in the ratio of 3 : 8, then the ratio of moduli of elasticity of the two materials is:
a)
7 : 4
b)
3 : 8
c)
8 : 3
d)
64 : 9
|
|
Sarthak Menon answered |
Solution:
Given,
Unit elongation of bar A to bar B = 3 : 8
Let,
A = Area of cross-section of the bar
L = Length of the bar
F = Tensile force applied on the bar
σA = Stress developed in bar A
σB = Stress developed in bar B
Ea = Modulus of elasticity of material A
Eb = Modulus of elasticity of material B
Formula used:
Stress = Force / Area
Unit elongation = Elongation / Original length
Stress = Modulus of elasticity × Strain
Calculation:
Let us assume that the original length of both bars is L.
Then, the elongation produced in bar A and bar B is 3L and 8L respectively.
According to the question,
Unit elongation of bar A to bar B = 3 : 8
∴ 3L / L = Ea / Eb
∴ Ea / Eb = 3 / 8
Stress developed in bar A = F / A
Stress developed in bar B = F / A
From the formula of stress,
Stress = Modulus of elasticity × Strain
∴ σA / Ea = 3L / L
∴ σA = 3Ea
Similarly,
σB / Eb = 8L / L
∴ σB = 8Eb
Now,
σA / σB = F / A × Eb / F × A / Ea
∴ σA / σB = Ea / Eb
From the above calculation,
Ea / Eb = 3 / 8
∴ σA / σB = 3 / 8
Therefore, the ratio of moduli of elasticity of the two materials is 8 : 3.
Hence, the correct option is (c) 8 : 3.
Given,
Unit elongation of bar A to bar B = 3 : 8
Let,
A = Area of cross-section of the bar
L = Length of the bar
F = Tensile force applied on the bar
σA = Stress developed in bar A
σB = Stress developed in bar B
Ea = Modulus of elasticity of material A
Eb = Modulus of elasticity of material B
Formula used:
Stress = Force / Area
Unit elongation = Elongation / Original length
Stress = Modulus of elasticity × Strain
Calculation:
Let us assume that the original length of both bars is L.
Then, the elongation produced in bar A and bar B is 3L and 8L respectively.
According to the question,
Unit elongation of bar A to bar B = 3 : 8
∴ 3L / L = Ea / Eb
∴ Ea / Eb = 3 / 8
Stress developed in bar A = F / A
Stress developed in bar B = F / A
From the formula of stress,
Stress = Modulus of elasticity × Strain
∴ σA / Ea = 3L / L
∴ σA = 3Ea
Similarly,
σB / Eb = 8L / L
∴ σB = 8Eb
Now,
σA / σB = F / A × Eb / F × A / Ea
∴ σA / σB = Ea / Eb
From the above calculation,
Ea / Eb = 3 / 8
∴ σA / σB = 3 / 8
Therefore, the ratio of moduli of elasticity of the two materials is 8 : 3.
Hence, the correct option is (c) 8 : 3.
The crippling load taken by a column with both ends hinged is 100 kN. The crippling load taken by the same column with one end fixed and other end free will be:- a)400 kN
- b)100 kN
- c)50 kN
- d)25 kN
Correct answer is option 'D'. Can you explain this answer?
The crippling load taken by a column with both ends hinged is 100 kN. The crippling load taken by the same column with one end fixed and other end free will be:
a)
400 kN
b)
100 kN
c)
50 kN
d)
25 kN
|
|
Anirudh Kulkarni answered |
The crippling load carrying capacity, 
if all other parameters remain constant
A point within the cross sectional plane of a beam through which the resultant of the external loading on the beam has to pass through to ensure pure bending without twisting of the cross-sectional of the beam is called:- a)Moment center
- b)Centroid
- c)Shear center
- d)Elastic center
Correct answer is option 'C'. Can you explain this answer?
A point within the cross sectional plane of a beam through which the resultant of the external loading on the beam has to pass through to ensure pure bending without twisting of the cross-sectional of the beam is called:
a)
Moment center
b)
Centroid
c)
Shear center
d)
Elastic center
|
Bijoy Chauhan answered |
Shear center is the correct answer.
Explanation:
The shear center is a point within the cross-sectional plane of a beam where the external loading on the beam must pass through in order to ensure pure bending without twisting of the cross-section. It is an important concept in structural engineering, particularly in the design of beams and other structural elements.
In order to understand the significance of the shear center, it is important to first understand the concepts of bending and twisting in a beam.
Bending refers to the deformation of a beam under load, which causes the beam to curve. This bending is typically caused by a moment applied to the beam. When a beam bends, the top and bottom surfaces of the beam experience tension and compression, respectively.
Twisting, on the other hand, refers to the deformation of a beam that causes it to rotate about its longitudinal axis. Twisting can occur when a shear force is applied to the beam, causing the top and bottom surfaces of the beam to experience shear stresses.
When a beam is subjected to both bending and twisting, it is important to ensure that the resulting deformation is pure bending without any twisting. This is particularly important in structural applications where twisting can lead to instability or failure of the structure.
The shear center is the point within the cross-sectional plane of the beam where the external loading must pass through in order to eliminate twisting and ensure pure bending. It is the point where the shear forces and bending moments due to the external loading are in equilibrium, resulting in pure bending without any twisting.
By ensuring that the external loading passes through the shear center, engineers can design beams and other structural elements that are resistant to twisting and have predictable and controllable behavior under load.
In summary, the shear center is the point within the cross-sectional plane of a beam where the external loading must pass through to ensure pure bending without twisting. It is an important concept in structural engineering and plays a critical role in the design of beams and other structural elements.
Explanation:
The shear center is a point within the cross-sectional plane of a beam where the external loading on the beam must pass through in order to ensure pure bending without twisting of the cross-section. It is an important concept in structural engineering, particularly in the design of beams and other structural elements.
In order to understand the significance of the shear center, it is important to first understand the concepts of bending and twisting in a beam.
Bending refers to the deformation of a beam under load, which causes the beam to curve. This bending is typically caused by a moment applied to the beam. When a beam bends, the top and bottom surfaces of the beam experience tension and compression, respectively.
Twisting, on the other hand, refers to the deformation of a beam that causes it to rotate about its longitudinal axis. Twisting can occur when a shear force is applied to the beam, causing the top and bottom surfaces of the beam to experience shear stresses.
When a beam is subjected to both bending and twisting, it is important to ensure that the resulting deformation is pure bending without any twisting. This is particularly important in structural applications where twisting can lead to instability or failure of the structure.
The shear center is the point within the cross-sectional plane of the beam where the external loading must pass through in order to eliminate twisting and ensure pure bending. It is the point where the shear forces and bending moments due to the external loading are in equilibrium, resulting in pure bending without any twisting.
By ensuring that the external loading passes through the shear center, engineers can design beams and other structural elements that are resistant to twisting and have predictable and controllable behavior under load.
In summary, the shear center is the point within the cross-sectional plane of a beam where the external loading must pass through to ensure pure bending without twisting. It is an important concept in structural engineering and plays a critical role in the design of beams and other structural elements.
A rectangular beam is 24 cm wide and 50 cm deep, its section modulus is given by:- a)1000 cm3
- b)50000 cm3
- c)10000 cm3
- d)100000 cm3
Correct answer is option 'C'. Can you explain this answer?
A rectangular beam is 24 cm wide and 50 cm deep, its section modulus is given by:
a)
1000 cm3
b)
50000 cm3
c)
10000 cm3
d)
100000 cm3
|
|
Aniket Mehta answered |
Section modulus is defined as the ratio of moment of inertia of a beam about its C.G to the maximum distance of extreme x-section of the beam (Ymax).
If the deflection and slope at the free end of a cantilever beam subjected to the uniformly distributed load are 15 mm and 0.02 radians respectively. The length of the beam is _______.- a)800 mm
- b)1000 mm
- c)1200 mm
- d)1500 mm
Correct answer is option 'B'. Can you explain this answer?
If the deflection and slope at the free end of a cantilever beam subjected to the uniformly distributed load are 15 mm and 0.02 radians respectively. The length of the beam is _______.
a)
800 mm
b)
1000 mm
c)
1200 mm
d)
1500 mm
|
|
Sarthak Kulkarni answered |
Deflection at the free end of a cantilever beam subjected to udl, 
Slope at free end subjected to udl, 
A circular shaft of 50 mm diameter is required to transmit a torque from one shaft to another. If shear stress is not to exceed 40 Mpa, the safe torque is:- a)0.882 kN-m
- b)0.982 kN-m
- c)1.982 kN-m
- d)2.00 kN-m
Correct answer is option 'B'. Can you explain this answer?
A circular shaft of 50 mm diameter is required to transmit a torque from one shaft to another. If shear stress is not to exceed 40 Mpa, the safe torque is:
a)
0.882 kN-m
b)
0.982 kN-m
c)
1.982 kN-m
d)
2.00 kN-m
|
|
Saptarshi Khanna answered |
Using the relation:
Where,
T = Torque applied to the shaft
IP = Polar section modulus
τ = shear stress in the material
R = radius of the shaft
= 0.982 kN-m
The major and minor principal stresses at a point are 3Mpa and -3Mpa respectively. The maximum shear stress at the point is- a)Zero
- b)3Mpa
- c)6Mpa
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The major and minor principal stresses at a point are 3Mpa and -3Mpa respectively. The maximum shear stress at the point is
a)
Zero
b)
3Mpa
c)
6Mpa
d)
None of these
|
Anshu Kumar answered |
Explanation:
In the theory of elasticity, the stress at a point can be resolved into three principal stresses: the major principal stress (σ1), the intermediate principal stress (σ2), and the minor principal stress (σ3). The major and minor principal stresses are the maximum and minimum values of stress at the point, respectively.
Given:
Major principal stress (σ1) = 3 MPa
Minor principal stress (σ3) = -3 MPa
To find the maximum shear stress (τmax), we can use the following formula:
τmax = (σ1 - σ3) / 2
Substituting the given values:
τmax = (3 MPa - (-3 MPa)) / 2
= (3 MPa + 3 MPa) / 2
= 6 MPa / 2
= 3 MPa
Therefore, the maximum shear stress at the point is 3 MPa.
In the theory of elasticity, the stress at a point can be resolved into three principal stresses: the major principal stress (σ1), the intermediate principal stress (σ2), and the minor principal stress (σ3). The major and minor principal stresses are the maximum and minimum values of stress at the point, respectively.
Given:
Major principal stress (σ1) = 3 MPa
Minor principal stress (σ3) = -3 MPa
To find the maximum shear stress (τmax), we can use the following formula:
τmax = (σ1 - σ3) / 2
Substituting the given values:
τmax = (3 MPa - (-3 MPa)) / 2
= (3 MPa + 3 MPa) / 2
= 6 MPa / 2
= 3 MPa
Therefore, the maximum shear stress at the point is 3 MPa.
Rankine theory is applicable to the ________.- a)Short strut/column
- b)Long column
- c)Both short and long column
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Rankine theory is applicable to the ________.
a)
Short strut/column
b)
Long column
c)
Both short and long column
d)
None of these
|
Anmol Roy answered |
For short or long columns Rankine’s Formula is used.
Where P is crippling load by Rankine formula; PC is crushing load; PE is crippling load by Euler’s formula.
For Short column: PE is very large so
will be very small and negligible as compared to 
so 
For Long column: PE is small so 
will be large as compared to 
will be large as compared to Hence the value of 
can be neglected. So
can be neglected. So
Hence the crippling load by Rankine’s formula for long column is approximately equal to the crippling load by Euler’s formula.
Modulus of rigidity is defined as the ratio of ________.- a)Longitudinal stress and longitudinal strain
- b)Volumetric stress and volumetric strain
- c)Lateral stress and lateral strain
- d)Shear stress and shear strain
Correct answer is option 'D'. Can you explain this answer?
Modulus of rigidity is defined as the ratio of ________.
a)
Longitudinal stress and longitudinal strain
b)
Volumetric stress and volumetric strain
c)
Lateral stress and lateral strain
d)
Shear stress and shear strain
|
|
Aniket Mehta answered |
The ratio of tensile stress or compressive stress to the corresponding strain is a constant. This ratio is known as Young’s Modulus or Modulus of Elasticity and is denoted by E.
The ratio of shear stress to the corresponding shear strain within the elastic limit, is known as Modulus of Rigidity or Shear Modulus. This is denoted by C or G or N.
When the shear force diagram is a parabolic curve between two points, it indicates that there is a- a)point load at the two points
- b)no loading between the two points
- c)uniformly distributed load between the two points
- d)uniformly varying load between the two points
Correct answer is option 'D'. Can you explain this answer?
When the shear force diagram is a parabolic curve between two points, it indicates that there is a
a)
point load at the two points
b)
no loading between the two points
c)
uniformly distributed load between the two points
d)
uniformly varying load between the two points
|
|
Jay Menon answered |
The general relationship between the shear force diagram, bending moment diagram and loading diagram will be:
1. Shear force diagram is 1o higher than loading diagram.
2. Bending moment diagram is 1o higher than shear force diagram.
3. For the uniformly varying load on the beam, the shear force diagram is parabolic in nature.
4. For the uniformly varying load on the beam, the bending moment diagram is also parabolic but is 1o higher than shear force diagram.
The ratio of crippling loads of a column having both the ends fixed to the column having both the ends hinged, is- a)4
- b)2
- c)1
- d)3
Correct answer is option 'A'. Can you explain this answer?
The ratio of crippling loads of a column having both the ends fixed to the column having both the ends hinged, is
a)
4
b)
2
c)
1
d)
3
|
Aditya Majumdar answered |
Crippling load is given by:
For both ends fixed, 
For both ends hinged, leff = l
What is the limit to Poisson's ratio?- a)0.1
- b)0.2
- c)0.3
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
What is the limit to Poisson's ratio?
a)
0.1
b)
0.2
c)
0.3
d)
None of these
|
Devika Tiwari answered |
Limit of poisons ratio varies between -1 to 0.5.
μ = 0.5 for rubber.
Match the following beams with their bending moment diagrams:
- a)A-1, b-3, c-4, d-2
- b)A-3, b-4, c-2, d-1
- c)A-4, b-3, c-1, d-2
- d)A-1, b-2, c-3, d-4
Correct answer is option 'B'. Can you explain this answer?
Match the following beams with their bending moment diagrams:
a)
A-1, b-3, c-4, d-2
b)
A-3, b-4, c-2, d-1
c)
A-4, b-3, c-1, d-2
d)
A-1, b-2, c-3, d-4
|
|
Divya Mehta answered |
Bending moment for two point load in simply supported beam will be trapezoidal.
Bending moment diagram for UDL in simply supported beam will be parabolic.
BMD for UDL in cantilever beam as shown in figure (ii)
A tensile test is performed on a round bar. After fracture it has been found that the diameter remains approximately same at fracture. The material under test was- a)Mild steel
- b)Cast iron
- c)Wrought iron
- d)Copper
Correct answer is option 'B'. Can you explain this answer?
A tensile test is performed on a round bar. After fracture it has been found that the diameter remains approximately same at fracture. The material under test was
a)
Mild steel
b)
Cast iron
c)
Wrought iron
d)
Copper
|
|
Alok Iyer answered |
In Brittle materials under tension test undergoes brittle fracture i.e there failure plane is 90° to the axis of load and there is no elongation in the rod that’s why the diameter remains same before and after the load. Example: Cast Iron, concrete etc
But in case of ductile materials, material first elongate and then fail, their failure plane is 45° to the axis of the load. After failure cup-cone failure is seen. Example Mild steel, high tensile steel etc.
The correct relation between Modulus of elasticity (E), Shear Modulus (G) and Bulk modulus (K) is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The correct relation between Modulus of elasticity (E), Shear Modulus (G) and Bulk modulus (K) is:
a)
b)
c)
d)
|
|
Sahil Chawla answered |
Relation between Modulus of elasticity (E), Shear Modulus (G) and Bulk modulus (K) is given by:
Now, rewriting the equation as follows:
3KE + GE = 9KG
3KE – 9KG = - GE
K(3E – 9G) = - GE
Maximum Shear strain Energy theory was postulated by:- a)Tresca
- b)St Venant
- c)Rankine
- d)Von-mises
Correct answer is option 'D'. Can you explain this answer?
Maximum Shear strain Energy theory was postulated by:
a)
Tresca
b)
St Venant
c)
Rankine
d)
Von-mises
|
Sanaya Sengupta answered |
1. Maximum Principal stress theory was postulated by Rankine. It is suitable for brittle materials.
2. Maximum Principal strain theory was postulated by St Venant. This theory is not accurate for brittle and ductile materials both.
3. Maximum Shear stress theory was postulated by Tresca. This theory is suitable for ductile materials. Its results are the safest.
4. Maximum shear strain energy theory was postulated by Von-mises. Its results in case of pure shear are the accurate for ductile materials.
The maximum deflection due to a uniformly distributed load w/unit length over entire span of a cantilever of length L and of flexural rigidity EI, is- a)wL3/3EI
- b)wL4/3EI
- c)wL4/8EI
- d)wL4/12EI
Correct answer is option 'C'. Can you explain this answer?
The maximum deflection due to a uniformly distributed load w/unit length over entire span of a cantilever of length L and of flexural rigidity EI, is
a)
wL3/3EI
b)
wL4/3EI
c)
wL4/8EI
d)
wL4/12EI
|
Aarav Sharma answered |
Formula Used:
The equation for the maximum deflection due to a uniformly distributed load w/unit length over the entire span of a cantilever is given by:
δmax = (5/384)wL^4/EI
where,
δmax = maximum deflection
w = uniformly distributed load
L = length of the cantilever
E = modulus of elasticity
I = moment of inertia
Solution:
Given, a cantilever of length L and of flexural rigidity EI, subjected to a uniformly distributed load w/unit length over the entire span.
Using the formula mentioned above, we can calculate the maximum deflection as:
δmax = (5/384)wL^4/EI
On solving this equation, we get:
δmax = wL^4/8EI
Therefore, the correct answer is option C, i.e., wL^4/8EI.
Conclusion:
The maximum deflection due to a uniformly distributed load w/unit length over the entire span of a cantilever of length L and of flexural rigidity EI is given by the equation δmax = wL^4/8EI. This equation can be easily derived using the basic principles of mechanics and is useful in determining the maximum deflection of cantilever beams under different loading conditions.
The equation for the maximum deflection due to a uniformly distributed load w/unit length over the entire span of a cantilever is given by:
δmax = (5/384)wL^4/EI
where,
δmax = maximum deflection
w = uniformly distributed load
L = length of the cantilever
E = modulus of elasticity
I = moment of inertia
Solution:
Given, a cantilever of length L and of flexural rigidity EI, subjected to a uniformly distributed load w/unit length over the entire span.
Using the formula mentioned above, we can calculate the maximum deflection as:
δmax = (5/384)wL^4/EI
On solving this equation, we get:
δmax = wL^4/8EI
Therefore, the correct answer is option C, i.e., wL^4/8EI.
Conclusion:
The maximum deflection due to a uniformly distributed load w/unit length over the entire span of a cantilever of length L and of flexural rigidity EI is given by the equation δmax = wL^4/8EI. This equation can be easily derived using the basic principles of mechanics and is useful in determining the maximum deflection of cantilever beams under different loading conditions.
The change in the slope between two points in a straight member under flexure is equal to the area of M/EI diagram between those two points. This statement is known as -- a)Conjugate beam theorem
- b)Macaulay’s theorem
- c)Moment area theorem
- d)Castigliano’s theorem
Correct answer is option 'C'. Can you explain this answer?
The change in the slope between two points in a straight member under flexure is equal to the area of M/EI diagram between those two points. This statement is known as -
a)
Conjugate beam theorem
b)
Macaulay’s theorem
c)
Moment area theorem
d)
Castigliano’s theorem
|
|
Aarav Sharma answered |
's methodc)Moment area methodd)Euler-Bernoulli beam theorye)None of the above
c) Moment area method
c) Moment area method
The Mohr’s circle given below corresponds to which one of the following stress condition.
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The Mohr’s circle given below corresponds to which one of the following stress condition.
a)
b)
c)
d)
|
|
Manasa Sen answered |
Radius of Mohr’s circle = 100 MPa
And centre of Mohr’s circle is at a distance,
from the origin. Here it is in origin.
both the normal stress may be zero which is a pure shear case or opposite in nature which don’t exist in any of the options. So it is the pure shear case, where the radius is
A simply supported beam with a gradually varying load from zero at B and w per unit length at A is shown in figure below, the shear force at B is equal to:
- a)
- b)
- c)wL
- d)
Correct answer is option 'A'. Can you explain this answer?
A simply supported beam with a gradually varying load from zero at B and w per unit length at A is shown in figure below, the shear force at B is equal to:
a)
b)
c)
wL
d)
|
|
Dipanjan Ghosh answered |
SFy = 0
RA + RB = 
Taking moment of all the forces about B = 0.
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