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All questions of Solid Mechanics for Civil Engineering (CE) Exam
The outside diameter of a hollow shaft is twice its inside diameter. The ratio of its torque carrying capacity to that a solid shaft of the same material and the same outside diameter is- a)15/16
- b)3/4
- c)1/2
- d)1/16
Correct answer is option 'A'. Can you explain this answer?
The outside diameter of a hollow shaft is twice its inside diameter. The ratio of its torque carrying capacity to that a solid shaft of the same material and the same outside diameter is
a)
15/16
b)
3/4
c)
1/2
d)
1/16
|
Sanvi Kapoor answered |
The torque carrying capacity Tis given by
Let outside diameter of hollow and solid shaft = 2d
∴ Inside diameter of hollow shaft = d
∴ Inside diameter of hollow shaft = d
In the engineering stress-strain curve for mild steel, the Ultimate Tensile Strength (UTS) refers to- a)Yield stress
- b)Proportional limit
- c)Maximum stress
- d)Fracture stress
Correct answer is option 'C'. Can you explain this answer?
In the engineering stress-strain curve for mild steel, the Ultimate Tensile Strength (UTS) refers to
a)
Yield stress
b)
Proportional limit
c)
Maximum stress
d)
Fracture stress
|
Rithika Reddy answered |
The engineering stress-strain curve for mild steel is
Ultimate tensile strength represents the maximum stress that a material can withstand without Fracture.
Ultimate tensile strength represents the maximum stress that a material can withstand without Fracture.
Which of the following is the most appropriate theory of failure for mild steel?
- a)Maximum principal stress theory
- b)Maximum principal strain theory
- c)Maximum shear stress theory
- d)Maximum shear strain energy theory
Correct answer is option 'D'. Can you explain this answer?
Which of the following is the most appropriate theory of failure for mild steel?
a)
Maximum principal stress theory
b)
Maximum principal strain theory
c)
Maximum shear stress theory
d)
Maximum shear strain energy theory
|
Arjun Menon answered |
Maximum principal stress theory (Rankine’s theory)
According to this theory, permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.
For design criterion, the maximum principal stress (σ1) must not exceed the working stress ‘σy’ for the material.
Note: For no shear failure τ ≤ 0.57 σy
Note: For no shear failure τ ≤ 0.57 σy
Graphical representation
For brittle material, which do not fail by yielding but fail by brittle fracture, this theory gives satisfactory result.
The graph is always square even for different values of σ1 and σ2.
Maximum principal strain theory (ST. Venant’s theory)
According to this theory, a ductile material begins to yield when the maximum principal strain reaches the strain at which yielding occurs in simple tension.
For no failure in uni – axial loading.
For no failure in tri – axial loading.
For design, Here, ϵ = Principal strainσ1, σ2 and σ3 = Principal stresses
Graphical Representation
This story over estimate the elastic strength of ductile material.
Maximum shear stress theory
(Guest & Tresca’s Theory)
According to this theory, failure of specimen subjected to any combination of load when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.
Graphical Representation
σ1 and σ2 are maximum and minimum principal stress respectively.
σ1 and σ2 are maximum and minimum principal stress respectively.
Here, τmax = Maximum shear stress
σy = permissible stress
This theory gives satisfactory result for ductile material.
Maximum strain energy theory (Haigh’s theory)
According to this theory, a body complex stress fails when the total strain energy at elastic limit in simple tension.
Graphical Representation.
This theory does not apply to brittle material for which elastic limit stress in tension and in compression are quite different.
This theory does not apply to brittle material for which elastic limit stress in tension and in compression are quite different.
Maximum shear strain energy / Distortion energy theory / Mises – Henky theory.
It states that inelastic action at any point in body, under any combination of stress begging, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point in a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension / compression test.
It gives very good result in ductile material.
It cannot be applied for material under hydrostatic pressure.
All theories will give same results if loading is uniaxial.
If the modulus of elasticity is zero, the material is said to be- a)rigid
- b)elastic
- c)flexible
- d)plastic
Correct answer is option 'D'. Can you explain this answer?
If the modulus of elasticity is zero, the material is said to be
a)
rigid
b)
elastic
c)
flexible
d)
plastic
|
|
Prerna Menon answered |
E =0 Slope of Stress- Strain Curve is straight horizontal line. It means at constant stress, strain keeps increasing which is the characteristic property of perfectly plastic material.
A beam with a square section of 80 mm x 80 mm is simply supported at its ends. A load W is applied at the centre of the beam. If the maximum shear stress developed in the beam section is 6 N/mm2. What is the magnitude of W?- a)2.56 kN
- b)25.6 kN
- c)-51.2kN
- d)5.12 kN
Correct answer is option 'C'. Can you explain this answer?
A beam with a square section of 80 mm x 80 mm is simply supported at its ends. A load W is applied at the centre of the beam. If the maximum shear stress developed in the beam section is 6 N/mm2. What is the magnitude of W?
a)
2.56 kN
b)
25.6 kN
c)
-51.2kN
d)
5.12 kN
|
Saptarshi Khanna answered |
Given data:
Dimensions of the square section of the beam = 80 mm x 80 mm
Maximum shear stress developed in the beam section = 6 N/mm2
We need to find the magnitude of the load W.
Formula used:
Maximum shear stress (τmax) = (4/3) x (Shear force (V) / Area of cross-section (A))
Calculation:
Area of cross-section (A) = 80 mm x 80 mm = 6400 mm2
Shear force (V) at the center of the beam = W/2
Substituting the given values in the formula, we get:
6 = (4/3) x (W/2) / 6400
W/2 = 6 x 6400 x 3 / 4
W/2 = 72,000
W = 2 x 72,000
W = -144,000 N
Therefore, the magnitude of the load W is -51.2 kN (Option C).
Explanation:
The given beam is simply supported at its ends and has a square cross-section of 80 mm x 80 mm. The load W is applied at the center of the beam. The maximum shear stress developed in the beam section is 6 N/mm2. To find the magnitude of the load W, we use the formula for maximum shear stress and substitute the given values. On solving the equation, we get the magnitude of the load W as -51.2 kN. The negative sign indicates that the load is acting downwards.
Dimensions of the square section of the beam = 80 mm x 80 mm
Maximum shear stress developed in the beam section = 6 N/mm2
We need to find the magnitude of the load W.
Formula used:
Maximum shear stress (τmax) = (4/3) x (Shear force (V) / Area of cross-section (A))
Calculation:
Area of cross-section (A) = 80 mm x 80 mm = 6400 mm2
Shear force (V) at the center of the beam = W/2
Substituting the given values in the formula, we get:
6 = (4/3) x (W/2) / 6400
W/2 = 6 x 6400 x 3 / 4
W/2 = 72,000
W = 2 x 72,000
W = -144,000 N
Therefore, the magnitude of the load W is -51.2 kN (Option C).
Explanation:
The given beam is simply supported at its ends and has a square cross-section of 80 mm x 80 mm. The load W is applied at the center of the beam. The maximum shear stress developed in the beam section is 6 N/mm2. To find the magnitude of the load W, we use the formula for maximum shear stress and substitute the given values. On solving the equation, we get the magnitude of the load W as -51.2 kN. The negative sign indicates that the load is acting downwards.
A material has identical properties in all directions, it is said to be- a)homogeneous
- b)isotropic
- c)elastic
- d)orthotropic
Correct answer is option 'B'. Can you explain this answer?
A material has identical properties in all directions, it is said to be
a)
homogeneous
b)
isotropic
c)
elastic
d)
orthotropic
|
|
Divya Kulkarni answered |
Isotropic- Identical properties in all directions
Orthotropic- Different properties in all three directions
Homogeneous- A material of uniform composition
Orthotropic- Different properties in all three directions
Homogeneous- A material of uniform composition
A beam of T-section has I = 1200 cm4 and depth = 12 cm. Flange of the section is in compression. If the maximum tensile stress is two times the maximum compressive stress, what is section modulus in compression?- a)300 cm3
- b)200 cm3
- c)100 cm3
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
A beam of T-section has I = 1200 cm4 and depth = 12 cm. Flange of the section is in compression. If the maximum tensile stress is two times the maximum compressive stress, what is section modulus in compression?
a)
300 cm3
b)
200 cm3
c)
100 cm3
d)
None of these
|
|
Sanvi Kapoor answered |
An overhang beam of uniform El is loaded as shown
The deflection at the free end will be- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
An overhang beam of uniform El is loaded as shown
The deflection at the free end will be
The deflection at the free end will be
a)
b)
c)
d)
|
Gate Funda answered |
The reaction RB at B is, ∑MC = 0
⇒
∴
∴ Total strain energy
So by Castigliano’s theorem, deflection at free end,
⇒
∴
∴ Total strain energy
So by Castigliano’s theorem, deflection at free end,
Young’s modulus of elasticity and Poisson’s ratio of a material are 1.25 x 105 MPa and 0.34 respectively. The modulus of rigidity of the material is- a)0.4025 x105 MPa
- b)0.4664 x 105 MPa
- c)0.8375 x105 MPa
- d)0.9469 x105 MPa
Correct answer is option 'B'. Can you explain this answer?
Young’s modulus of elasticity and Poisson’s ratio of a material are 1.25 x 105 MPa and 0.34 respectively. The modulus of rigidity of the material is
a)
0.4025 x105 MPa
b)
0.4664 x 105 MPa
c)
0.8375 x105 MPa
d)
0.9469 x105 MPa
|
|
Sanya Agarwal answered |
= 0.4664 x 105 MPa
Match List-I with List-II and select the correct answer using the code given below the lists.List-I (Loaded Beam)A.
B.
C.
List-II (Maximum Bending moment)
1. 
2. 
3. 
4. 
- a)A-3, B-2, C-4
- b)A-4, B-1, C-3
- c)A-3, B-1, C-4
- d)A-4, B-2, C-3
Correct answer is option 'C'. Can you explain this answer?
Match List-I with List-II and select the correct answer using the code given below the lists.
List-I (Loaded Beam)
A.
B.
C.
List-II (Maximum Bending moment)
1.
2.
3.
4.
1.
2.
3.
4.
a)
A-3, B-2, C-4
b)
A-4, B-1, C-3
c)
A-3, B-1, C-4
d)
A-4, B-2, C-3
|
Lavanya Menon answered |
A-3, B-1, C-4
Beam A is cantilever so maximum bending moment occurs at the fixed support
Beam B is simply supported so maximum bending moment occurs at midspan
Beam C is cantilever so maximum bending moment occurs at fixed support
MC=
Beam B is simply supported so maximum bending moment occurs at midspan
Beam C is cantilever so maximum bending moment occurs at fixed support
MC=
The principal stresses at a point in a critical - section of a machine component are σ1 = 60 MPa, σ2 = 5 MPa and σ3 = – 40 MPa. For the material of the component, the tensile yield strength is σy = 200 MPa. According to the maximum shear stress theory, the factor of safety is
- a)1.67
- b)3.6
- c)4
- d)2
Correct answer is option 'D'. Can you explain this answer?
The principal stresses at a point in a critical - section of a machine component are σ1 = 60 MPa, σ2 = 5 MPa and σ3 = – 40 MPa. For the material of the component, the tensile yield strength is σy = 200 MPa. According to the maximum shear stress theory, the factor of safety is
a)
1.67
b)
3.6
c)
4
d)
2
|
|
Pallabi Tiwari answered |
The radius of Mohr’s circle of stress of a strained element is 20 N/mm2 and minor principal tensile stress is 10 N/mm2; The major principal stress is- a)30N/mm2
- b)50 N/mm2
- c)60 N/mm2
- d)100 N/mm2
Correct answer is option 'B'. Can you explain this answer?
The radius of Mohr’s circle of stress of a strained element is 20 N/mm2 and minor principal tensile stress is 10 N/mm2; The major principal stress is
a)
30N/mm2
b)
50 N/mm2
c)
60 N/mm2
d)
100 N/mm2
|
|
Anand Mehta answered |
Minor principal stress = C - r = 10
Centre of circle (C) = 10 + 20 =30 N/mm2
Major principal stress = C + r = 30 + 20 = 50 N/mm2
Centre of circle (C) = 10 + 20 =30 N/mm2
Major principal stress = C + r = 30 + 20 = 50 N/mm2
If a beam is subjected to a constant bending moment along its length then the shear force will
- a)also have a constant value every where along its length
- b)be zero at all sections along the beam
- c)be maximum at the centre and zero at the ends
- d)be maximum at the ends and zero at the centre
Correct answer is option 'B'. Can you explain this answer?
If a beam is subjected to a constant bending moment along its length then the shear force will
a)
also have a constant value every where along its length
b)
be zero at all sections along the beam
c)
be maximum at the centre and zero at the ends
d)
be maximum at the ends and zero at the centre
|
|
Milan Ghosh answered |
The relation between shear force (V) and bending moment (M) is:
dM/dx = V
it means the slope of a bending moment diagram will represent the magnitude of shear force at that section.
Since the bending moment is constant along the length, therefore its derivative i.e. shear force is equal to zero at all sections along the beam.
There is no shear force between the loads and the bending moment is constant for that section along the length and vice-versa.
There is no shear force between the loads and the bending moment is constant for that section along the length and vice-versa.
Additional Information
The relation between shear force (V) and loading rate (w) is:
dV/dx = w
it means a positive slope of the shear force diagram represents an upward loading rate.
The relation between loading rate and shear force can be written as:
An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square : (σcirtical)circular’ is- a)1 : 4
- b)3 : 4
- c)4 : 3
- d)4 : 1
Correct answer is option 'C'. Can you explain this answer?
An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square : (σcirtical)circular’ is
a)
1 : 4
b)
3 : 4
c)
4 : 3
d)
4 : 1
|
Shail Rane answered |
PRINCIPLE: (Same for question no. 11 and 12)A "contingent contract" is a contract to do or not to do something contingent on the occurrence of a particular event. Contingent contracts are permissible in law only if the events that they refer to are possible.FACT: A agrees to pay B a sum of money if B marries C. C marries D. Is this contract enforceable in law?- a)No, Contingency referred to is now impossible
- b)Yes. There is a possibility of C and D getting divorced or D dying
- c)Yes. Marriages are not permanent
- d)No. The law must assume that marriages are sacrosanct and in the eyes of law, the contingency is now impossible.
Correct answer is option 'D'. Can you explain this answer?
PRINCIPLE: (Same for question no. 11 and 12)
A "contingent contract" is a contract to do or not to do something contingent on the occurrence of a particular event. Contingent contracts are permissible in law only if the events that they refer to are possible.
FACT: A agrees to pay B a sum of money if B marries C. C marries D. Is this contract enforceable in law?
a)
No, Contingency referred to is now impossible
b)
Yes. There is a possibility of C and D getting divorced or D dying
c)
Yes. Marriages are not permanent
d)
No. The law must assume that marriages are sacrosanct and in the eyes of law, the contingency is now impossible.
|
Sonam joshi answered |
As per principal the Contingent contracts are permissible in law only if the events that they refer to are possible.
A solid steel shaft is surrounded by a copper shaft, such that, Isteel = 1/2 Icopper. If Gcopper = 1/2 Gsteel, what is the ratio of Ts/Tc if the composite shaft is subjected to a twisting moment?- a)2
- b)1
- c)0.5
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
A solid steel shaft is surrounded by a copper shaft, such that, Isteel = 1/2 Icopper. If Gcopper = 1/2 Gsteel, what is the ratio of Ts/Tc if the composite shaft is subjected to a twisting moment?
a)
2
b)
1
c)
0.5
d)
None of these
|
Sankar Dasgupta answered |
Both shaft will rotate through same degree.
Hence 
A hollow shaft of inner radius 30 mm and outer radius 50 mm is subjected to a twisting moment. If the shear stress developed at inner radius of shaft is 60 N/mm2. What is the maximum shear stress in shaft?- a)60 N/mm2
- b)75 N/mm2
- c)100 N/mm2
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
A hollow shaft of inner radius 30 mm and outer radius 50 mm is subjected to a twisting moment. If the shear stress developed at inner radius of shaft is 60 N/mm2. What is the maximum shear stress in shaft?
a)
60 N/mm2
b)
75 N/mm2
c)
100 N/mm2
d)
None of these
|
Ishita Patel answered |
Given:
Inner radius of shaft (r1) = 30 mm
Outer radius of shaft (r2) = 50 mm
Shear stress at inner radius (τ1) = 60 N/mm^2
To find:
Maximum shear stress in the shaft
Formula:
The shear stress in a hollow shaft can be calculated using the torsion formula:
τ = (T * r) / J
Where:
τ = Shear stress
T = Twisting moment
r = Radial distance from the center
J = Polar moment of inertia
Analysis:
In a hollow shaft, the maximum shear stress occurs at the outer radius (r2). We need to calculate the shear stress at the outer radius using the given shear stress at the inner radius.
Step-by-step solution:
1. Convert the radii from millimeters to meters.
r1 = 30 mm = 0.03 m
r2 = 50 mm = 0.05 m
2. Calculate the polar moment of inertia (J) for a hollow shaft.
J = (π/2) * (r2^4 - r1^4)
3. Since we have the shear stress at the inner radius (τ1) and need to find the maximum shear stress at the outer radius (τ2), we can use the ratio of radii to relate the two stresses.
τ1 / τ2 = r2 / r1
Substitute the given values:
60 / τ2 = 0.05 / 0.03
Cross-multiply and solve for τ2:
τ2 = (60 * 0.03) / 0.05
= 36 N/mm^2
Therefore, the maximum shear stress in the shaft is 36 N/mm^2.
Hence, the correct answer is option C) 36 N/mm^2.
Inner radius of shaft (r1) = 30 mm
Outer radius of shaft (r2) = 50 mm
Shear stress at inner radius (τ1) = 60 N/mm^2
To find:
Maximum shear stress in the shaft
Formula:
The shear stress in a hollow shaft can be calculated using the torsion formula:
τ = (T * r) / J
Where:
τ = Shear stress
T = Twisting moment
r = Radial distance from the center
J = Polar moment of inertia
Analysis:
In a hollow shaft, the maximum shear stress occurs at the outer radius (r2). We need to calculate the shear stress at the outer radius using the given shear stress at the inner radius.
Step-by-step solution:
1. Convert the radii from millimeters to meters.
r1 = 30 mm = 0.03 m
r2 = 50 mm = 0.05 m
2. Calculate the polar moment of inertia (J) for a hollow shaft.
J = (π/2) * (r2^4 - r1^4)
3. Since we have the shear stress at the inner radius (τ1) and need to find the maximum shear stress at the outer radius (τ2), we can use the ratio of radii to relate the two stresses.
τ1 / τ2 = r2 / r1
Substitute the given values:
60 / τ2 = 0.05 / 0.03
Cross-multiply and solve for τ2:
τ2 = (60 * 0.03) / 0.05
= 36 N/mm^2
Therefore, the maximum shear stress in the shaft is 36 N/mm^2.
Hence, the correct answer is option C) 36 N/mm^2.
For metallic minerals creep becomes an important consideration at- a)500°C
- b)550°C
- c)half of the melting point temperature on absolute scale
- d)any temperature
Correct answer is option 'C'. Can you explain this answer?
For metallic minerals creep becomes an important consideration at
a)
500°C
b)
550°C
c)
half of the melting point temperature on absolute scale
d)
any temperature
|
|
Nitya Nambiar answered |
The temperature at which the creep becomes an important consideration is called HOMOLOGOUS TEMPERATURE and this temperature is nearly half of the melting point temperature.
In the creep test, the following type of stress is applied to the specimen- a)uniaxial compression
- b)uniaxial tension
- c)biaxial compression or tension
- d)alternating stress
Correct answer is option 'B'. Can you explain this answer?
In the creep test, the following type of stress is applied to the specimen
a)
uniaxial compression
b)
uniaxial tension
c)
biaxial compression or tension
d)
alternating stress
|
Avinash Mehta answered |
In the creep test, uniaxial tension is applied to the specimen. The purpose of the test is to measure the deformation of a material under a sustained load, which is often known as creep deformation, over a certain period of time. The test is typically used to evaluate the behavior of a material when subjected to long-term loads and it can provide information on the strength and ductility of the material, as well as its ability to maintain its properties under sustained loads. The test requires a specialized testing machine, which can apply a constant load to the specimen while monitoring the deformation over time. The results are typically presented in the form of a creep curve, which plots the deformation of the specimen over time.
A beam is of i-section with flanges 200 mm x 10 mm and web 180 mm x 10 mm. Due to the bending moment applied on the beam section, maximum stress developed in the beam section is 100 MPa, what is the stress developed at inner edge of the flange?- a)110 MPa
- b)100 MPa
- c)90 MPa
- d)120 MPa
Correct answer is option 'C'. Can you explain this answer?
A beam is of i-section with flanges 200 mm x 10 mm and web 180 mm x 10 mm. Due to the bending moment applied on the beam section, maximum stress developed in the beam section is 100 MPa, what is the stress developed at inner edge of the flange?
a)
110 MPa
b)
100 MPa
c)
90 MPa
d)
120 MPa
|
|
Partho Jain answered |
A shaft subjected to pure torsion is to be designed which of the following theories gives the largest diameter of shaft?- a)Maximum principal stress
- b)Maximum shear stress theory
- c)Strain energy theory
- d)All the above theories give same diameter
Correct answer is option 'B'. Can you explain this answer?
A shaft subjected to pure torsion is to be designed which of the following theories gives the largest diameter of shaft?
a)
Maximum principal stress
b)
Maximum shear stress theory
c)
Strain energy theory
d)
All the above theories give same diameter
|
|
Kirti Sharma answered |
It is clear from the relation T/J = (Shear Stress)/ (Radial Distance) that a shaft subjected to pure torsion is to be designed for maximum shear stress theory.
At a point in a steel member, major principal stress is 1000 kg/cm2, and minor principal stress is compressive. If uniaxial tensile yield stress is 1500 kg/cm2, then the magnitude of minor principal stress at which yielding will commence, according to maximum shearing stress theory is- a)500 kg/cm2
- b)250 kg/cm2
- c)1000 kg/cm2
- d)0
Correct answer is option 'A'. Can you explain this answer?
At a point in a steel member, major principal stress is 1000 kg/cm2, and minor principal stress is compressive. If uniaxial tensile yield stress is 1500 kg/cm2, then the magnitude of minor principal stress at which yielding will commence, according to maximum shearing stress theory is
a)
500 kg/cm2
b)
250 kg/cm2
c)
1000 kg/cm2
d)
0
|
|
Pallabi Chavan answered |
But σ2 is compressive
A simply supported beam of uniform flexural rigidity is loaded as shown in the given figure. The rotation at the end A is
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A simply supported beam of uniform flexural rigidity is loaded as shown in the given figure. The rotation at the end A is
a)
b)
c)
d)
|
|
Rajdeep Gupta answered |
The 
diagram of beam is,
diagram of beam is,
The slope at mid span is zero.
The difference between the slope at two points is 
area between these points.
area between these points.So slope at A,
A cantilever of uniform strength σ, having rectangular section of constant breadth b but variable depth d is subjected to a UDL throughout its length. If the depth of the section is 150 mm at the fixed end, then what is the depth of the middle of the length of cantilever- a)150 mm
- b)100 mm
- c)75 mm
- d)125 mm
Correct answer is option 'C'. Can you explain this answer?
A cantilever of uniform strength σ, having rectangular section of constant breadth b but variable depth d is subjected to a UDL throughout its length. If the depth of the section is 150 mm at the fixed end, then what is the depth of the middle of the length of cantilever
a)
150 mm
b)
100 mm
c)
75 mm
d)
125 mm
|
|
Aditi Sarkar answered |
A cantilever of uniform strength is a beam that is fixed at one end and extends horizontally, with a cross-sectional area that remains constant throughout its length. This means that the beam has the same strength and stiffness at every point along its length, making it ideal for supporting loads that are distributed evenly across its surface.
The strength of a cantilever beam is determined by its cross-sectional area, material properties, and length. The longer the beam, the greater the bending moment that it can withstand before it fails. The material properties of the beam, such as its modulus of elasticity and yield strength, also play a significant role in determining its strength.
A cantilever beam of uniform strength is often used in construction applications, such as in the design of bridges, buildings, and other structures. These beams can be made from a variety of materials, including steel, concrete, and wood, depending on the specific requirements of the project.
In summary, a cantilever of uniform strength is a strong and stable beam that can support loads evenly across its surface, making it a popular choice for many construction applications.
The strength of a cantilever beam is determined by its cross-sectional area, material properties, and length. The longer the beam, the greater the bending moment that it can withstand before it fails. The material properties of the beam, such as its modulus of elasticity and yield strength, also play a significant role in determining its strength.
A cantilever beam of uniform strength is often used in construction applications, such as in the design of bridges, buildings, and other structures. These beams can be made from a variety of materials, including steel, concrete, and wood, depending on the specific requirements of the project.
In summary, a cantilever of uniform strength is a strong and stable beam that can support loads evenly across its surface, making it a popular choice for many construction applications.
If ∈1 and ∈2 ( ∈1 > ∈2) are the m axim um and minimum strains in the neighbourhood of a point in a stressed material of Young’s modulus E and Poisson’s ratio p, then the maximum principal stress will be given by- a)E∈1
- b)E(∈1 + ∈2)
- c)E(∈1 + μ∈2)/(1-μ2)
- d)E(∈2 + μ∈1)/(1-μ2)
Correct answer is option 'C'. Can you explain this answer?
If ∈1 and ∈2 ( ∈1 > ∈2) are the m axim um and minimum strains in the neighbourhood of a point in a stressed material of Young’s modulus E and Poisson’s ratio p, then the maximum principal stress will be given by
a)
E∈1
b)
E(∈1 + ∈2)
c)
E(∈1 + μ∈2)/(1-μ2)
d)
E(∈2 + μ∈
1
)/(1-μ2)|
|
Manasa Sen answered |
Materials having elongation less than 5% are considered brittle. In such cases, factor of safety is based on- a)yield stress
- b)endurance limit
- c)limit of proportionality
- d)ultimate stress
Correct answer is option 'C'. Can you explain this answer?
Materials having elongation less than 5% are considered brittle. In such cases, factor of safety is based on
a)
yield stress
b)
endurance limit
c)
limit of proportionality
d)
ultimate stress
|
|
Neha Kumar answered |
-he limit of proportionality refers to the point beyond which Hooke's law is no longer true when stretching a material.
-When an elastic object is stretched beyond its limit of proportionality, the object does not return to its original length when the force is removed.
A thin cylinder of unit length, thickness ‘t’ and radius ‘r’ is subjected to internal pressure ‘p’. What is the circumferential stress?- a)pr/2Et
- b)pr/2t
- c)pr/t
- d)2pr/t
Correct answer is option 'C'. Can you explain this answer?
A thin cylinder of unit length, thickness ‘t’ and radius ‘r’ is subjected to internal pressure ‘p’. What is the circumferential stress?
a)
pr/2Et
b)
pr/2t
c)
pr/t
d)
2pr/t
|
|
Anisha Chakraborty answered |
Circumferential stress is nothing but hoop stress.
For a cylindrical bar of 30 mm dia& 900 mm length, during a tension test, it is found that longitudinal strain is 4 times of lateral strain. The bulk modulus for the bar is (X) × 105 N/mm2 (if E = 3 × 105 N/mm2) The value if X is ___________
Correct answer is '2'. Can you explain this answer?
For a cylindrical bar of 30 mm dia& 900 mm length, during a tension test, it is found that longitudinal strain is 4 times of lateral strain. The bulk modulus for the bar is (X) × 105 N/mm2 (if E = 3 × 105 N/mm2) The value if X is ___________
|
|
Amar Desai answered |
E = 2 × 105 N/mm2 & E = 3k (1 – 2μ) ____________________(1)
From (1) we get K = E/3(1−2μ) [∴μ=0.25 given]
put μ=0.25 in equation 1,we get
k=2*105N/mm2.
From (1) we get K = E/3(1−2μ) [∴μ=0.25 given]
put μ=0.25 in equation 1,we get
k=2*105N/mm2.
In metal forming operation when the material has just started yielding, the principal stresses are σ1 = +180 MPa, σ2 = -100 MPa, σ3 = 0. Following the von Mises criterion, the yield stress is ________ MPa.
Correct answer is between '245,246'. Can you explain this answer?
In metal forming operation when the material has just started yielding, the principal stresses are σ1 = +180 MPa, σ2 = -100 MPa, σ3 = 0. Following the von Mises criterion, the yield stress is ________ MPa.
|
|
Prerna Kaur answered |
According to von Mises theory, for yielding
σyt = 245.76 MPa
σyt = 245.76 MPa
Hoop stress and longitudinal stress in a boiler shell under internal pressure are 100 MN/m2 and 50 MN/m2respectively. Young’s modulus of elasticity and Poisson’s ratio of the shell material are 200 GN/m2 and 0.3 respectively. The hoop strain in boiler shell is- a)0.425 x 10-3
- b)0.500 x 10-3
- c)0.575 x 10-3
- d)0.750 x 10-3
Correct answer is option 'A'. Can you explain this answer?
Hoop stress and longitudinal stress in a boiler shell under internal pressure are 100 MN/m2 and 50 MN/m2respectively. Young’s modulus of elasticity and Poisson’s ratio of the shell material are 200 GN/m2 and 0.3 respectively. The hoop strain in boiler shell is
a)
0.425 x 10-3
b)
0.500 x 10-3
c)
0.575 x 10-3
d)
0.750 x 10-3
|
|
Kirti Sharma answered |
If the magnitude of shear force is constant, then the magnitude of the slope of bending moment curve is
- a)zero
- b)increasing
- c)constant
- d)decreasing
Correct answer is option 'C'. Can you explain this answer?
If the magnitude of shear force is constant, then the magnitude of the slope of bending moment curve is
a)
zero
b)
increasing
c)
constant
d)
decreasing
|
|
Garima Kulkarni answered |
Explanation:
The following point should always be kept in mind while drawing SFD and BMD:
1. The rate of change of the shear force at any point on the axis of the beam is equal to the negative of the intensity of the distributed load at that same point.
i.e. w=−dV/dx
Here, negative sign represents that the loading is acting downward.
2. The rate of change of the bending moment at any point on the axis of a beam is equal to the shear force at that same point.
i.e. V=dM/dx
Calculation: -
From (2),
We have,
dM/dx=V=constant
Thus,
The magnitude of slope of moment diagram is also constant.
The maximum bending moment under a particular point load among a train of point loads crossing a simply Supported girder occurs at the location when that load is at - a)at mid span
- b)so placed that load point and the CG of the train of loads coincides
- c)at one-quarter span
- d)so placed that load point and the CG of the train of loads equi-distant from the mid span.
Correct answer is option 'D'. Can you explain this answer?
The maximum bending moment under a particular point load among a train of point loads crossing a simply Supported girder occurs at the location when that load is at
a)
at mid span
b)
so placed that load point and the CG of the train of loads coincides
c)
at one-quarter span
d)
so placed that load point and the CG of the train of loads equi-distant from the mid span.
|
Vertex Academy answered |
Position of Maximum Bending moment under a particular point load among a train of point loads:
Conclusion: The maximum bending moment under a particular point load among a train of point loads crossing a simply Supported girder occurs at the location when that load is at a point so that the load and the resultant should be equidistant from the mid-span.
Conclusion: The maximum bending moment under a particular point load among a train of point loads crossing a simply Supported girder occurs at the location when that load is at a point so that the load and the resultant should be equidistant from the mid-span.
Euler’s formula for a mild steel long column hinged at both end is not valid for slenderness ratio- a)greater than 80
- b)less than 80
- c)greater than 180
- d)greater than 120
Correct answer is option 'B'. Can you explain this answer?
Euler’s formula for a mild steel long column hinged at both end is not valid for slenderness ratio
a)
greater than 80
b)
less than 80
c)
greater than 180
d)
greater than 120
|
|
Lekshmi Rane answered |
Hence, when the slenderness ratio is less than this limit for mild steel columns , Euler’s formula will not be valid.
A cantilever beam of span ‘l’ and uniform flexural rigidity ‘EI’ is loaded with an upward force ‘W’ at the mid point and downward force ‘P’ at the free end. The deflection at the free end will be zero if - a)W = 3P/2
- b)W = 2P
- c)W = 16P/5
- d)W = 5P
Correct answer is option 'C'. Can you explain this answer?
A cantilever beam of span ‘l’ and uniform flexural rigidity ‘EI’ is loaded with an upward force ‘W’ at the mid point and downward force ‘P’ at the free end. The deflection at the free end will be zero if
a)
W = 3P/2
b)
W = 2P
c)
W = 16P/5
d)
W = 5P
|
|
Ameya Deshpande answered |
Upward deflection at B due to load W,
Downward deflection at S due to load P,
For no deflection at B,
⇒ W = 16P/5
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