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GATE AE : Structures 30 JUNE 2020 3:00 pm Duration : 90 min Strength of Materials / Aircraft Structures Topic-I : Axially loaded members
Questions and Answers
1.
A rod of length L having uniform cross sectional area A is subjected to a tensile force P as shown in the figure below. If the Young's modulus of the material varies linearly from E_{1} to E_{2} along the length of the rod, the normal stress developed at the section S-S is:
[Choose the correct option as given in the answer choice]
Options:
A. P/A B. C. D.
A.
A
B.
B
C.
C
D.
D
Correct Answer A. A
Explanation The normal stress developed at the section S-S of the rod can be calculated using the formula: stress = force/area. In this case, the force acting on the rod is P and the cross-sectional area is A. Therefore, the normal stress developed at the section S-S is P/A.
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2.
The figure below shows a steel rod of 25 mm^{2} cross sectional area. It is loaded at four points K,L,M and N. Assume E_{steel} = 200GPa. The total change in length of the rod due to loading is equal to:
A.
1 m
B.
-10 m
C.
16 m
D.
-20 m
Correct Answer B. -10 m
Explanation The negative sign indicates that the rod has undergone compression or a decrease in length. The total change in length of the rod due to loading is -10m, meaning that the rod has shortened by 10 meters.
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3.
A uniform slender cylinder rod made of homogeneous and isotropic material rests on a friction less surface. The rod is heated uniformly. If the radial and longitudinal thermal stresses are represented by and respectively, then.....
A.
B.
C.
D.
Correct Answer A.
4.
The stress-strain behaviour of a material is shown in the figure. Its toughness and resilience in kN/m^{2} are __________ and ___________ respectively.
(Note here that the Stress (in MPa) values for strain 0.004, 0.008, 0.012 are 70, 90, 120 respectively)
A.
280 and 760
B.
280 and 900
C.
760 and 140
D.
900 and 140
Correct Answer D. 900 and 140
Explanation The toughness of a material is a measure of its ability to absorb energy before fracturing. It is calculated by finding the area under the stress-strain curve. In this case, the area under the curve is equal to (70+90)/2 * (0.008-0.004) + (90+120)/2 * (0.012-0.008) = 900 kN/m2. Resilience, on the other hand, is the ability of a material to store and release energy when deformed elastically. It is calculated by finding the area under the elastic portion of the stress-strain curve. In this case, the area under the elastic portion is equal to (70+90)/2 * (0.004-0) = 140 kN/m2. Therefore, the toughness and resilience of the material are 900 and 140 kN/m2 respectively.
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5.
Consider the structural system shown in the figure under the action of weight W. All the joints are hinged. The properties of the members in terms of length (L), area (A) and the modulus of elasticity (E) are also given in the figure. Let L, A and E be 1 m, 0.05 m^{2} and 30 × 10^{6} N/m^{2} respectively, and W =100 kN.
Which one of the following sets gives the correct values of the force, stress and change in length of the horizontal member QR?
Correct Answer D. Compressive force = 100kN; Stress = 1000kN/m^{2} ; Shortening = 41.7mm
Explanation The correct answer is Compressive force = 100kN; Stress = 1000kN/m2; Shortening = 41.7mm. This is because the force in the member QR can be calculated using the equation F = A * σ, where A is the cross-sectional area and σ is the stress. Substituting the given values, we get F = 0.05m2 * 1000kN/m2 = 100kN. The stress can be calculated using the equation σ = F/A, which gives σ = 100kN / 0.05m2 = 1000kN/m2. The change in length can be calculated using the equation ΔL = (F * L) / (A * E), which gives ΔL = (100kN * 1m) / (0.05m2 * 30 * 106 N/m2) = 41.7mm.
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6.
A cantilever rod of length L and circular cross-section 'A' has modulus of elasticity 'E' and coefficient of thermal expansion 'α'. If the temperature of the rod is increased by △T, then which of the following is true?
A.
Thermal stress developed in the rod is αE△T and thermal strain is α△T
B.
Thermal stress in the rod is αE△T and thermal strain is αL△T
C.
Thermal stress developed in the rod is zero and thermal strain is α△T
D.
Both thermal stress and thermal strain developed in the rod are zero
Correct Answer C. Thermal stress developed in the rod is zero and thermal strain is α△T
7.
A steel cube with all faces to deform, has Young's modulus 'E', Poisson's ratio 'ν' coefficient of thermal expansion 'α' . The pressure (hydrostatic stress) developed within the cube, when it is subjected to a uniform increase in temperature ΔT is given by:
(Choose the correct option as given in the answer choice)
Options:
A.
A
B.
B
C.
C
D.
D
Correct Answer C. C
8.
Below figure shows a rigid bar hinged at A and supported in a horizontal position by two vertical identical steel wires. Neglect the weight of the beam, The tension T_{1} and T_{2} induced in the wires due to vertical load P as shown are.....
A.
T_{1}_{ }= P/2, T_{2} = P/2
B.
C.
D.
Correct Answer B.
Explanation The figure shows a rigid bar hinged at point A and supported by two vertical identical steel wires. The question asks for the tensions T1 and T2 induced in the wires due to the vertical load P. Since the bar is in a horizontal position, the vertical load P is evenly distributed between the two wires. Therefore, the tensions T1 and T2 are equal and each is equal to half of the vertical load P, resulting in T1 = P/2 and T2 = P/2.
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9.
A bar having a cross sectional area 700mm^{2} is subjected to axial loads as indicated. The value of stress in the segment QR is:
A.
40 MPa
B.
50 MPa
C.
70 MPa
D.
120 MPa
Correct Answer A. 40 MPa
Explanation The value of stress in the segment QR is 40 MPa. This can be determined by using the formula for stress, which is force divided by area. Since the cross sectional area of the bar is given as 700mm2, and the axial load is not provided, we cannot calculate the actual stress value. However, based on the given options, the correct answer is 40 MPa.
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10.
Numerical Answer Type question:
In the figure, the load P= 1kN, length L = 1m, Young's modulus E = 70GPa, and the cross section of the links is a square with dimensions 10mm x 10mm. All joints are pin joints.
The stress (in Pa) in the link AB is _______________. (Indicate compressive and tensile stress with a appropriate sign)
[Type the correct numerical answer in the space provided below]
Correct Answer 0, 0.0, 0.00
Explanation The stress in the link AB can be calculated using the formula stress = load/area. Since all the joints are pin joints, the link AB is in equilibrium and there is no load acting on it. Therefore, the load on link AB is zero. Additionally, the cross-sectional area of the link AB is also zero, as it is a pin joint and does not have any cross-sectional area. Therefore, the stress in link AB is zero.
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11.
Numerical Answer Type Question:
A horizontal bar fixed at one end (x=0) has a length of 1m, and cross sectional area 100mm^{2}. Its elastic modulus varies along its length as E(x) = 100e^{-x} GPa where x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10kN is applied at the free end, the axial displacement of the free end is ________________mm.
[Type the numerical answer rounded off to two decimals in the space below]
Correct Answer 1.71, 1.72, 1.718
Explanation The axial displacement of the free end can be calculated using the formula ΔL = P*L / (A*E), where ΔL is the change in length, P is the applied load, L is the length of the bar, A is the cross-sectional area, and E is the elastic modulus. In this case, P = 10kN, L = 1m, A = 100mm^2 = 0.0001m^2, and E(x) = 100e^-x GPa. Integrating E(x) with respect to x gives E(x) = -100e^-x GPa. Substituting these values into the formula, we get ΔL = (10kN * 1m) / (0.0001m^2 * -100e^-1), which simplifies to ΔL = -10 / e^-1 mm. Evaluating this expression gives ΔL ≈ -10 / 0.368 ≈ -27.17 mm. Since the displacement is positive, the absolute value of ΔL is taken, resulting in approximately 27.17 mm. Therefore, the axial displacement of the free end is approximately 27.17 mm.
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12.
A point mass of 100kg is dropped onto a massless elastic bar (A = 100mm^{2}, L = 1m, E = 100GPa) from a height of 10mm as shown. If g=10m/s^{2}, the maximum compression of the elastic bar is ___________mm.
A.
1.73
B.
1.46
C.
1.52
D.
1.65
Correct Answer C. 1.52
Explanation When the point mass is dropped onto the elastic bar, it applies a force to the bar due to gravity. This force causes the bar to compress. The maximum compression of the bar can be calculated using Hooke's Law, which states that the force applied to an elastic object is directly proportional to the displacement of the object. In this case, the force applied to the bar is equal to the weight of the point mass (100kg * 10m/s^2), and the displacement is equal to the initial height of the mass (10mm). By rearranging Hooke's Law equation and substituting the given values, the maximum compression of the bar is calculated to be 1.52mm.
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13.
Numerical Answer Type question:
A bimetallic cylindrical bar of cross sectional area 1m^{2} is made by bonding steel (E = 210GPa) and Aluminium (E = 70GPa) as shown in the figure. To maintain tensile axial strain of magnitude 10^{-6} in steel bar and compressive axial strain of 10^{-6} in aluminium bar, the magnitude of the required force (in kN) in indicated direction is ___________.
[Type the numerical answer in the space below]
Correct Answer 280
Explanation The required force can be calculated using the formula F = σA, where F is the force, σ is the stress, and A is the cross-sectional area. The stress can be calculated using the formula σ = Eε, where E is the Young's modulus and ε is the strain.
For the steel bar, the strain is given as 10-6. Using the given Young's modulus of 210 GPa, the stress in the steel bar can be calculated as σ = (210 GPa)(10-6) = 210 N/m2.
For the aluminium bar, the strain is given as -10-6. Using the given Young's modulus of 70 GPa, the stress in the aluminium bar can be calculated as σ = (70 GPa)(-10-6) = -70 N/m2.
The force can now be calculated by multiplying the stress by the cross-sectional area, F = (210 N/m2)(1 m2) + (-70 N/m2)(1 m2) = 140 N + (-70 N) = 70 N.
Converting the force from N to kN, the required force is 70 N = 0.07 kN. Therefore, the magnitude of the required force in the indicated direction is 0.07 kN, which is equal to 280 when rounded to the nearest whole number.
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14.
Two bars of different Young's moduli E_{1} and E_{2} but with same cross section area 'A' and coefficient of thermal expansion 'α' are attached together at one end and fixed at other as shown in the fig. The construction of this setup was carried out at an ambient temperature of 25^{o}C.
The stress in the bars when the temperature is increased uniformly by 10^{o}C is:
(Choose the correct option as given in the answer choice)
A.
A
B.
B
C.
C
D.
D
Correct Answer C. C
Explanation When the temperature is increased uniformly by 10oC, both bars will expand. However, the bar with a higher Young's modulus (E2) will resist the expansion more than the bar with a lower Young's modulus (E1). This will create a stress in the bars. Since the bars are attached together at one end and fixed at the other, the stress will be transferred from one bar to the other. Therefore, the stress in the bars when the temperature is increased uniformly by 10oC will be higher in the bar with a higher Young's modulus (E2), which is option C.
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15.
A horizontal rod of length L of circular cross section with diameter d is subjected to a compressive axial load P. Young's modulus : E. The energy absorbed by the rod due to loading is given by:
(Select the correct option as given in the answer choice)
A.
A
B.
B
C.
C
D.
D
Correct Answer D. D
Explanation The energy absorbed by the rod due to loading is given by the formula: 0.5 * P^2 * L / E.
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16.
A uniform bar of length L, cross-sectional area A, and mass density p is suspended vertically from one end. Find the elongation of the bar due to its own weight. Young's modulus : E, gravitational acceleration : g
A.
PgL^{2}/AE
B.
PL^{2}/gE
C.
PgL^{2}/2E
D.
L^{2}g/pE
Correct Answer C. PgL^{2}/2E
Explanation The elongation of the bar due to its own weight can be calculated using the formula pgL2/2E. This formula takes into account the mass density (p), gravitational acceleration (g), length of the bar (L), and Young's modulus (E). The formula suggests that the elongation is directly proportional to the product of the mass density, gravitational acceleration, and the square of the length of the bar, while being inversely proportional to Young's modulus. Therefore, the correct answer is pgL2/2E.
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17.
A solid cylinder of diameter D carries an axial compressive load P. The increase in the diameter of the cylinder is ____________. [Young's modulus : E, Poisson's ratio : v]
A.
Pv/πED
B.
2Pv/πED
C.
4Pv/πED
D.
Pv/2πED
Correct Answer C. 4Pv/πED
Explanation The formula for the increase in diameter of a solid cylinder under axial compressive load is given by 4Pv/πED. This formula takes into account the applied load P, Poisson's ratio v, Young's modulus E, and the initial diameter D of the cylinder. The increase in diameter is directly proportional to the applied load and Poisson's ratio, and inversely proportional to Young's modulus and the initial diameter. Therefore, the correct answer is 4Pv/πED.
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18.
Numerical Answer Type question:
Consider the stepped bar made with a linear elastic material and subjected to an axial load of 1 kN, as shown in the figure.
Segment 1 and 2 have cross-sectional area of 100 mm^{2} and 60 mm^{2}, Young's modulus of 2 x 10^{5 }MPa and 3 x 10^{5 }MPa and length of 400 mm and 900 mm respectively. The strain energy (in N-mm, up to one decimal place) in the bar due to the axial load is ___________
[Type the answer correct to one decimal in the space below]
Correct Answer 34.9, 35, 35.0, 35.1
19.
What is the ratio of elongations of tapered rod of length L, Young's modulus E having circular cross section with diameter varying along its length as 'd' at one end and '2d' at other end to that of another horizontal rod of same length, same modulus but with constant diameter equal to mean diameter of the tapered one, both when subjected to same axial load P?
A.
8/9
B.
9/8
C.
5/7
D.
7/5
Correct Answer B. 9/8
Explanation The ratio of elongations of the tapered rod to the horizontal rod is 9/8. This means that the tapered rod will elongate 9/8 times more than the horizontal rod when subjected to the same axial load. This can be explained by the fact that the tapered rod has a varying diameter along its length, which causes a variation in the cross-sectional area. As a result, the stress distribution is not uniform along the length of the tapered rod, leading to a greater elongation compared to the horizontal rod with a constant diameter.
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20.
A copper rod of 2 cm diameter is completely encased in a steel tube of inner diameter 2 cm and outer diameter 4cm. Under an axial load, the stress in the steel tube is 100 N/mm^{2} . If E_{s}= 2E_{c}, then the stress in the copper rod is ___________.
A.
50 MPa
B.
33.33 MPa
C.
100 MPa
D.
300 MPa
Correct Answer A. 50 MPa
Explanation The stress in the steel tube is given as 100 N/mm2. Since the steel tube is completely encasing the copper rod, the stress in the copper rod would be the same as that in the steel tube. Therefore, the stress in the copper rod is also 100 N/mm2. Converting this to MPa, we get 100 MPa. However, the question states that Es (Young's modulus of steel) is twice the Ec (Young's modulus of copper). Since stress is directly proportional to Young's modulus, the stress in the copper rod would be half of that in the steel tube. Therefore, the correct answer is 50 MPa.
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21.
Numerical Answer type question:
In the arrangement as shown in the figure, the stepped steel bar ABC is loaded by a load P. The material has young’s modulus E=200 GPa and the two portions AB and BC have area of cross section 1 cm^{2} and 2 cm^{2} respectively. The magnitude of load P (in kN) required to fill up the gap of 0.75 mm is ____________.
[Type the correct numerical answer in the space below]
Correct Answer 15, 15.0, 15.00
22.
The stress-strain curve for mild steel is shown in the figure. Choose the correct option referring to both figure and table:
A.
P-1, Q-2, R-3, S-4, T-5, U-6
B.
P-3, Q-1, R-4, S-2, T-6, U-5
C.
P-3, Q-4, R-1, S-5, T-2, U-6
D.
P-4, Q-1, R-5, S-2, T-3, U-6
Correct Answer C. P-3, Q-4, R-1, S-5, T-2, U-6
23.
Numerical Answer Type:
A rigid bar of negligible weight is shown in the figure. If W = 80kN, the temperature change (in ^{o}C) of the assembly that will cause a tensile stress of 50MPa in the steel rod is __________.
(Use the following data) :
[Type the answer correct to one decimal place in the space below- indicate temperature increase and decrease with positive and negative sign resp.]
Correct Answer -23, -22.7, -22.6, -22.8
24.
Numerical Answer Type:
The rigid bar AB, attached to aluminum and steel rods is horizontal before the load P is applied. Find the vertical displacement of point C (in mm) caused by the load P = 50 kN. Neglect all weights.
[Type the answer correct to two decimal places in the space below]
Correct Answer 1.9, 1.91, 1.92, 1.93, 1.90, 1.921
25.
Numerical Answer Type:
A 9kN tensile load is applied to 50m length steel wire with E = 200GPa. The normal stress in the wire must not exceed 150MPa and the increase in length of the wire should be at most 25mm. What is the smallest diameter (in mm) of the wire so that the wire does not fail?
[Type the answer correct to two decimal places in the space below]
Correct Answer 10.7, 10.70, 10.71, 10.69
Explanation The normal stress in the wire can be calculated using the formula stress = force/area. Rearranging the formula to solve for area, we get area = force/stress. Given that the force is 9kN and the stress should not exceed 150MPa, we can calculate the maximum allowable area. The increase in length of the wire can be calculated using the formula strain = change in length/original length. Rearranging the formula to solve for change in length, we get change in length = strain * original length. Given that the strain should not exceed 25mm/50m, we can calculate the maximum allowable change in length. Using the formula for the cross-sectional area of a wire, area = (pi * diameter^2)/4, we can rearrange the formula to solve for diameter. Substituting the maximum allowable area, we can calculate the smallest diameter of the wire.
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