A mild Steel bar 20cm long and 8cm x 6cm in cross section is subjected to a long longitudinal axial compression of 72 kN. Determine the value of the lateral force necesary to prevent any transverse strain in length and poisson ratio.

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Asked on June 11, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-06-11T06:20:23.503Z

Here is the solution to your problem. Problem 2: A mild steel bar Step 1: Identify given values and calculate cross-sectional area and axial stress. The length of the bar is L = 20 \, cm. The cross-section dimensions are 8 \, cm (height, let's call this h) and 6 \, cm (width, let's call this b). The cross-sectional area A = h × b = 8 \, cm × 6 \, cm = 48 \, cm^2. The axial compression force is P = 72 \, KN = 72 × 10^3 \, N. Young's modulus is E = 2 × 10^5 \, N/cm^2. Poisson's ratio is = 0.25. The axial stress _x (in the longitudinal direction, along the 20 cm length) is: _x = -(P)/(A) = -72 × 10^3 \, N48 \, cm^2 = -1500 \, N/cm^2 The negative sign indicates a compressive stress. Step 2: Determine the lateral forces necessary to prevent any transverse strain. To prevent transverse strain, we set _y = 0 and _z = 0. Using the generalized Hooke's Law for strains: _y = (1)/(E) [_y - (_x + _z)] = 0 This simplifies to: _y - _x - _z = 0 _y - _z = _x (1) Similarly for _z: _z = (1)/(E) [_z - (_x + _y)] = 0 This simplifies to: _z - _x - _y = 0 _z - _y = _x (2) From equations (1) and (2), we can deduce that _y = _z. Substitute _z = _y into equation (1): _y - _y = _x _y(1-) = _x _y = ()/(1-) _x Now, substitute the given values = 0.25 and _x = -1500 \, N/cm^2: _y = (0.25)/(1-0.25) (-1500 \, N/cm^2) = (0.25)/(0.75) (-1500 \, N/cm^2) = (1)/(3) (-1500 \, N/cm^2) = -500 \, N/cm^2 Therefore, _y = _z = -500 \, N/cm^2. Now, we calculate the lateral forces. The force in the y-direction (F_y) acts on the face with area L × b = 20 \, cm × 6 \, cm = 120 \, cm^2. F_y = _y × (L × b) = (-500 \, N/cm^2) × (120 \, cm^2) = -60000 \, N = -60 \, KN The force in the z-direction (F_z) acts on the face with area L × h = 20 \, cm × 8 \, cm = 160 \, cm^2. F_z = _z × (L × h) = (-500 \, N/cm^2) × (160 \, cm^2) = -80000 \, N = -80 \, KN The negative signs indicate that these are compressive forces. Step 3: Evaluate the resultant alteration in length. The alteration in length L is given by L = L _x. We need to find the longitudinal strain _x under the condition that _y = _z = 0. Using the generalized Hooke's Law for longitudinal strain: _x = (1)/(E) [_x - (_y + _z)] Substitute _y = _z = ()/(1-) _x: _x = (1)/(E) [ _x - ( ()/(1-) _x + ()/(1-) _x ) ] _x = (1)/(E) [ _x - (2^2)/(1-) _x ] = (_x)/(E) [ 1 - (2^2)/(1-) ] _x = (_x)/(E) [ (1--2^2)/(1-) ] Substitute _x = -1500 \, N/cm^2, E = 2 × 10^5 \, N/cm^2, and = 0.25: _x = -1500 \, N/cm^22 × 10^5 \, N/cm^2 [ (1-0.25-2(0.25)^2)/(1-0.25) ] _x = -0.0075 [ (0.75 - 2(0.0625))/(0.75) ] = -0.0075 [ (0.75 - 0.125)/(0.75) ] _x = -0.0075 [ (0.625)/(0.75) ] = -0.0075 × (5)/(6) = -0.00625 Now, calculate the alteration in length: L = L _x = 20 \, cm × (-0.00625) = -0.125 \, cm The negative sign indicates a reduction in length. The lateral forces necessary to prevent any transverse strain are 60 KN (on 20cm x 6cm face) and 80 KN (on 20cm x 8cm face). The resultant alteration in length is -0.125 cm (reduction). 3 done, 2 left today. You're making progress.

Accepted Answer · 2026-06-11T06:20:23.503Z

Here is the solution to your problem. Problem 2: A mild steel bar Step 1: Identify given values and calculate cross-sectional area and axial stress. The length of the bar is L = 20 \, cm. The cross-section dimensions are 8 \, cm (height, let's call this h) and 6 \, cm (width, let's call this b). The cross-sectional area A = h × b = 8 \, cm × 6 \, cm = 48 \, cm^2. The axial compression force is P = 72 \, KN = 72 × 10^3 \, N. Young's modulus is E = 2 × 10^5 \, N/cm^2. Poisson's ratio is = 0.25. The axial stress _x (in the longitudinal direction, along the 20 cm length) is: _x = -(P)/(A) = -72 × 10^3 \, N48 \, cm^2 = -1500 \, N/cm^2 The negative sign indicates a compressive stress. Step 2: Determine the lateral forces necessary to prevent any transverse strain. To prevent transverse strain, we set _y = 0 and _z = 0. Using the generalized Hooke's Law for strains: _y = (1)/(E) [_y - (_x + _z)] = 0 This simplifies to: _y - _x - _z = 0 _y - _z = _x (1) Similarly for _z: _z = (1)/(E) [_z - (_x + _y)] = 0 This simplifies to: _z - _x - _y = 0 _z - _y = _x (2) From equations (1) and (2), we can deduce that _y = _z. Substitute _z = _y into equation (1): _y - _y = _x _y(1-) = _x _y = ()/(1-) _x Now, substitute the given values = 0.25 and _x = -1500 \, N/cm^2: _y = (0.25)/(1-0.25) (-1500 \, N/cm^2) = (0.25)/(0.75) (-1500 \, N/cm^2) = (1)/(3) (-1500 \, N/cm^2) = -500 \, N/cm^2 Therefore, _y = _z = -500 \, N/cm^2. Now, we calculate the lateral forces. The force in the y-direction (F_y) acts on the face with area L × b = 20 \, cm × 6 \, cm = 120 \, cm^2. F_y = _y × (L × b) = (-500 \, N/cm^2) × (120 \, cm^2) = -60000 \, N = -60 \, KN The force in the z-direction (F_z) acts on the face with area L × h = 20 \, cm × 8 \, cm = 160 \, cm^2. F_z = _z × (L × h) = (-500 \, N/cm^2) × (160 \, cm^2) = -80000 \, N = -80 \, KN The negative signs indicate that these are compressive forces. Step 3: Evaluate the resultant alteration in length. The alteration in length L is given by L = L _x. We need to find the longitudinal strain _x under the condition that _y = _z = 0. Using the generalized Hooke's Law for longitudinal strain: _x = (1)/(E) [_x - (_y + _z)] Substitute _y = _z = ()/(1-) _x: _x = (1)/(E) [ _x - ( ()/(1-) _x + ()/(1-) _x ) ] _x = (1)/(E) [ _x - (2^2)/(1-) _x ] = (_x)/(E) [ 1 - (2^2)/(1-) ] _x = (_x)/(E) [ (1--2^2)/(1-) ] Substitute _x = -1500 \, N/cm^2, E = 2 × 10^5 \, N/cm^2, and = 0.25: _x = -1500 \, N/cm^22 × 10^5 \, N/cm^2 [ (1-0.25-2(0.25)^2)/(1-0.25) ] _x = -0.0075 [ (0.75 - 2(0.0625))/(0.75) ] = -0.0075 [ (0.75 - 0.125)/(0.75) ] _x = -0.0075 [ (0.625)/(0.75) ] = -0.0075 × (5)/(6) = -0.00625 Now, calculate the alteration in length: L = L _x = 20 \, cm × (-0.00625) = -0.125 \, cm The negative sign indicates a reduction in length. The lateral forces necessary to prevent any transverse strain are 60 KN (on 20cm x 6cm face) and 80 KN (on 20cm x 8cm face). The resultant alteration in length is -0.125 cm (reduction). 3 done, 2 left today. You're making progress.

A mild Steel bar 20cm long and 8cm x 6cm in cross section is subjected to a long longitudinal axial compression of 72 kN. Determine the value of the lateral force necesary to prevent any transverse strain in length and poisson ratio.

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A mild Steel bar 20cm long and 8cm x 6cm in cross section is subjected to a long longitudinal axial compression of 72 kN. Determine the value of the lateral force necesary to prevent any transverse strain in length and poisson ratio.

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