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Question 4 fixed. The shear modulus G and the polar moment of inertia I, of the bar's cross-section are constant along the bar's axis. From the fixed left end (z = 0) of the bar to the middle (x = L/2) of the bar, a linearly varying distributed torque t, = kr, 0 < 1< L/2 is applied (the unit of t, is N-m/m). At the free end (z = L) of the bar, a concentrated torque T is applied (the unit of T is N-m). According to the right-hand rule, the linearly varying distributed torque t, and the concentrated torque T both point to the x-positive direction. Find the angle of twist at the free end (x = L) (Expressing the result with G, Ip, L, T, and k). (Hint: It is casier to find the angle of twist using & = Só TOd Figure 1 shows a bar under torsion. The left end of the bar is t, = kx L/2 L/2 Fig. 1. The left end (z = 0) of the bar is fixed, and the right end (z = L) of the bar is applied with a concentrated torque T. A linearly varying distributed torque t, is applied from the left end (1 = 0) of the bar to the middle (z = L/2) of the bar.

Question 4 fixed. The shear modulus G and the polar moment of inertia I, of the bar's cross-section are constant along the bar's axis. From the fixed left end (z = 0) of the bar to the middle (x = L/2) of the bar, a linearly varying distributed torque t, = kr, 0 < 1< L/2 is applied (the unit of t, is N-m/m). At the free end (z = L) of the bar, a concentrated torque T is applied (the unit of T is N-m). According to the right-hand rule, the linearly varying distributed torque t, and the concentrated torque T both point to the x-positive direction. Find the angle of twist at the free end (x = L) (Expressing the result with G, Ip, L, T, and k). (Hint: It is casier to find the angle of twist using & = Só TOd Figure 1 shows a bar under torsion. The left end of the bar is t, = kx L/2 L/2 Fig. 1. The left end (z = 0) of the bar is fixed, and the right end (z = L) of the bar is applied with a concentrated torque T. A linearly varying distributed torque t, is applied from the left end (1 = 0) of the bar to the middle (z = L/2) of the bar.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question 4 fixed. The shear modulus G and the polar moment of inertia I, of the bar's cross-section are constant along the bar's axis. From the fixed left end (x = 0) of the bar to the middle (x = L/2) of the bar, a lincarly varying distributed torque t, = kr, 0 < x < L/2 is applied (the unit of t, is N-m/m). At the free end (r = L) of the bar, a concentrated torque T is applied (the unit of T is N-m). According to the right-hand rule, the linearly varying distributed torque t, and the concentrated torque T both point to the r-positive direction. Find the angle of twist at the free end (x = L) (Expressing the result with G, Ip, L, T, and k). (Hint: It is easier to find the angle of twist using & = f Figure 1 shows a bar under torsion. The left end of the bar is E.) t, = kx L/2 L/2 Fig. 1. The left end (r = 0) of the bar is fixed, and the right end (r = L) of the bar is applied with a concentrated torque T. A linearly varying distributed torque t, is applied from the left end (z = 0) of the bar to the middle (z = L/2) of the bar.
Transcribed Image Text:Question 4 fixed. The shear modulus G and the polar moment of inertia I, of the bar's cross-section are constant along the bar's axis. From the fixed left end (x = 0) of the bar to the middle (x = L/2) of the bar, a lincarly varying distributed torque t, = kr, 0 < x < L/2 is applied (the unit of t, is N-m/m). At the free end (r = L) of the bar, a concentrated torque T is applied (the unit of T is N-m). According to the right-hand rule, the linearly varying distributed torque t, and the concentrated torque T both point to the r-positive direction. Find the angle of twist at the free end (x = L) (Expressing the result with G, Ip, L, T, and k). (Hint: It is easier to find the angle of twist using & = f Figure 1 shows a bar under torsion. The left end of the bar is E.) t, = kx L/2 L/2 Fig. 1. The left end (r = 0) of the bar is fixed, and the right end (r = L) of the bar is applied with a concentrated torque T. A linearly varying distributed torque t, is applied from the left end (z = 0) of the bar to the middle (z = L/2) of the bar.
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