Question

m_a = 1146 n.m m_b = m_c = 350 n.m m_d = 446 n.m

Explanation:

Step1: Identify the problem type

This is a problem related to torsion in shafts, likely involving the calculation of torque in different segments. We need to analyze the torques at different sections (1-1, 2-2, 3-3) based on the given bending moments (actually torques here, maybe a notation mix-up).

Step2: Analyze section 1-1 (between B and C)

For section 1-1, the torque \( T_1 \) is equal to \( M_B \) (since it's in equilibrium, the torque at section 1-1 should balance \( M_B \)). So \( T_1 = M_B = 350 \, N.m \) (but wait, direction: since \( M_B \) is a counter - clockwise torque on B, the torque in the shaft between B and C (section 1 - 1) should be equal to \( M_B \) in magnitude, considering equilibrium.

Step3: Analyze section 2-2 (between C and A)

For section 2-2, we need to consider the torques from both \( M_B \) and \( M_C \). Since \( M_B \) and \( M_C \) are in opposite directions (from the diagram, \( M_B \) is on B, \( M_C \) is on C, and \( M_A \) is on A). Let's assume the direction of \( M_A \) is clockwise (from the diagram, \( M_A \) is a clockwise torque on A). So the torque at section 2 - 2, \( T_2 \) should be \( M_A - M_C - M_B \)? Wait, no, let's do equilibrium. The sum of torques on the left - hand side of section 2 - 2 should be zero. The torques acting are \( M_B \) (counter - clockwise), \( M_C \) (clockwise, since it's on C, opposite to \( M_B \)) and the torque at section 2 - 2 \( T_2 \). Wait, maybe a better approach: the torque in segment CA (between C and A) is \( T_2 = M_A - M_C - M_B \)? Wait, no, let's look at the directions. Let's take the right - hand rule. If \( M_B = 350 \, N.m \) (counter - clockwise), \( M_C = 350 \, N.m \) (clockwise, as per the diagram, the arrow for \( M_C \) is opposite to \( M_B \)), and \( M_A = 1146 \, N.m \) (clockwise). So for section 2 - 2, the torque \( T_2 \) is \( M_A - M_C - M_B \)? Wait, no, let's consider the equilibrium of the shaft segment. Let's take the segment from C to A. The torques acting on this segment are \( M_C \) (clockwise), \( M_B \) (counter - clockwise) and the torque at section 2 - 2 \( T_2 \), and \( M_A \) (clockwise) at A. Wait, maybe I messed up the notation. Actually, these are torques (twisting moments) applied to the shafts. So for the shaft between B and C (segment BC), the torque \( T_{BC}=M_B = 350 \, N.m \) (assuming \( M_B \) is the torque applied to B, so the shaft BC has a torque of \( 350 \, N.m \)).

For the shaft between C and A (segment CA), we have to consider the torques from B, C, and A. Let's sum the torques. The torque at section 2 - 2, \( T_{CA}\) should satisfy \( T_{CA}+M_B - M_C - M_A = 0 \)? Wait, no, let's do free - body diagram. Take the part of the shaft to the left of section 2 - 2. The torques acting on it are \( M_B \) (counter - clockwise), \( M_C \) (clockwise), and the torque \( T_{2}\) (the torque in the shaft at section 2 - 2, which is the internal torque). For equilibrium, the sum of torques about the shaft axis should be zero. So \( T_{2}+M_B - M_C = 0 \)? No, that's not right. Wait, maybe the diagram shows that \( M_A \) is a clockwise torque on A, \( M_B \) is a counter - clockwise torque on B, \( M_C \) is a counter - clockwise torque on C? Wait, the diagram has \( M_B \) as a counter - clockwise arrow on B, \( M_C \) as a counter - clockwise arrow on C, and \( M_A \) as a clockwise arrow on A, \( M_D \) as a clockwise arrow on D.

So for segment BC (between B and C), the internal torque \( T_1 \) (at section 1 - 1) is equal to \( M_B \) (since only \( M_B \) is acting on this segment). So \( T_1 = M_B = 350 \, N.m \).

For segment CA (between C and A), the internal torque \( T_2 \) (at section 2 - 2) is \( M_A - M_B - M_C \)? Wait, no. Let's consider the equilibrium of the shaft from B to A. The torques are \( M_B \) (counter - clockwise), \( M_C \) (counter - clockwise), and \( M_A \) (clockwise). So the internal torque at section 2 - 2 (between C and A) should be \( T_2 = M_A - M_B - M_C \). Wait, no, equilibrium: sum of torques = 0. Let's take the direction of clockwise as positive. Then \( M_A-(M_B + M_C)+T_2 = 0 \)? No, maybe I got the directions wrong. Let's use the right - hand rule: if a torque is clockwise when viewed from the right, it's negative, counter - clockwise is positive.

Given \( M_A = 1146 \, N.m \) (clockwise, so negative), \( M_B = 350 \, N.m \) (counter - clockwise, positive), \( M_C = 350 \, N.m \) (counter - clockwise, positive), \( M_D = 446 \, N.m \) (clockwise, negative).

For section 1 - 1 (BC segment):
The torque \( T_1 \) is equal to \( M_B \) (since only \( M_B \) is acting on this segment). So \( T_1 = 350 \, N.m \).

For section 2 - 2 (CA segment):
The torques acting on the left - hand side of section 2 - 2 are \( M_B \) (positive), \( M_C \) (positive), and the torque \( T_2 \) (internal torque), and the torque \( M_A \) (negative) on the right. Wait, no, the shaft is from B to C to A to D. So from B to C: torque is \( M_B \). From C to A: we have \( M_B \), \( M_C \), and \( M_A \). So the internal torque at section 2 - 2 (between C and A) is \( T_2 = M_A - M_B - M_C \)? Wait, \( M_A = 1146 \), \( M_B = 350 \), \( M_C = 350 \). So \( T_2=1146 - 350 - 350=446 \, N.m \)? Wait, no, that can't be. Wait, maybe the other way: the torque in segment CA is \( M_A-(M_B + M_C)\)? Wait, \( 1146-(350 + 350)=1146 - 700 = 446 \, N.m \). Wait, but \( M_D = 446 \, N.m \).

For segment AD (between A and D), the internal torque \( T_3 \) (at section 3 - 3) is equal to \( M_D \) (since only \( M_D \) is acting on this segment, considering equilibrium with \( M_A \), \( M_B \), \( M_C \)). Wait, no, let's check equilibrium for the whole shaft. The sum of torques should be zero: \( M_A-(M_B + M_C+M_D)=1146-(350 + 350 + 446)=1146 - 1146 = 0 \), which checks out.

So for section 1 - 1 (BC): \( T_1 = M_B = 350 \, N.m \)

For section 2 - 2 (CA): \( T_2 = M_A - M_B - M_C=1146 - 350 - 350 = 446 \, N.m \)

For section 3 - 3 (AD): \( T_3 = M_D = 446 \, N.m \) (wait, no, \( T_3 \) should be equal to \( M_D \) in magnitude, but direction: if \( M_D \) is clockwise, \( T_3 \) is counter - clockwise, but magnitude is 446 \( N.m \))

Wait, maybe the question is to find the torques at each section. Let's confirm:

- Section 1 - 1 (between B and C): The torque \( T_1 \) is equal to \( M_B \) (since the only torque acting on this segment is \( M_B \)). So \( T_1 = 350 \, N.m \)

- Section 2 - 2 (between C and A): We have to consider the torques from B, C, and A. The net torque at section 2 - 2 is \( M_A - M_B - M_C \) (using equilibrium: sum of torques on one side of the section should be zero). So \( T_2=1146 - 350 - 350 = 446 \, N.m \)

- Section 3 - 3 (between A and D): The torque \( T_3 \) is equal to \( M_D \) (since the only torque acting on this segment from the right is \( M_D \), and from the left, the net torque up to A is \( M_A - M_B - M_C = M_D \), so \( T_3 = M_D = 446 \, N.m \)

Answer:

Torques at sections: \( T_1 = 350 \, N.m \), \( T_2 = 446 \, N.m \), \( T_3 = 446 \, N.m \) (depending on the specific question, if it's to find torques at each section)