Question

plot the diagram of the axial force of the rod
6kn a i b ii c iii d
10kn 8kn 4kn
i ii iii

Explanation:

Step1: Analyze Segment I (A - B)

The axial force in segment I (from A to B) is constant. The force at A is 6 kN (compressive, but we consider magnitude for diagram or use sign convention: let's assume tension positive, compression negative. So axial force \( F_{N1} = -6\) kN (or 6 kN compressive).

Step2: Analyze Segment II (B - C)

To find axial force in B - C, we sum forces from A to any point in B - C. Let's take a section in B - C. The forces acting: 6 kN (compressive) and 10 kN (tensile? Wait, no, the direction: the 10 kN is applied at B, direction? Wait, the diagram: at B, there is a 10 kN force (let's see the direction: the 6 kN is to the left at A, 10 kN is to the right? Wait, maybe better to use equilibrium. For segment B - C, the axial force \( F_{N2} = -6 + 10 = 4\) kN? Wait, no, maybe sign convention: if we take positive as tensile (pulling away from the section), then at A, force is 6 kN to the left (compressive, so \( F_{N1} = -6\) kN). Then at B, a 10 kN force to the right (tensile). So for segment B - C, the axial force is \( F_{N2} = -6 + 10 = 4\) kN (tensile). Then at C, there is an 8 kN force to the right? Wait, no, the 8 kN is applied at C, direction? Wait the diagram: A---I---B---II---C---III---D. Forces: 6 kN left at A, 10 kN right at B, 8 kN right at C, 4 kN right at D. Wait, no, maybe the 10 kN is to the left? Wait, the problem is to plot axial force diagram. Let's use the method of sections.

For segment I (A - B): consider a section between A and B. The axial force \( F_{N1}\) is equal to the force at A, but direction: if we take the left part, force at A is 6 kN left, so axial force in the rod (compressive) is 6 kN. So \( F_{N1} = -6\) kN (compression).

For segment II (B - C): take a section between B and C. The forces acting on the left part (A - section) are 6 kN left and 10 kN right. So net force: \( -6 + 10 = 4\) kN (tensile, so \( F_{N2} = 4\) kN).

For segment III (C - D): take a section between C and D. Forces on left part: 6 kN left, 10 kN right, 8 kN right. So net force: \( -6 + 10 + 8 = 12\) kN? Wait, no, wait the 4 kN at D is right. Wait, no, maybe the 10 kN is to the left, 8 kN to the left, 4 kN to the right? Wait, the diagram is a bit unclear, but let's re - examine.

Wait, the axial force diagram steps:

1. Start at A: axial force is 6 kN compressive (so at A, \( F_N = -6\) kN).

2. At B, there is a 10 kN force (let's assume it's a tensile force, i.e., pulling the rod to the right). So the axial force changes by + 10 kN (since it's a tensile load). So at B, the axial force becomes \( -6 + 10 = 4\) kN (tensile).

3. At C, there is an 8 kN force (tensile, pulling right), so axial force becomes \( 4 + 8 = 12\) kN? Wait, no, that can't be, because at D, there is a 4 kN force to the right, and the axial force at D should be equal to the force at D? Wait, no, the rod is in equilibrium, so the sum of forces should be zero: 6 (left) + 10 (right) + 8 (right) + 4 (right) = 6 - 10 - 8 - 4 = -16? No, that's not zero. Wait, maybe the 10 kN and 8 kN are to the left? Let's correct the force directions.

Assume all external forces: 6 kN left at A, 10 kN left at B, 8 kN left at C, 4 kN right at D. Then sum of forces: 6 + 10 + 8 - 4 = 20 kN left, which is not zero. So wrong.

Alternative: 6 kN left at A, 10 kN right at B, 8 kN left at C, 4 kN right at D. Sum: -6 + 10 - 8 + 4 = 0. Ah, that works! So forces: A: 6 kN left (compression), B: 10 kN right (tension), C: 8 kN left (compression), D: 4 kN right (tension).

Now, axial force diagram:

- Segment I (A - B): axial force \( F_{N1}\). Take section between A and B: left part has 6 kN left, so axial force in the rod is 6 kN compressive (\( F_{N1} = -6\) kN).

- Segment II (B - C): take section between B and C. Left part has 6 kN left and 10 kN right. So net force: \( -6 + 10 = 4\) kN (tensile, \( F_{N2} = 4\) kN).

- Segment III (C - D): take section between C and D. Left part has 6 kN left, 10 kN right, 8 kN left. So net force: \( -6 + 10 - 8 = -4\) kN (compressive, \( F_{N3} = -4\) kN). Then at D, the 4 kN right force: \( -4 + 4 = 0\), which makes sense (equilibrium).

Wait, that sums to zero. So:

- Axial force in A - B: constant at -6 kN (compressive).

- Axial force in B - C: constant at 4 kN (tensile).

- Axial force in C - D: constant at -4 kN (compressive).

Now, to plot the axial force diagram:

- For segment A - B (length I), draw a horizontal line at \( F_N = -6\) kN (or 6 kN compressive).

- At B, the axial force jumps by + 10 kN (from -6 to 4) due to the 10 kN right (tensile) force.

- For segment B - C (length II), draw a horizontal line at \( F_N = 4\) kN (tensile).

- At C, the axial force jumps by -8 kN (from 4 to -4) due to the 8 kN left (compressive) force.

- For segment C - D (length III), draw a horizontal line at \( F_N = -4\) kN (compressive).

- At D, the axial force jumps by + 4 kN (from -4 to 0) due to the 4 kN right (tensile) force, which closes the diagram (equilibrium).

So the axial force diagram has three segments:

1. A - B: \( F_N = -6\) kN (compressive).

2. B - C: \( F_N = 4\) kN (tensile).

3. C - D: \( F_N = -4\) kN (compressive).

To plot, we can use a coordinate system where the x - axis is the length of the rod (divided into segments I, II, III) and the y - axis is the axial force (with positive upwards for tensile, negative for compressive).

Answer:

The axial force diagram has three constant - force segments:

• Segment \( A - B \): Axial force \( F_N=-6\) kN (compressive).
• Segment \( B - C \): Axial force \( F_N = 4\) kN (tensile).
• Segment \( C - D \): Axial force \( F_N=-4\) kN (compressive).

To plot, draw horizontal lines for each segment at these force values, with jumps at \( B \) (jump of \( + 10\) kN) and \( C \) (jump of \( - 8\) kN) corresponding to the applied forces.