QUESTION IMAGE
Question
- a circular rod which the diameter d=70mm is subjected to the forces as shown in the figure, f=4kn, a=0.6m the permissible stress is σ₀=120mpa try to check the strength of the rod with the third strength theory calculate the bending moment and draw the diagram.
Step1: Analyze Support Reactions
The rod is symmetrically loaded, so reactions at A and B: \( R_A = R_B = F = 4\,\text{kN} \) (vertical equilibrium: \( R_A + R_B = 2F \), symmetry gives \( R_A = R_B = F \)).
Step2: Shear Force (SF) and Bending Moment (BM) in Segments
- Segment A - C (0 to \( a \)):
Shear Force: \( V = R_A = 4\,\text{kN} \) (constant).
Bending Moment: \( M(x) = R_A \cdot x = 4x \) (linear, at \( x = 0 \), \( M = 0 \); at \( x = a \), \( M = 4a \)).
- Segment C - D ( \( a \) to \( 2a \)):
Shear Force: \( V = R_A - F = 0 \) (constant, since two equal downward forces cancel the reaction).
Bending Moment: \( M(x) = R_A \cdot x - F \cdot (x - a) = 4x - 4(x - 0.6) = 2.4\,\text{kN·m} \) (constant, as SF = 0).
- Segment D - B ( \( 2a \) to \( 3a \)):
Shear Force: \( V = R_A - 2F = -4\,\text{kN} \) (constant).
Bending Moment: \( M(x) = R_A \cdot x - F \cdot (x - a) - F \cdot (x - 2a) = 4x - 4(x - 0.6) - 4(x - 1.2) = -4x + 7.2 \) (linear, at \( x = 2a \), \( M = 2.4\,\text{kN·m} \); at \( x = 3a \), \( M = 0 \)).
Step3: Calculate Bending Moment Magnitudes
Given \( a = 0.6\,\text{m} \):
- At \( x = a \) (C) and \( x = 2a \) (D): \( M = 4 \times 0.6 = 2.4\,\text{kN·m} \) (maximum BM, constant in C-D segment).
Step4: Draw Bending Moment Diagram
- A-C (0 to \( 0.6\,\text{m} \)): Straight line from \( (0, 0) \) to \( (0.6, 2.4) \).
- C-D (0.6 to \( 1.2\,\text{m} \)): Horizontal line at \( M = 2.4\,\text{kN·m} \).
- D-B (1.2 to \( 1.8\,\text{m} \)): Straight line from \( (1.2, 2.4) \) to \( (1.8, 0) \).
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Bending Moment Values:
- At \( x = 0 \) (A): \( M = 0 \)
- At \( x = a \) (C) and \( x = 2a \) (D): \( M = 2.4\,\text{kN·m} \) (maximum)
- At \( x = 3a \) (B): \( M = 0 \)
Bending Moment Diagram:
- Linear increase from A to C, constant from C to D, linear decrease from D to B, with maximum \( M = 2.4\,\text{kN·m} \) in the middle segment (C-D).