Question

draw the shear and moment diagram and identify the maximum bending moment and maximum shear force.

1.

Explanation:

Calculate reaction forces

Using the Static Equilibrium knowledge point

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Determine shear force equations

Using the Shear and Moment Diagrams knowledge point

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At \(x = 1^+\), \(V = 36\text{ kN}\). At \(x = 4\), \(V = -24\text{ kN}\).
The shear force is zero at \(36 - 20(x - 1) = 0 \implies x = 2.8\text{ m}\).

Determine bending moment equations

Using the Shear and Moment Diagrams knowledge point

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At \(x = 1\), \(M = -30\text{ kN}\cdot\text{m}\).
At \(x = 2.8\), \(M_{\text{max, local}} = -30(2.8) + 66(1.8) - 10(1.8)^2 = 2.4\text{ kN}\cdot\text{m}\).
At \(x = 4\), \(M = -48\text{ kN}\cdot\text{m}\).
At \(x = 5^-\), \(M = -24\text{ kN}\cdot\text{m}\).
At \(x = 5^+\), \(M = 36\text{ kN}\cdot\text{m}\).
At \(x = 6\), \(M = 0\text{ kN}\cdot\text{m}\).

Identify maximum values

Using the Shear and Moment Diagrams knowledge point

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Answer:

The reaction forces are:

$$ R_1 = 66\text{ kN}, \quad R_2 = 24\text{ kN} $$

The shear force values at key points along the beam (from left \(x=0\) to right \(x=6\text{ m}\)):

• \(x = 0\text{ m}\) to \(1^-\text{ m}\): \(V = -30\text{ kN}\)
• \(x = 1^+\text{ m}\): \(V = 36\text{ kN}\)
• \(x = 2.8\text{ m}\): \(V = 0\text{ kN}\)
• \(x = 4\text{ m}\) to \(6\text{ m}\): \(V = -24\text{ kN}\)

The bending moment values at key points:

• \(x = 0\text{ m}\): \(M = 0\text{ kN}\cdot\text{m}\)
• \(x = 1\text{ m}\): \(M = -30\text{ kN}\cdot\text{m}\)
• \(x = 2.8\text{ m}\): \(M = 2.4\text{ kN}\cdot\text{m}\) (local maximum)
• \(x = 4\text{ m}\): \(M = -48\text{ kN}\cdot\text{m}\)
• \(x = 5^-\text{ m}\): \(M = -24\text{ kN}\cdot\text{m}\)
• \(x = 5^+\text{ m}\): \(M = 36\text{ kN}\cdot\text{m}\)
• \(x = 6\text{ m}\): \(M = 0\text{ kN}\cdot\text{m}\)

Maximum Shear Force:

$$ |V_{\text{max}}| = 36\text{ kN} \quad (\text{at } x = 1^+\text{ m}) $$

Maximum Bending Moment:

$$ |M_{\text{max}}| = 48\text{ kN}\cdot\text{m} \quad (\text{at } x = 4\text{ m}) $$