Question: Problem 7.56 Part A Draw the shear diagram for the beam.…

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Anu Bhardwaj · Accepted Answer

Given: Beam ABC A : pinned support C : roller support Distributed load: 1.5 kN/m over the right 4 m span (from B to C) Lengths: AB = 2 m BC = 4 m Total span = 6 m Step 1: Support Reactions Let’s find vertical reactions at supports A and C . Let: R A R_A ​ = reaction at A (vertical) R C R_C  = reaction at C (vertical) Total distributed load: w = 1.5 kN/m w = 1.5 \, 	ext{kN/m}  over 4 m ⇒ Total load = 1.5 × 4 = 6 kN 1.5 	imes 4 = 6 \, 	ext{kN} = This load acts at the center of the 4 m segment , i.e., 2 m to the left of point C , or 4 m from point A . ➤ Moment about A: ∑ M A = 0 = − R C ⋅ 6 + 6 ⋅ 4 ⇒ R C = 4 kN \sum M_A = 0 = -R_C \cdot 6 + 6 \cdot 4 \Rightarrow R_C = 4 \, 	ext{kN} ➤ Vertical force equilibrium: ∑ F y = 0 = R A + R C − 6 ⇒ R A = 6 − 4 = 2 kN \sum F_y = 0 = R_A + R_C - 6 \Rightarrow R_A = 6 - 4 = 2 \, 	ext{kN} 📉 Part A: Shear Force Diagram (V) Break the beam into regions: 1. From A to B (0–2 m): No load → constant shear V = R A = 2 kN (horizontal line) V = R_A = 2 \, 	ext{kN} \quad 	ext{(horizontal line)} 2. From B to C (2–6 m): Uniform distributed load of 1.5 kN/m → linear decrease in shear. At x = 2 x = 2  m: V = 2 kN V = 2 \, 	ext{kN} At x = 6 x = 6 : V = 2 − 1.5 ⋅ ( 6 − 2 ) = 2 − 6 = − 4 kN V = 2 - 1.5 \cdot (6 - 2) = 2 - 6 = -4 \, 	ext{kN} At point C , support reaction R C = 4 kN R_C = 4 \, 	ext{kN}  brings shear back to zero: − 4 + 4 = 0 -4 + 4 = 0 ✅ Shear Diagram Summary: Horizontal line at 2 kN from A to B (0–2 m) Linear drop from 2 kN to -4 kN from B to C (2–6 m) Jump up 4 kN at C → ends at 0 Part B: Bending Moment Diagram (M) Use areas under the shear diagram. At A (x = 0): M = 0 M = 0 From A to B (0–2 m): Rectangle area under shear curve: M B = 2 ⋅ 2 = 4 kN . m M_B = 2 \cdot 2 = 4 \, 	ext{kN·m} From B to C (2–6 m): Area under triangle: Area = 1 2 ⋅ ( 4 ) ⋅ ( 2 + 4 ) = 1 2 ⋅ 4 ⋅ ( 2 − 4 ) = − 12 kN . m 	ext{Area} = \frac{1}{2} \cdot (4) \cdot (2 + 4) = \frac{1}{2} \cdot 4 \cdot (2 - 4) = -12 \, 	ext{kN·m} Moment at C = 4 − 12 = − 8 kN . m 4 - 12 = -8 \, 	ext{kN·m} ✅ Moment Diagram Summary: Increases linearly from A to B (0 to 4 kN·m) Decreases parabolically from B to C (4 to -8 kN·m) The moment at C is not zero due to roller support. 📊 Final Answers ✔️ Shear Diagram: Horizontal from 0–2 m at +2 kN Linear decline to -4 kN from 2–6 m Jump to 0 at C ✔️ Moment Diagram: Linear rise to 4 kN·m at B Parabolic drop to -8 kN·m at C

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Question: Problem 7.56 Part A Draw the shear diagram for the beam. Follow the sign convention. (Figure 1) Part B Draw the moment diagram for the beam. Follow the sign convention.

Anu Bhardwaj

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