For the beam loaded as shown in figure, Calculate the support reactions at A and B.[Figure: beam AF, 9 m long. A (left end) is a hinge support; B is a roller support 6 m from A. Dimensions from A: 2 m (point of 40 kN load and 20 kN·m couple), 1 m to…

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Q.6 (B) For the beam loaded as shown in figure, Calculate the support reactions at A and B. [Figure: beam AF, 9 m long. A (left end) is a hinge support; B is a roller support 6 m from A. Dimensions from A: 2 m (point of 40 kN load and 20 kN·m couple), 1 m to D (x = 3 m), 3 m to B (x = 6 m), 1 m to E (x = 7 m), 2 m to the free end F (x = 9 m). Loads: 40 kN vertical downward at x = 2 m; a clockwise couple of 20 kN·m at the same point (x = 2 m); a triangular (uniformly varying) load increasing from zero at D to 24 kN/m (printed as “24 kNm”) at B; a vertical downward load P at E. No numerical value of P is printed in the paper.]

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Note: the magnitude of P is not printed in the paper, so the reactions are obtained in terms of P (substitute the value of P if supplied). The 20 kN·m couple is read as clockwise from the arrow in the figure; the UVL intensity “24 kNm” is taken as 24 kN/m. UVL: resultant = 1/2 × 3 × 24 = 36 kN, acting at 2/3 × 3 = 2 m from D, i.e. at x = 5 m from A. Horizontal: no horizontal load, so H_A = 0 . Σ M_A = 0 (anticlockwise +): R_B(6) - 40(2) - 20 - 36(5) - P(7) = 0 R_B = 280 + 7P/6 = 46.67 + 1.167P kN (↑) Σ F_y = 0 : R_A + R_B = 40 + 36 + P ⇒ R_A = 76 + P - 46.67 - 1.167P = 29.33 - 0.167P kN (↑) Check ( Σ M_B = 0 ): -R_A(6) + 40(4) - 20 + 36(1) - P(1) = -176 + P + 160 - 20 + 36 - P = 0 . ✔ Answer: R_A = (29.33 - P/6) kN ↑, H_A = 0 ; R_B = (46.67 + 7P/6) kN ↑. (For example, if P were zero: R_A = 29.33 kN, R_B = 46.67 kN.)

For the beam loaded as shown in figure, Calculate the support reactions at A and B.[Figure: beam AF, 9 m long. A (left end) is a hinge support; B is a roller support 6 m from A. Dimensions from A: 2 m (point of 40 kN load and 20 kN·m couple), 1 m to…

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