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AFD, SFD, BMD of frame with triangular load

Short answer

Reactions: A_x=30 kN left, A_y=19.83 kN up, D_y=40.17 kN up. AB column has constant moment 150 kN·m above mid-height. Beam BC: maximum sagging moment 171.7 kN·m at 2.32 m from B (shear zero), moment at C is 15 kN·m balanced by applied couple; column CD has no moment.

Draw axial force, shear force and bending moment diagram and indicate the salient features if any for the given frame loaded as shown in the figure below. [Figure: frame A-B-C-D with pin at A and roller at D; column A-B 5 m + 5 m with 30 kN horizontal load at mid-height; beam B-C 8 m + 2 m carrying a triangular load of up to 10 kN/m, a 20 kN point load, and a 15 kN·m moment at C; column C-D 4 m.]

Reading of the figure: A is a hinge at (0, 0) and D is a roller at (10, 6). Column AB is 10 m high (5 + 5), with a 30 kN horizontal load to the right at mid-height (5 m). Beam B–C is 10 m long: a triangular load from 10 kN/m at B decreasing to zero at 8 m, a 20 kN point load at 8 m, and a 15 kN·m couple at C (taken counter-clockwise). Column CD is 4 m long, with the roller D at its foot. The roller gives a vertical reaction only.

Support reactions

Triangular load = ½ × 10 × 8 = 40 kN, acting at 8/3 = 2.667 m from B. Moments about A (anticlockwise positive): the 30 kN load at 5 m height pushes to the right, giving −30 × 5 = −150; the 40 kN triangular load gives −40 × 2.667 = −106.67; the 20 kN gives −20 × 8 = −160; the couple gives +15; D_y at 10 m gives +10 D_y.

ΣM_A = 0:  10 D_y − 150 − 106.67 − 160 + 15 = 0
           D_y = 401.67/10 = 40.17 kN ↑

ΣFx = 0: A_x + 30 = 0, so A_x = 30 kN ← ΣFy = 0: A_y + 40.17 − 40 − 20 = 0 → A_y = 19.83 kN ↑

Member forces

Column AB (0 to 10 m)

  • N = −19.83 kN (compression), constant.
  • V = 30 kN from A to mid-height, 0 from mid-height to B.
  • M = 30y from A to mid-height: 0 → 150 kN·m at 5 m; constant 150 kN·m up to B (no shear).

Beam BC (x measured from B; w(x) = 10(1 − x/8))

  • N = 0 (no horizontal force).
  • V(x) = 19.83 − 10x + 0.625x², so V_B = 19.83 kN; V = 0 at x = 2.32 m; just left of the 20 kN load V(8) = −20.17 kN; just right of it −40.17 kN, constant to C.
  • M(x) = 150 + 19.83x − 5x² + 0.2083x³
    • M_B = 150.00 kN·m
    • Maximum M = 171.70 kN·m at x = 2.32 m
    • M(4 m) = 162.67, M(8 m) = 95.33 kN·m
    • M_C (beam end) = 95.33 − 40.17 × 2 = 15.00 kN·m, balanced by the applied 15 kN·m couple, so the column has no moment.

Column CD (4 m)

  • N = −40.17 kN (compression); V = 0; M = 0 (the roller reaction acts along the column).

Values at key points

PointN (kN)V (kN)M (kN·m)
A−19.83300
Mid AB (5 m)−19.8330 → 0150
B−19.83 / 00 / 19.83150
2.32 m from B00171.70 (max)
8 m from B0−20.17 / −40.1795.33
C0−40.1715 → 0
D−40.1700
BMD (kN·m)
        150 ----(171.7)---- 95.3 \ 15
         |                         C|
         |                          | (zero)
        150 (constant)              |
         |                          |
        0 A                       D (0)

Salient features: the maximum bending moment is 171.7 kN·m in the beam at 2.32 m from B, where the shear is zero; the moment in the column is constant at 150 kN·m between mid-height and B because the shear there is zero; the shear in the beam changes sign once; there is a sudden jump of 20 kN in the SFD at the point load, and the 15 kN·m couple at C produces a drop of 15 kN·m in the moment from the beam end to the column; the BMD in the beam is cubic (triangular load) and linear under no load.

✓ Answer checked by Claude Opus 5.5 (high effort)

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