Skip to main content

Frame AFD, SFD, BMD (15 kN/m, 75 kN)

Short answer

The frame has a hinge at mid-BC. Support reactions: H_A=25.55 kN→, V_A=91.94 kN↑, H_D=49.45 kN→, V_D=13.06 kN↑. Axial force constant on each member (e.g., AB -91.94 kN). Shear constant on AB (-25.55 kN), linear on BC, steps at 75 kN load on CD. Maximum bending moment 204.38 kN·m at B; local maximum on BC 34.44 kN·m. BMD is drawn on tension side.

Draw axial force, shear force and bending moment diagram for the given frame. [Figure: left column A-B of height 8 m with A hinged at the base; inclined member B-C rising 2 m over a horizontal span of 7 m and carrying a 15 kN/m UDL; right column C-D of height 8 m (D hinged at the base); a 75 kN horizontal force acts on the right column at 3 m from the base; overall width 7 m.]

Assumptions (taken from the figure):

  • A(0,0) hinged; AB = 8 m, so B = (0, 8). B to C (7, 10) is inclined (rises 2 m in 7 m, length 7.28 m). The right column CD is 8 m high, so the hinged support D = (7, 2) is 2 m above the level of A.
  • With two hinged supports and no internal hinge the frame is indeterminate, so a hinge is assumed at the middle of BC, (3.5, 9), giving a three-hinged frame.
  • 15 kN/m acts downward per metre of horizontal projection of BC (total 105 kN); the 75 kN horizontal force acts to the left on CD, 3 m above D, i.e. at (7, 5).

Sign convention: N is positive in tension; V is positive when the forces on the part before the section (travelling A → end) point along the left-hand normal of the member axis (upward on a left-to-right beam); M is quoted as a magnitude with the tension face named. Support reactions are drawn on the free-body of the whole frame.

               ---C
          -----   |
   -------        |
B--               |
|                 |
|                 |
|                 |
|                 |
|                 |
|                 D  (2 m above A)
|
|
A

Support reactions

Moments about A: 7VD−2HD−105(3.5)+75(5)=07V_D - 2H_D - 105(3.5) + 75(5) = 0. Hinge at the middle of BC (left part): −3.5VA+9HA+52.5(1.75)=0-3.5V_A + 9H_A + 52.5(1.75) = 0. With ΣFx=0\Sigma F_x = 0 (HA+HD=75H_A + H_D = 75) and ΣFy=0\Sigma F_y = 0 (VA+VD=105V_A + V_D = 105):

  • HAH_A = 25.55 kN →
  • VAV_A = 91.94 kN ↑
  • HDH_D = 49.45 kN →
  • VDV_D = 13.06 kN ↑

Check: moments of the right part about the hinge: 13.06(3.5)+49.45(7)−75(4)−52.5(1.75)≈013.06(3.5) + 49.45(7) - 75(4) - 52.5(1.75) \approx 0 ✓

Axial force, shear force and bending moment at the key sections

SectionN (kN)V (kN)M (kN·m)
AB at A-91.94-25.550.00
AB at B-91.94-25.55-204.38 (left face)
BC at B-49.8281.39-204.38 (top)
BC at hinge (3.64 m from B)-35.4030.910.00
BC at C-20.98-19.5720.63 (bottom)
CD at C-13.0625.5520.63 (left face)
CD just above the 75 kN load-13.0625.55148.36 (left face)
CD just below the 75 kN load-13.06-49.45148.36 (left face)
CD at D-13.06-49.450.00

Shape of the diagrams

  • AB: V constant, M linear
  • BC: V linear, M parabolic
  • CD: V constant on each side of the 75 kN load (it steps by 75 kN there), M linear on each side
  • AFD: constant between loads on a member, changing only where a load component acts along the member.
  • The BMD is drawn on the tension side of each member; it is continuous at rigid joints (moment is transferred) and zero at hinges and at free ends.

Salient features

  • Maximum bending moment = 204.38 kN·m at joint B (left face of AB, top of BC)
  • V = 0 at 5.87 m from B along BC (5.64 m horizontally): M = 34.44 kN·m (bottom), the local maximum
  • The moment on BC changes sign only at the internal hinge (3.64 m from B), where M = 0
  • On CD the moment is 20.63 kN·m at C and 148.36 kN·m at the 75 kN load, falling to 0 at D (tension on the left, inner face)

✓ Answer checked by Claude Opus 5.5 (high effort)

All Strength of Materials chapter 1 questions →