- 2070 Chaitra · 15 marks
Force method: frame, SFD and BMD
Short answer
Using force method with redundants at hinge E, horizontal reaction H_E = 1.93 kN left, vertical V_E = 47.63 kN up. Bending moments: M_A = 44.35 kNm, M_B = 27.93 kNm, M_C = -47.72 kNm, M_D = 0; zero moment in BC at 2.86 m from B. Shear forces: AB = 18.07 kN, BC at B = -2.63 kN, at C = -27.63 kN, CD = 20 kN, CE = 1.93 kN.
Draw shear force and bending moment diagrams for the frame given below. Use the force method. [Figure: frame; beam B-C-D with 5 kN/m UDL on BC (5 m) and a 2 m overhang CD with 20 kN at D; 20 kN horizontal at B; columns EI: AB (A fixed) of height 4 m, middle column CE hinged at E, right side with a 2 m dimension marked.]
Reading of the figure: left column AB (4 m, A fixed); beam B-C-D with BC = 5 m carrying 5 kN/m and a 2 m overhang CD with 20 kN at D; a middle column CE (4 m) hinged at E, below C; 20 kN horizontal at B (towards the right). is constant for all members.
, so . Release the hinge at E and take = horizontal reaction (), = vertical reaction () at E as redundants. The primary structure is fixed at A.
20 kN-> B ============ C ====== D 5 kN/m on BC
| | v 20 kN
| 4 m | 4 m
| o E (hinge)
A (fixed)
|<--- 5 m --->|<- 2 ->|
Unit and load moments (+ tension on the inner/lower face)
is the moment in the primary structure under the real loads (20 kN at B, 5 kN/m on BC, 20 kN at D), with the column CE unstressed; and are the moments due to unit forces at E. Their values at the key sections are given in the table below.
Final bending moments (kNm)
| Section | ||||
|---|---|---|---|---|
| A | -282.50 | 0.00 | 5.00 | -44.35 |
| B (column top) | -202.50 | 4.00 | 5.00 | 27.93 |
| mid BC | -105.63 | 4.00 | 2.50 | 5.73 |
| C (beam, left) | -40.00 | 4.00 | 0.00 | -47.72 |
| C (overhang) | -40.00 | 0.00 | 0.00 | -40.00 |
| D | 0.00 | 0.00 | 0.00 | 0.00 |
| C (column) | 0.00 | 4.00 | 0.00 | -7.72 |
| E | 0.00 | 0.00 | 0.00 | 0.00 |
At the joint C the member moments balance: -47.72 = -40.00 + (-7.72).
Shear force (kN)
| Member | SF |
|---|---|
| AB | +18.07 |
| BC (at B / at C) | -2.63 / -27.63 |
| CD (overhang) | +20.00 |
| CE | +1.93 |
Reactions
- E (hinge): kN (towards the left), kN ()
- A (fixed): kN (towards the left), kN (downward), kNm
Check: : kN; : .
In BC the moment falls from kNm at B through zero at 2.86 m from B to -47.72 kNm at C.
Answer: BMD ordinates: kNm, kNm, kNm (beam side), and the BM in BC changes sign at 2.86 m from B; the SFD values are in the table above.
✓ Answer checked by Claude Opus 5.5 (high effort)
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