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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). EIEI is constant for all members.

r=3+2=5r = 3 + 2 = 5, so Ds=2D_s = 2. Release the hinge at E and take X1X_1 = horizontal reaction (→\rightarrow), X2X_2 = vertical reaction (↑\uparrow) 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)

M0M_0 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; m1m_1 and m2m_2 are the moments due to unit forces at E. Their values at the key sections are given in the table below.

f11=122.667EIf12=90.000EIf22=141.667EIΔ10=−4050.0EIΔ20=−6574.0EI\begin{aligned} f_{11} &= \frac{122.667}{EI} \\ f_{12} &= \frac{90.000}{EI} \\ f_{22} &= \frac{141.667}{EI} \\ \Delta_{10} &= \frac{-4050.0}{EI} \\ \Delta_{20} &= \frac{-6574.0}{EI} \end{aligned} 122.667 X1+90.000 X2=4050.090.000 X1+141.667 X2=6574.0\begin{aligned} 122.667\,X_1 + 90.000\,X_2 &= 4050.0 \\ 90.000\,X_1 + 141.667\,X_2 &= 6574.0 \end{aligned} X1=HE=−1.93 kN,X2=VE=47.63 kNX_1 = H_E = -1.93\ \text{kN},\qquad X_2 = V_E = 47.63\ \text{kN}

Final bending moments (kNm)

SectionM0M_0m1m_1m2m_2M=M0+∑XmM=M_0+\sum X m
A-282.500.005.00-44.35
B (column top)-202.504.005.0027.93
mid BC-105.634.002.505.73
C (beam, left)-40.004.000.00-47.72
C (overhang)-40.000.000.00-40.00
D0.000.000.000.00
C (column)0.004.000.00-7.72
E0.000.000.000.00

At the joint C the member moments balance: -47.72 = -40.00 + (-7.72).

Shear force (kN)

MemberSF
AB+18.07
BC (at B / at C)-2.63 / -27.63
CD (overhang)+20.00
CE+1.93

Reactions

  • E (hinge): HE=1.93H_E = 1.93 kN (towards the left), VE=47.63V_E = 47.63 kN (↑\uparrow)
  • A (fixed): HA=18.07H_A = 18.07 kN (towards the left), VA=2.63V_A = 2.63 kN (downward), MA=44.35M_A = 44.35 kNm

Check: ΣV\Sigma V: −2.63+47.63=5×5+20=45-2.63 + 47.63 = 5\times5 + 20 = 45 kN; ΣH\Sigma H: 20−18.07−1.93=020 - 18.07 - 1.93 = 0.

In BC the moment falls from +27.93+27.93 kNm at B through zero at 2.86 m from B to -47.72 kNm at C.

Answer: BMD ordinates: MA=44.35M_A = 44.35 kNm, MB=27.93M_B = 27.93 kNm, MC=−47.72M_C = -47.72 kNm (beam side), MD=0M_D = 0 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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