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Flexibility matrix: reactions at E and A, BMD

Short answer

Release hinge at E, take horizontal and vertical reactions as redundants; flexibility matrix compatibility yields H_E = 74.68 kN (left), V_E = 369.24 kN (up). At fixed base A: H_A = 54.68 kN (right), V_A = 380.76 kN (up), M_A = 175.80 kNm (clockwise).

Determine the reactions at support E and A and draw bending moment diagram of the frame shown in figure below by using the flexibility matrix method (force method). [Figure: frame; column AB (2EI) of 10 m with A fixed, 20 kN horizontal at B, with a 7 m dimension marked; beam BC-D (2EI) of 10 m carrying 75 kN/m UDL; column DE (EI) of 5 m, E hinged.]

Reading of the figure: A (0, 0) is fixed; column AB is 10 m high (2EI2EI) with a 20 kN horizontal load (towards the right) at 7 m above A; beam BD is 10 m long (2EI2EI) with 75 kN/m over its whole length; column DE is 5 m long (EIEI), so E is 5 m above the level of A, and E is hinged.

Ds=3+2−3=2D_s = 3 + 2 - 3 = 2. Release the hinge at E: X1X_1 = horizontal reaction (→\rightarrow), X2X_2 = vertical reaction (↑\uparrow). The primary structure is a cantilever fixed at A.

 B ================= D    75 kN/m
 |                   |
 |  20 kN ->         | 5 m
 |                   E (hinge)
 | 10 m
 A (fixed)

Primary-structure moments (+ tension on the inner face)

  • M0M_0: at B, −75×10×5=−3750-75\times10\times5 = -3750 kNm; at the 20 kN level, −3750-3750; at A, −(3750+20×7)=−3890-(3750 + 20\times7) = -3890 kNm; the beam falls parabolically from −3750-3750 at B to 0 at D.
  • m1m_1 (unit horizontal force at E): DE from 0 at E to 5 at D; beam constant 5; column from 5 at B to −5-5 at A.
  • m2m_2 (unit vertical force at E): beam from 10 at B to 0 at D; column constant 10.

Flexibility matrix [F][F] and load vector {Δ0}\{\Delta_0\} (in 1/EI1/EI units)

f11=208.333EIf12=125.000EIf22=666.667EIΔ10=−30596.7EIΔ20=−236825.0EI\begin{aligned} f_{11} &= \frac{208.333}{EI} \\ f_{12} &= \frac{125.000}{EI} \\ f_{22} &= \frac{666.667}{EI} \\ \Delta_{10} &= \frac{-30596.7}{EI} \\ \Delta_{20} &= \frac{-236825.0}{EI} \end{aligned}

Compatibility: [F]{X}=−{Δ0}[F]\{X\} = -\{\Delta_0\}

208.333 X1+125.000 X2=30596.7125.000 X1+666.667 X2=236825.0\begin{aligned} 208.333\,X_1 + 125.000\,X_2 &= 30596.7 \\ 125.000\,X_1 + 666.667\,X_2 &= 236825.0 \end{aligned} X1=HE=−74.68 kN,X2=VE=369.24 kNX_1 = H_E = -74.68\ \text{kN},\qquad X_2 = V_E = 369.24\ \text{kN}

Final bending moments M=M0+X1m1+X2m2M = M_0 + X_1m_1 + X_2m_2 (kNm)

SectionM0M_0m1m_1m2m_2M=M0+∑XmM=M_0+\sum X m
A-3890.00-5.0010.00175.80
Q (20 kN level)-3750.002.0010.00-206.96
B (column)-3750.005.0010.00-431.00
mid BD-937.505.005.00535.30
D (beam)0.005.000.00-373.40
E0.000.000.000.00

Maximum sagging moment in the beam: 535.52 kNm at 5.08 m from B.

Reactions

  • E (hinge): HE=74.68H_E = 74.68 kN (towards the left), VE=369.24V_E = 369.24 kN (↑\uparrow)
  • A (fixed): HA=54.68H_A = 54.68 kN (towards the right), VA=380.76V_A = 380.76 kN (↑\uparrow), MA=175.80M_A = 175.80 kNm (clockwise)

Check: ΣH\Sigma H: 20+54.68−74.68=020 + 54.68 - 74.68 = 0; ΣV\Sigma V: 380.76+369.24=75×10=750380.76 + 369.24 = 75\times10 = 750 kN.

Answer: at E, HE=74.68H_E = 74.68 kN (towards the left) and VE=369.24V_E = 369.24 kN (upward); at A, HA=54.68H_A = 54.68 kN (towards the right), VA=380.76V_A = 380.76 kN (upward) and MA=175.80M_A = 175.80 kNm (clockwise).

✓ Answer checked by Claude Opus 5.5 (high effort)

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