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Force method: frame reactions and BMD

Short answer

Release hinge at D, treat horizontal and vertical reactions as redundants X1, X2. From compatibility, X1 = –31.44 kN (31.44 kN left) and X2 = 121.79 kN upward.

Determine the horizontal and vertical reaction at the hinged support and also draw BMD using the force method. [Figure: portal frame; beam of 2EI, span 5 m, with 30 kN/m UDL; columns EI of height 4 m; left base fixed, right base hinged; 100 kN horizontal at the top-left joint.]

Reading of the figure: portal frame ABCD, columns AB and CD are 4 m high (EIEI), the beam BC is 5 m long (2EI2EI) with 30 kN/m, a 100 kN horizontal load acts at B (towards C), A is fixed and D is hinged.

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

 100 kN
  ->  B ============ C   30 kN/m
      |              |
      | 4 m          | 4 m
      |              |
     ///A            o D (hinge)
      |<---- 5 m --->|

Primary structure (cantilever fixed at A)

  • M0M_0 (+ tension on the inner face): at B, −30×5×2.5=−375-30\times5\times2.5 = -375 kNm (the 100 kN load acts at the joint B); at A, −(375+100×4)=−775-(375 + 100\times4) = -775 kNm; the beam falls parabolically from −375-375 at B to 0 at C; column CD has no moment.
  • m1m_1 (unit horizontal force at D): CD from 0 at D to 4 at C, beam constant 4, column from 4 at B down to 0 at A.
  • m2m_2 (unit vertical force at D): beam from 5 at B to 0 at C, column AB constant 5.

Flexibility coefficients (in 1/EI1/EI units)

f11=82.667EIf12=65.000EIf22=120.833EIΔ10=−5316.7EIΔ20=−12671.9EI\begin{aligned} f_{11} &= \frac{82.667}{EI} \\ f_{12} &= \frac{65.000}{EI} \\ f_{22} &= \frac{120.833}{EI} \\ \Delta_{10} &= \frac{-5316.7}{EI} \\ \Delta_{20} &= \frac{-12671.9}{EI} \end{aligned}

Compatibility (ΔDx=ΔDy=0\Delta_{Dx} = \Delta_{Dy} = 0)

82.667 X1+65.000 X2=5316.765.000 X1+120.833 X2=12671.9\begin{aligned} 82.667\,X_1 + 65.000\,X_2 &= 5316.7 \\ 65.000\,X_1 + 120.833\,X_2 &= 12671.9 \end{aligned} X1=HD=−31.44 kN,X2=VD=121.79 kNX_1 = H_D = -31.44\ \text{kN},\qquad X_2 = V_D = 121.79\ \text{kN}

Final bending moments (kNm)

SectionM0M_0m1m_1m2m_2M=M0+∑XmM=M_0+\sum X m
A-775.000.005.00-166.07
B (column)-375.004.005.00108.15
mid BC-93.754.002.5084.94
C (beam)0.004.000.00-125.78
D0.000.000.000.00

The maximum sagging moment in the beam is 121.42 kNm at 0.94 m from B.

Reactions

  • At D (hinge): HD=31.44H_D = 31.44 kN (towards the left), VD=121.79V_D = 121.79 kN (↑\uparrow)
  • At A (fixed): HA=68.56H_A = 68.56 kN (towards the left), VA=28.21V_A = 28.21 kN (↑\uparrow), MA=166.07M_A = 166.07 kNm (anticlockwise)

Check: ΣH\Sigma H: 100−68.56−31.44=0100 - 68.56 - 31.44 = 0; ΣV\Sigma V: 28.21+121.79=30×5=15028.21 + 121.79 = 30\times5 = 150 kN.

Answer: at the hinged support D, the horizontal reaction is 31.4431.44 kN (towards the left) and the vertical reaction is 121.79121.79 kN (upward). End moments: MA=166.07M_A = 166.07 kNm, MB=108.15M_B = 108.15 kNm, MC=−125.78M_C = -125.78 kNm (signs as in the table).

✓ Answer checked by Claude Opus 5.5 (high effort)

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