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Force method: reactions at D, frame with 110 kN

Short answer

Using force method with hinge D released, horizontal reaction H_D = 33.72 kN (left), vertical reaction V_D = 126.32 kN (up). Fixed end A: H_A = 76.28 kN left, V_A = 23.68 kN up, M_A = 183.40 kNm anticlockwise.

Determine horizontal and vertical reactions at support D of the frame shown in figure below using the force method. [Figure: portal frame ABCD; beam BC (2EI) of 5 m with 30 kN/m UDL; columns AB and CD (EI) 4 m high; 110 kN horizontal at B; A fixed, D hinged.]

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 110 kN horizontal load acts at B (towards C), A is fixed and D is hinged.

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

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

Moment diagrams in the primary structure (+ tension on the inner face)

  • M0M_0: in the beam, −375-375 kNm at B (UDL resultant 150×2.5150\times2.5) falling parabolically to 0 at C; in the column AB it varies linearly from −375-375 at B to −(375+110×4)=−815-(375 + 110\times4) = -815 kNm at A.
  • m1m_1 (unit horizontal force at D): CD from 0 to 4, beam constant 4, column AB 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=−5423.3EIΔ20=−13071.9EI\begin{aligned} f_{11} &= \frac{82.667}{EI} \\ f_{12} &= \frac{65.000}{EI} \\ f_{22} &= \frac{120.833}{EI} \\ \Delta_{10} &= \frac{-5423.3}{EI} \\ \Delta_{20} &= \frac{-13071.9}{EI} \end{aligned}

Compatibility

82.667 X1+65.000 X2=5423.365.000 X1+120.833 X2=13071.9\begin{aligned} 82.667\,X_1 + 65.000\,X_2 &= 5423.3 \\ 65.000\,X_1 + 120.833\,X_2 &= 13071.9 \end{aligned} X1=HD=−33.72 kN,X2=VD=126.32 kNX_1 = H_D = -33.72\ \text{kN},\qquad X_2 = V_D = 126.32\ \text{kN}

Final bending moments (kNm)

SectionM0M_0m1m_1m2m_2M=M0+∑XmM=M_0+\sum X m
A-815.000.005.00-183.40
B (column)-375.004.005.00121.72
mid BC-93.754.002.5087.17
C (beam)0.004.000.00-134.88
D0.000.000.000.00

Maximum sagging moment in BC: 131.07 kNm at 0.79 m from B.

Reactions

  • D (hinge): HD=33.72H_D = 33.72 kN (towards the left), VD=126.32V_D = 126.32 kN (↑\uparrow)
  • A (fixed): HA=76.28H_A = 76.28 kN (towards the left), VA=23.68V_A = 23.68 kN (↑\uparrow), MA=183.40M_A = 183.40 kNm (anticlockwise)

Check: ΣH\Sigma H: 110−76.28−33.72=0110 - 76.28 - 33.72 = 0; ΣV\Sigma V: 23.68+126.32=30×5=15023.68 + 126.32 = 30\times5 = 150 kN.

Answer: at the hinged support D, the horizontal reaction is 33.7233.72 kN (towards the left) and the vertical reaction is 126.32126.32 kN (upward).

✓ Answer checked by Claude Opus 5.5 (high effort)

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