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Consistent deformation: frame, BMD, SFD, normal thrust

Short answer

Releasing hinge at D, choosing H_D and V_D as redundants, consistent deformation gives H_D = 3.09 kN (left) and V_D = 10.84 kN (up). Fixed-end reactions: H_A = 26.91 kN (left), V_A = 4.16 kN (up), M_A = 31.63 kNm (anticlockwise).

Using the consistent deformation method analyse the frame shown in figure and draw bending moment, shear force and normal thrust diagram. [Figure: frame; left column 2EI with 10 kN/m UDL horizontally, height 3 m + 1 m...; beam EI of 2 m + 2 m with 15 kN downward at mid-span; right column 2EI.]

Reading of the figure (as assumed): portal frame of height 4 m and span 4 m. The left column AB (2EI2EI) is fixed at A and carries a horizontal UDL of 10 kN/m over its lower 3 m (towards the right); the beam BC (EIEI, 2 m + 2 m) carries 15 kN downward at mid-span F; the right column CD (2EI2EI) is hinged at D.

r=3+2=5r = 3 + 2 = 5, so Ds=2D_s = 2. By the consistent deformation method, release the hinge at D and take the reactions X1X_1 (horizontal, →\rightarrow) and X2X_2 (vertical, ↑\uparrow) at D as redundants. The displacements of D in the primary structure must be cancelled by the redundants, that is, the total displacement at D along each redundant is zero.

  B ----F---- C
  |   15 kN   |
 10->         | 4 m
 kN/m         |
  |           o D
  A (fixed)

Consistent deformation equations

Δ10+f11X1+f12X2=0Δ20+f21X1+f22X2=0\begin{aligned} \Delta_{10} + f_{11}X_1 + f_{12}X_2 &= 0 \\ \Delta_{20} + f_{21}X_1 + f_{22}X_2 &= 0 \end{aligned}

where Δi0=∫M0 miEIdx\Delta_{i0} = \int \dfrac{M_0\,m_i}{EI}dx (displacements of D in the primary structure) and fij=∫mimjEIdxf_{ij} = \int\dfrac{m_im_j}{EI}dx.

f11=85.333EIf12=48.000EIf22=53.333EIΔ10=−256.9EIΔ20=−430.0EI\begin{aligned} f_{11} &= \frac{85.333}{EI} \\ f_{12} &= \frac{48.000}{EI} \\ f_{22} &= \frac{53.333}{EI} \\ \Delta_{10} &= \frac{-256.9}{EI} \\ \Delta_{20} &= \frac{-430.0}{EI} \end{aligned} 85.333 X1+48.000 X2=256.948.000 X1+53.333 X2=430.0\begin{aligned} 85.333\,X_1 + 48.000\,X_2 &= 256.9 \\ 48.000\,X_1 + 53.333\,X_2 &= 430.0 \end{aligned} X1=HD=−3.09 kN,X2=VD=10.84 kNX_1 = H_D = -3.09\ \text{kN},\qquad X_2 = V_D = 10.84\ \text{kN}

Final bending moments M=M0+X1m1+X2m2M = M_0 + X_1m_1 + X_2m_2 (kNm, + tension on the inside)

SectionM0M_0m1m_1m2m_2M=M0+∑XmM=M_0+\sum X m
A-75.000.004.00-31.63
Q (top of UDL)-30.003.004.004.10
B-30.004.004.001.01
F (load)0.004.002.009.33
C0.004.000.00-12.35
D0.000.000.000.00

Shear force (kN)

MemberAt lower endAt upper end
AQ (UDL zone)+26.91-3.09
QB-3.09-3.09
BF+4.16+4.16
FC-10.84-10.84
CD+3.09+3.09

(Column shears are horizontal forces; positive means towards the right on the left face.)

Normal thrust (kN, compression negative)

MemberAxial force
AB (left column)-4.16
BC (beam)-3.09
CD (right column)-10.84

Reactions

  • D (hinge): HD=3.09H_D = 3.09 kN (towards the left), VD=10.84V_D = 10.84 kN (↑\uparrow)
  • A (fixed): HA=26.91H_A = 26.91 kN (towards the left), VA=4.16V_A = 4.16 kN (↑\uparrow), MA=31.63M_A = 31.63 kNm (anticlockwise)

Check: ΣH\Sigma H: 30−26.91−3.09=030 - 26.91 - 3.09 = 0 (UDL resultant =10×3=30= 10\times3 = 30 kN); ΣV\Sigma V: 4.16+10.84=154.16 + 10.84 = 15 kN.

Answer: HD=3.09H_D = 3.09 kN, VD=10.84V_D = 10.84 kN; HA=26.91H_A = 26.91 kN, VA=4.16V_A = 4.16 kN, MA=31.63M_A = 31.63 kNm. The BMD, SFD and thrust values are tabulated above.

✓ Answer checked by Claude Opus 5.5 (high effort)

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