Force method: fixed portal frame, 30 kN/m

Portal frame: span 4 m, height 4 m, both bases fixed, EI constant, UDL $w = 30$ kN/m on the beam BC. $r = 6$, so $D_s = 3$.

Release the fixed support at D and take $X_1 = H_D$ ($\rightarrow$), $X_2 = V_D$ ($\uparrow$) and $X_3 = M_D$ (anticlockwise) as redundants. The primary structure is a cantilever fixed at A.

   30 kN/m
 B vvvvvvvvvv C
 |            |
 | 4 m        | 4 m
 |            |
A///        ///D   (both fixed)
 |<-- 4 m -->|

Unit and load moment diagrams (+ tension on the inside)

($M_0 = -30\times4\times2 = -240$ kNm at A and B, falling parabolically to 0 at C.)

Flexibility coefficients (in $1/EI$ units)

$$[f] = \begin{bmatrix} 106.67 & 64 & 32 \\ 64 & 85.33 & 24 \\ 32 & 24 & 12 \end{bmatrix},\qquad \{\Delta_0\} = \begin{Bmatrix} -3200 \\ -4800 \\ -1280 \end{Bmatrix}$$

Compatibility ($[f]\{X\} = -\{\Delta_0\}$)

$$\begin{aligned} 106.67X_1 + 64X_2 + 32X_3 &= 3200 \\ 64X_1 + 85.33X_2 + 24X_3 &= 4800 \\ 32X_1 + 24X_2 + 12X_3 &= 1280 \end{aligned}$$ $$X_1 = H_D = -10\ \text{kN},\qquad X_2 = V_D = 60\ \text{kN},\qquad X_3 = M_D = 13.33\ \text{kNm}$$

(The symmetric structure and load give $V_A = V_D = 60$ kN and $|H| = 10$ kN inwards at both supports.)

Reactions

Bending moment diagram (kNm, + tension inside)

The BM in each column is linear, $M = 13.33 - 10y$, zero at $y = 1.33$ m above the base. In the beam $M = -26.67 + 60x - 15x^2$, zero at $0.51$ m and $3.49$ m from B, with the maximum $+33.33$ kNm at mid-span.

Shear force diagram (kN)

• Columns: constant $10$ kN (horizontal), acting inwards at the base.
• Beam: $+60$ kN at B, falling linearly through zero at mid-span to $-60$ kN at C.

Normal thrust diagram (kN)

• Columns: $-60$ (compression) throughout.
• Beam: $-10$ (compression) throughout.

Answer: $H = 10$ kN, $V = 60$ kN and $M = 13.33$ kNm at each base; $M_B = M_C = -26.67$ kNm; mid-span $M = +33.33$ kNm.