What is the compatibility statement of the flexibility method here and what is released?

[F]{X} + {Delta0} = 0, after releasing the hinge D and taking its horizontal and vertical reactions as redundants

[F]{X} = {Delta0} with A released instead

[K]{X} = {P}, solving for joint displacements

{X} = [F]{Delta0} with the fixed end A released

How is M0 at A found for the cantilever primary structure?

-(50 x 2 + 25 x 8 x 4) = -900 kNm

Why is f12 = f21 = 0.01067 in the matrix [F]?

Maxwell's reciprocal theorem makes the flexibility matrix symmetric

Because of the equilibrium equations at A

X1 = -42.23 and X2 = 104.55. What are the reactions at D and the check?

H_D = 42.23 kN to the left and V_D = 104.55 kN up; the check is sum H = 50 - 7.77 - 42.23 = 0 and sum V = 200 kN

H_D = 42.23 kN to the right and V_D = 104.55 kN down

H_D = 7.77 kN and V_D = 95.45 kN

H_D = 104.55 kN and V_D = 42.23 kN

2078 Bhadra · 10 marks

Flexibility matrix method: portal frame

Short answer

Use the flexibility method with redundants at the hinge D. Compatibility gives X1 = 42.23 kN (left) and X2 = 104.55 kN (upward), so HD = 42.23 kN left and VD = 104.55 kN upward.

Analyse the portal frame ABCD shown in the figure below using the flexibility matrix method. Take $EI = 12000\ \text{kN/m}^2$ . [Figure: frame with column AB of height 4 m (2 m + 2 m) with 50 kN horizontal at mid-height, beam BC of 8 m carrying 25 kN/m UDL, column CD of height 3 m; A fixed, D supported.]

Reading of the figure: A (0, 0) fixed; column AB is 4 m high with a 50 kN horizontal load (towards the right) at mid-height E; beam BC is 8 m long with 25 kN/m; column CD is 3 m high, so D is 1 m above the level of A, and D is hinged. $EI = 12000$ kNm $^2$ is constant.

$r = 3 + 2 = 5$ , so $D_s = 2$ . Release the hinge at D and take the reactions at D as the redundants: $X_1$ = horizontal ( $\rightarrow$ ), $X_2$ = vertical ( $\uparrow$ ). The primary structure is a cantilever fixed at A. In the flexibility method the compatibility condition is $[F]\{X\} + \{\Delta_0\} = 0$ .

Moments in the primary structure (sign: + tension on the inner face)

Loads only ( $M_0$ ): at A, $-(50\times2 + 25\times8\times4) = -900$ ; at E, $-800$ ; at B, $-800$ ; the beam falls parabolically to 0 at C.
Unit $X_1$ ( $m_1$ ): $m_1 =$ lever arm of the unit horizontal force; at C and along BC it is $3$ , at B it is $3$ , at A it is $-1$ .
Unit $X_2$ ( $m_2$ ): at B it is $8$ , falling linearly to 0 at C; along AB it is $8$ .

Flexibility matrix and load vector

\begin{aligned} f_{11} &= \frac{90.333}{EI} \\ f_{12} &= \frac{128.000}{EI} \\ f_{22} &= \frac{426.667}{EI} \\ \Delta_{10} &= \frac{-9566.7}{EI} \\ \Delta_{20} &= \frac{-39200.0}{EI} \end{aligned}

In units of m/kN with $EI = 12000$ :

[F] = \begin{bmatrix} 0.00753 & 0.01067 \\ 0.01067 & 0.03556 \end{bmatrix},\qquad \{\Delta_0\} = \begin{Bmatrix} -0.79722 \\ -3.26667 \end{Bmatrix}\ \text{m}

Compatibility equations

\begin{aligned} 90.333\,X_1 + 128.000\,X_2 &= 9566.7 \\ 128.000\,X_1 + 426.667\,X_2 &= 39200.0 \end{aligned}

\{X\} = -[F]^{-1}\{\Delta_0\} \ \Rightarrow\ X_1 = -42.23\ \text{kN},\quad X_2 = 104.55\ \text{kN}

Final bending moments (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-900.00	-1.00	8.00	-21.41
E	-800.00	1.00	8.00	-5.87
B	-800.00	3.00	8.00	-90.34
mid BC	-200.00	3.00	4.00	91.48
C	0.00	3.00	0.00	-126.70
D	0.00	0.00	0.00	0.00

Maximum sagging moment in BC: 91.89 kNm at 3.82 m from B.

Reactions

At D (hinge): $H_D = 42.23$ kN ( $\leftarrow$ ), $V_D = 104.55$ kN ( $\uparrow$ )
At A: $H_A = 7.77$ kN ( $\leftarrow$ ), $V_A = 95.45$ kN ( $\uparrow$ ), $M_A = 21.41$ kNm (anticlockwise)

Check: $\Sigma V = 200.00 = 25 \times 8 = 200$ kN; $\Sigma H = 50 - 7.77 - 42.23 = 0$ .

Answer: $X_1 = H_D = 42.23$ kN towards the left and $X_2 = V_D = 104.55$ kN upward.

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