Why is the 1x1 flexibility matrix enough for this truss?

m = 6, r = 3, j = 4 gives D_s = 1, so one redundant (diagonal BD)

Only the reactions are indeterminate by 3

Which expressions give f11 and Delta_10 for equal EA?

f11 = sum S1^2 L/EA, Delta_10 = sum S0 S1 L/EA

f11 = sum S0 L/EA, Delta_10 = sum S1 L/EA

f11 = sum S1 L, Delta_10 = sum S0 L

f11 = sum S0^2 L/EA, Delta_10 = sum S1^2 L/EA

2068 Chaitra · 8 marks

Truss: flexibility matrix method

Short answer

Using flexibility matrix with diagonal BD as redundant, the member forces (kN) are: AB 35.36, BC -14.64, CD -14.64, DA -64.64, AC 20.71, BD -50.00 (compression).

Use the force method (flexibility matrix) to solve the truss as shown in figure below. [Figure: square truss 4 m x 4 m with both diagonals; 100 kN downward and 50 kN horizontal at the top-left joint; bottom-left hinged, bottom-right roller.]

Assumptions: square truss ABCD, 4 m x 4 m: A bottom-left (hinge), B bottom-right (roller, vertical reaction only), C top-right, D top-left. 100 kN acts downward and 50 kN horizontally (to the right) at D. Both diagonals AC and BD are present; EA is the same for all members.

Degree of indeterminacy

$m=6$ , $r=3$ , $j=4$ : $D_s=6+3-8=1$ . Choose the diagonal BD as redundant $X_1$ (coordinate 1). The flexibility matrix is $1\times1$ : $[F]=[f_{11}]$ .

Reactions (statically determinate)

Taking moments about A (the 50 kN force acts 4 m above A, the 100 kN force passes through A's vertical line):

R_B\times4=50\times4\ \Rightarrow\ R_B=50\ \text{kN}\uparrow

$\sum F_x=0$ : $H_A=50$ kN to the left. $\sum F_y=0$ : $V_A=100-50=50$ kN upward.

Member forces by joint method

$S_0$ : forces in the released truss (BD removed) due to the loads.
$S_1$ : forces due to a unit tension pair applied along BD.

Member	L (m)	$S_0$ (kN)	$S_1$	$S_0S_1L$	$S_1^2L$	Final $S$ (kN)
AB	4.000	0.00	-0.707	0.00	2.000	35.36
BC	4.000	-50.00	-0.707	141.42	2.000	-14.64
CD	4.000	-50.00	-0.707	141.42	2.000	-14.64
DA	4.000	-100.00	-0.707	282.84	2.000	-64.64
AC	5.657	70.71	1.000	400.00	5.657	20.71
BD	5.657	0.00	1.000	0.00	5.657	-50.00
Sum				965.69	19.314

Flexibility and compatibility

f_{11}=\frac{\sum S_1^2L}{EA}=\frac{19.314}{EA},\qquad \Delta_{10}=\frac{\sum S_0S_1L}{EA}=\frac{965.69}{EA}

f_{11}X_1+\Delta_{10}=0\ \Rightarrow\ X_1=-\frac{965.69}{19.314}=-50.00\ \text{kN}

The diagonal BD carries 50.00 kN (compression).

Final forces ( $S=S_0+X_1S_1$ )

AB = 35.36, BC = -14.64, CD = -14.64, DA = -64.64, AC = 20.71, BD = -50.00 kN (+ tension, - compression).

Answer: BD = -50.00 kN, AC = 20.71 kN, AB = 35.36 kN, BC = -14.64 kN, CD = -14.64 kN, DA = -64.64 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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