Question

Use Direct Stiffness Method to calculate

1 OkN/m А B ΕΙ 5m ΕΙ 32kN ΕΙ 5m 2.5m. 2.5m

Member stiffness Matrices

Global Stiffness matrix

Displacement

Answer 1

anical rigid. lokN/m EI X= horizontal displacement at 2. Sol Assuming the members are (AE=00) for each member. y de teto de beb I. EI 5m 32kN EI 깊 10017 x 5m 2.5m 2.5m r = rotation at ( Y = Vestical displacement at 1.

Ya 1 EI D EI 24 Yu X3 11 ***2 Brate 3 EI 777 displacement at Each joint o U- Y 0 X2 displacement vector. Yz & ry X2 o r Y3 Urs fr V3 Yu r3 roq ru Y, 20, apo se due to rigid forces at a X2 = 7₂ = *3=Y3 so y due to anial rigid members forles. . 20.8333 S+ + 1oxs 12 -20 Po +20 - 3245 8 + 32x5

	member stiffness matrix
		y1	r1	y2	r2
		0.096	0.24	-0.096	0.24	y1
k1	EI	0.24	0.8	-0.24	0.4	r1
		-0.096	-0.24	0.096	-0.24	y2
		0.24	0.4	-0.24	0.8	r2
		x2	r2	x3	r3
		0.096	0.24	-0.096	0.24	x2
k2	EI	0.24	0.8	-0.24	0.4	r2
		-0.096	-0.24	0.096	-0.24	x3
		0.24	0.4	-0.24	0.8	r3
		y3	r3	y4	r4
		0.096	0.24	-0.096	0.24	y3
k3	EI	0.24	0.8	-0.24	0.4	r3
		-0.096	-0.24	0.096	-0.24	y4
		0.24	0.4	-0.24	0.8	r4

					global stiffness matrix
			y1	r1	x2	y2	r2	x3	y3	r3	y4	r4
y1			0.096	0.24	0	-0.096	0.24	0	0	0	0	0
r1			0.24	0.8	0	-0.24	0.4	0	0	0	0	0
x2			0	0	0.096	0	0.24	-0.096	0	0.24	0	0
y2			-0.096	-0.24	0	0.096	-0.24	0	0	0	0	0
r2	k	EI	0.24	0.4	0.24	-0.24	1.6	-0.24	0	0.4	0	0
x3			0	0	-0.096	0	-0.24	0.096	0	-0.24	0	0
y3			0	0	0	0	0	0	0.096	0.24	-0.096	0.24
r3			0	0	0.24	0	0.4	-0.24	0.24	1.6	-0.24	0.4
y4			0	0	0	0	0	0	-0.096	-0.24	0.096	-0.24
r4			0	0	0	0	0	0	0.24	0.4	-0.24	0.8

			reduced stiffness matrix
			r	r2	r3
r			1.6	0.4	0
r2	Kr	EI	0.4	1.6	0.4
r3			0	0.4	0.8

Reduced stiffness Matrin! a 1.6 0.4 kra I 2 = 1.6 0.4 Och 810 Ooh Equilibrium equation' - k Vr = Pro 20.833 91 0.4 le 20 106 Ellory och re 20 ooh 0.8 autor 17.79167 - en ب- 스 -27.0833 التر اشتاح E I 83 38.54167