Question

The frame shown below is fixed at A and C, and is supported by a roller at B. Use the numbering shown for the members and joints and determine the support reactions at all supports of the frame using the Stiffness Method. The 10 kN force is applied at the middle of the beam, and the 12 kN/m load is uniformly distributed on the column Take E = 200 GPa, 1 = 300(109) mm+ and A = 10(10-) mm2 for all. Y (10 KN 2 1 B + 1 2 12 kN/m 4.0 m 8 9 4.0 m + 12 EI/L 6EI/L2= 4E1/L = 2 EI L = AEL =

Answer 1

Lets analyze the problem with Moment distribution method

Point load in member AB,P=10kN

UDL in member BC, w=12kN/m

L_AB=4 m, L_BC=4 m

Fixed end moments

Member AB

FEM_AB= -PL/8= -5 kNm

FEM_BA= + PL/8= +5 kNm

Member BC

FEM_BC= -wL²/12= -12*16/12= -16 kNm

FEM_CB= +wL²/12= +12*16/12= +16 kNm

Joint	Member	Stiffness	Total Stiffness	DF
B	BA	4EI/4	EI	0.5
BC	4EI/4	EI	0.5

		A	B	C
			0.5	0.5	0
FEM	-5	5	-16	16
Balancing correction		5.5	5.5
Carry over	2.75			2.75
Final End moments	-2.25	10.5	-10.5	18.75

Support Reaction Member AB WA Hari RA JRA 2.25kN 10.5kNm Consider the force body diagram of member AB I FC HA = He I fyza RA + Rza RAx4 -2.25 +10.5 = 0 » R =-2.06 kom (W) R.HR RB = +2-06 Km 10.5kpchen Consider free body din grend BC 2/3=0 18.75 18.15 - Heth-10.5 = 0 اجرا He-2.06 len How He = 2.06 kan Roza R = 2.06kw.m

10.5 Kar 3 10.5 kw. 2.25 INN BUD C 18.75 2.06kn. - 2.06 km 2.25 kNm support 2.06 kr. Reactions 18.75 km 2.0 kn 2.06, IN