SD TRUSS PROBLEM WITH THE METHOD OF JOINTS
The truss structure ABCD is loaded with a concentrated force P = 2 kN applied at a 45° angle at joint D as indicated in the figure. All bars in the truss have equal cross sectional area and are made of the same homogeneous linear elastic material. The dimensions are H=3 m and L=8 m. Use the method of joints to obtain the axial forces in the bars and the Cartesian components of the reactions at the supports A and C.
1_A, 1_B, 1_C
1_A : -1.17, -1.17, 1.4142, 0.936, 2.35
1_B : -1.414, 0.702, 0.702
1_C:-4.05, -4.05,2.59,6.49
Anyone for HW4_2: Part I?
4_2 part IIA RxA=4, RyA=0
4_2 part IIB RxE=4, RyE=2
RxA= -4, sorry
problem 4_2 part :1
A load W=2 kN is applied vertically to joint C of truss ABCDE as indicated.
plz help
Please HW4_2: Part I?
Anybody for Problem: HW4_2: Part I please?
Has somebody formula for HW4_2: Part I?
N_AB N_BC N_BE N_BD N_CD N_ED
-1 1 -0,7071 0 0 0
0 0 -0,7071 -1 0 0
0 -1 0 0 -0,7071 0
0 0 0 0 -0,7071 0
0 0 0 0 0,7071 -1
0 0 0 1 0,7071 0
b
0
0
0
2
0
0
X = a\b
N AB 4,000
N BC 2,000
N BE -2,828
N BD 2,000
N CD -2,828
N ED -2,000
Simonsay thanks. Something is wrong with the expression, could you please check?
I don't get right answer.
-1,1,-0.7071,0,0,0
0,0,-0.7071,-1,0,0
0,-1,0,0,-0.7071,0
0,0,0,0,-0.7071,0
0,0,0,0,0.7071,-1
0,0,0,1,0.7071,0
1/sqrt(2) = 1/1.4142 = 0.7071
Simonsay, is that like this below? I have just tried it does not give right answer and says:
Expression or statement is incomplete or incorrect.
ANy help please and did someone manage?
% Define the coefficient matrix a
a =
N_AB -1,1,-0.7071,0,0,0
N_BC 0,0,-0.7071,1,0,0
N_BE 0,-1,0,0,-0.7071,0
N_BD 0,0,0,0,-0.7071,0
N_CD 0,0,0,0,0.7071,-1
N_ED 0,0,0,1,0.7071,0
% Define the right-hand-side vector b
b =
0
0
0
2
0
0
% Solve for solution vector
X = a\b
N_AB 4,000
N_BC 2,000
N_BE -2,828
N_BD 2,000
N_CD -2,828
N_ED -2,000
Please help simonsay! Have problem with expresion too.
try C/P this
% Define the coefficient matrix a
a =[-1 1 -sqrt(2)/2 0 0 0; 0 0 -sqrt(2)/2 -1 0 0; 0 -1 0 0 -sqrt(2)/2 0; 0 0 0 0 -sqrt(2)/2 0;
0 0 0 0 sqrt(2)/2 -1; 0 0 0 1 sqrt(2)/2 0];
% Define the right-hand-side vector b
b = [0; 0 ; 0; 2; 0; 0];
% Solve for solution vector
X = a\b