Friday

July 25, 2014

July 25, 2014

Posted by **helpless** on Tuesday, May 14, 2013 at 7:38am.

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

- Physics -
**simonsay**, Wednesday, May 15, 2013 at 10:14pm1_A : -1.17, -1.17, 1.4142, 0.936, 2.35

1_B : -1.414, 0.702, 0.702

- Physics -
**FLu**, Thursday, May 16, 2013 at 10:57am1_C:-4.05, -4.05,2.59,6.49

Anyone for HW4_2: Part I?

- Physics -
**simonsay**, Thursday, May 16, 2013 at 5:18pm4_2 part IIA RxA=4, RyA=0

4_2 part IIB RxE=4, RyE=2

- Physics -
**simonsay**, Thursday, May 16, 2013 at 5:24pmRxA= -4, sorry

- Physics -
**faryia**, Friday, May 17, 2013 at 10:42amproblem 4_2 part :1

A load W=2 kN is applied vertically to joint C of truss ABCDE as indicated.

plz help

- Physics -
**Any**, Saturday, May 18, 2013 at 3:32pmPlease HW4_2: Part I?

- Physics -
**Maga**, Sunday, May 19, 2013 at 10:15amAnybody for Problem: HW4_2: Part I please?

- Physics -
**FLu**, Sunday, May 19, 2013 at 1:31pmHas somebody formula for HW4_2: Part I?

- Physics -
**Yu**, Sunday, May 19, 2013 at 2:14pmPlease HW4_2: Part I?

- Physics -
**simonsay**, Monday, May 20, 2013 at 8:18amN_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

- Physics -
**FLu**, Monday, May 20, 2013 at 11:31amSimonsay thanks. Something is wrong with the expression, could you please check?

I don't get right answer.

- Physics -
**simonsay**, Monday, May 20, 2013 at 12:54pm-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

- Physics -
**FLu**, Monday, May 20, 2013 at 1:31pmSimonsay, 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

- Physics -
**Help**, Monday, May 20, 2013 at 1:32pmPlease help simonsay! Have problem with expresion too.

- Physics -
**regz**, Monday, May 20, 2013 at 1:48pmtry 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

- Physics -
**simonsay**, Monday, May 20, 2013 at 8:38pmyou must type the numbers, curly N just to indicate where is aplied

- Physics -
**simonsay**, Monday, May 20, 2013 at 8:38pmyou must type the numbers, curly N just to indicate where is aplied

- Physics -
**FLu**, Monday, May 20, 2013 at 9:45pmThanks regz and simonsay!

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