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SE 130B Homework 2

due 7/24/2012

Problem 1 (30 points)

We are given the truss structure and the loads shown in Figure 1.

Figure 1

a) Number the degrees of freedom for the structure (the free DOFs

must be numbered first).

b) Obtain the member fixed-end force vectors, {FF(i)}, of all three

members.

c) Determine the [Kff] matrix (using the member stiffness matrices in

the global coordinate system and the corresponding ID arrays).

d) Solve the structure (obtain the vector {Uf}).

d) Determine the member axial force diagram.

(1) (2)

(3)

T = 10-5 oC-1

5 kips

3 in. 3 in.

4 in.

4 in. T = 20 oC-1

0.1 in

E = 10000 ksi, A = 1 in2 for all members

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e) Repeat the calculations of steps a)-d), only this time do not include

the effect of the support displacements in the fixed-end force vector.

Instead, include the effect of the support displacements by means of

the equation:

[ ]{ } [ ]{ } { } { }ff f fd d f foK U K U P P+ =

where { }foP includes the effect of the fixed-end forces for loadsother than the support displacements.

f) Determine the member end forces, {F(i)}, for all three members, as

well as the support reactions, using the equation:

{ } ( ) ( ){ }( )elN T

i i

d bd

i=1

P A F =

Problem 2 (10 points)

Obtain the fifth column of the [k] matrix for a frame (Euler-

Bernoulli) member with constant E, I and A, by solving an appropriate

differential equation problem.

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Problem 3 (30 points)

For the structure shown in Figure 2:

a) Number the degrees of freedom (the free DOFs must be numbered

first).

b) Determine the [RBM], [ROT] and [Abf] matrices, as well as the ID

arrays required to assemble [Kff], for all three members.

c) Determine the k' and [k] matrices for all members.

d) Determine the matrix [Kff] and the vector {Pf}.

e) Solve for {Uf}, obtain the element basic forces, { }F' , the element

end forces {F}, and the reactions { }( ) ( ){ }( )

elN Ti i

d bd

i=1

P A F = .

f) Determine the deformed shape and the internal force diagrams, [M]

and [N], for all members.

X

Y

z10 in.

6 in. (1)

(2)

(3)

E = 29000 ksi, I = 10 in4 , A = 2 in2 for all

members

5 kips

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Note: positive internal force convention for internal member forces:

Problem 4 (30 points)

a. Develop a Matlab routine, called frame.m.

The routine must be able to make all the computations pertaining

to a frame member. More specifically, the routine must take as input

arguments the 2-dimensional vectors X and Y, containing the nodal

abscissas and ordinates of the member ends in the global coordinate

system, the 6-dimensional vector U, containing the nodal displacements

in the global coordinate system (that is, U is the vector Delta that we

have defined in class), and the values E, I and A. Additionally, another

parameter, itask, must be given as input in the member. This integer

will determine what kind of task the element routine must conduct. If

itask = 1, the routine must calculate the element stiffness matrix inthe GLOBAL coordinate system. If itask = 2, the routine must calculate

the element end forces in the GLOBAL coordinate system, {F}, as well

as the element basic deformation vector, du and basic force vector,

dF. In summary, a call command for the routine in Matlab must read:

x

x

Bending moments:

Axial forces:

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[k,F,du,dF]=frame(X,Y,U,E,I,A,itask)

If itask = 1, then the routine returns k, while the vectors F, du and Nwill have zero values (that is, they are NOT CALCULATED!).

If itask =2, then [k] will have zero values (we do NOT calculate it!),

while the routine will return vectors F, du and dF.

b. Use the routine to calculate the stiffness matrices [k] for the

following two frame members:

c. Finally, determine the basic deformation vector and basic forcevector of the above two members, corresponding to the following

displacement vector in the global coordinate system:

X

E = 29000 ksi

A = 1 in2

I = 10 in4

YNode 1

X1

= 16 in

Y1 = 3 in

Node 2

X2 = 16 in

Y2 = 13 in

E = 29000 ksi

A = 1 in2

I = 10 in4

Node 1

X1 = 7 in

Y1 = 3 in

Node 2X2 = 17 in

Y2

= 3 in

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{ }

1

2

3

4

5

6

0

0.1 in

0.05 rad

0.01 in

0.02 in

0.01 rad

= =