Hibbeler R.C. Structural Analysis

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14.7 Nodal Coordinates

On occasion a truss can be supported by a roller placed on an incline,

and when this occurs the constraint of zero deflection at the support

(node) cannot be directly defined using a single horizontal and vertical

global coordinate system. For example, consider the truss in Fig. 14–12a.

The condition of zero displacement at node ① is defined only along the

axis, and because the roller can displace along the axis this node

will have displacement components along both global coordinate axes, x,

y. For this reason we cannot include the zero displacement condition at

this node when writing the global stiffness equation for the truss using

x, y axes without making some modifications to the matrix analysis

procedure.

To solve this problem, so that it can easily be incorporated into a

computer analysis, we will use a set of nodal coordinates located

at the inclined support. These axes are oriented such that the reactions

and support displacements are along each of the coordinate axes,

Fig. 14–12a. In order to determine the global stiffness equation for the

truss, it then becomes necessary to develop force and displacement

transformation matrices for each of the connecting members at this

support so that the results can be summed within the same global x, y

coordinate system. To show how this is done, consider truss member 1

in Fig. 14–12b, having a global coordinate system x, y at the near node

, and a nodal coordinate system at the far node When

displacements D occur so that they have components along each of these

axes as shown in Fig. 14–12c, the displacements d in the direction

along the ends of the member become

= D

ﬂ

cos u

ﬂ

+ D

ﬂ

cos u

ﬂ

= D

cos u

+ D

cos u

x¿

F .y–x–,

y–x–,

x–y–

560

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

x–

y–

(a)

Fig. 14–12

14.7 NODAL COORDINATES 561

These equations can be written in matrix form as

Likewise, forces q at the near and far ends of the member, Fig. 14–12d,

have components Q along the global axes of

which can be expressed as

The displacement and force transformation matrices in the above

equations are used to develop the member stiffness matrix for this

situation. Applying Eq. 14–15, we have

Performing the matrix operations yields,

(14–24)

This stiffness matrix is then used for each member that is connected to

an inclined roller support, and the process of assembling the matrices to

form the structure stiffness matrix follows the standard procedure. The

following example problem illustrates its application.

k =

-l

ﬂ

-l

ﬂ

-l

ﬂ

-l

ﬂ

-l

ﬂ

-l

ﬂ

-l

ﬂ

-l

ﬂ

k = D

0 l

ﬂ

0 l

ﬂ

1 -1

-11

0 0

0 0 l

ﬂ

k = T

k¿T

ﬂ

T= D

0 l

ﬂ

0 l

ﬂ

= q

cos u

ﬂ

= q

cos u

ﬂ

= q

cos u

= q

cos u

d= c

00l

ﬂ

y¿

x–

y–

x¿

global coordinates

nodal

coordinates

local

coordinates

(b)

x–

y–

x¿

global

coordinates

(c)

–

Fx–

cos u

Fx–

cos u

x–

Fy–

cos u

y–

x–

y–

x–

x¿

(d)

Fx–

Fy–

x–

y–

EXAMPLE 14.6

Determine the support reactions for the truss shown in Fig. 14–13a.

SOLUTION

Notation. Since the roller support at ② is on an incline, we must use

nodal coordinates at this node. The joints and members are numbered

and the global x, y axes are established at node ③, Fig. 14–13b. Notice

that the code numbers 3 and 4 are along the axes in order to use

the condition that

Member Stiffness Matrices. The stiffness matrices for members 1

and 2 must be developed using Eq. 14–24 since these members have

code numbers in the direction of global and nodal axes. The stiffness

matrix for member 3 is determined in the usual manner.

Member 1. Fig. 14–13c,

Member 2. Fig. 14–13d,

Member 3.

= AE

5612

0.128 0.096 -0.128 -0.096

0.096 0.072 -0.096 -0.072

-0.128 -0.096 0.128 0.096

-0.096 -0.072 0.096 0.072

= 0.6l

= 0.8,

= AE

12 3 4

00 0 0

0 0.3333 -0.2357 -0.2357

0 -0.2357 0.1667 0.1667

ﬂ

=-0.707l

ﬂ

=-0.707,l

=-1,l

= 0,

= AE

56 3 4

0.25 0 -0.17675 0.17675

000 0

-0.17675 0 0.125 -0.125

0.17675 0 -0.125 0.125

ﬂ

=-0.707l

ﬂ

= 0.707,l

= 0,l

= 1,

= 0.

y–x–,

562

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

Fig. 14–13

4 m

(a)

3 m

30 kN

(b)

45

30 kN

y–

x–

y¿

x¿

y–

x–

 45

(c)

y–

 135

(d)

x¿

y–

x–

y¿

x–

 135

y–

 135

Structure Stiffness Matrix. Assembling these matrices to determine

the structure stiffness matrix, we have

(1)F

V= AE F

0.128 0.096 0 0 -0.128 -0.096

0.096 0.4053 -0.2357 -0.2357 -0.096 -0.072

0 -0.2357 0.2917 0.0417 -0.17675 0

0 -0.2357 0.0417 0.2917 0.17675 0

-0.128 -0.096 -0.17675 0.17675 0.378 0.096

-0.096 -0.072 0 0 0.096 0.072

Carrying out the matrix multiplication of the upper partitioned matrices,

the three unknown displacements D are determined from solving the

resulting simultaneous equations, i.e.,

The unknown reactions Q are obtained from the multiplication of

the lower partitioned matrices in Eq. (1). Using the computed

displacements, we have,

Ans.

Ans.=-22.5 kN

=-0.0961352.52- 0.0721-157.52+ 01-127.32

=-7.5 kN

=-0.1281352.52- 0.0961-157.52- 0.17675 1-127.32

= 31.8 kN

= 01352.52- 0.23571-157.52+ 0.04171-127.32

-127.3

-157.5

352.5

14.7 NODAL COORDINATES 563

14.8 Trusses Having Thermal Changes

and Fabrication Errors

If some of the members of the truss are subjected to an increase or

decrease in length due to thermal changes or fabrication errors, then it is

necessary to use the method of superposition to obtain the solution. This

requires three steps. First, the fixed-end forces necessary to prevent

movement of the nodes as caused by temperature or fabrication are

calculated. Second, the equal but opposite forces are placed on the truss

at the nodes and the displacements of the nodes are calculated using

the matrix analysis. Finally, the actual forces in the members and the

reactions on the truss are determined by superposing these two results.

This procedure, of course, is only necessary if the truss is statically

indeterminate. If the truss is statically determinate, the displacements at

the nodes can be found by this method; however, the temperature

changes and fabrication errors will not affect the reactions and the

member forces since the truss is free to adjust to these changes of length.

Thermal Effects. If a truss member of length L is subjected to a

temperature increase the member will undergo an increase in length

of where is the coefficient of thermal expansion. A

compressive force applied to the member will cause a decrease in

the member’s length of If we equate these two

displacements, then This force will hold the nodes of the

member fixed as shown in Fig. 14–14, and so we have

Realize that if a temperature decrease occurs, then becomes negative

and these forces reverse direction to hold the member in equilibrium.

We can transform these two forces into global coordinates using

Eq. 14–10, which yields

(14–25)

Fabrication Errors. If a truss member is made too long by an

amount before it is fitted into a truss, then the force needed to

keep the member at its design length L is and so for the

member in Fig. 14–14, we have

AE¢L

= AE¢L>L,

¢L

T= D

0 l

TAEa¢Tc

-1

d= AEa¢TD

-l

¢T

=-AEa¢T

= AEa¢T

= AEa¢T.

¢L¿=q

L>AE.

a¢L = a¢TL,

¢T,

564

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

Fig. 14–14

y¿

x¿

)

14.8 TRUSSES HAVING THERMAL CHANGES AND FABRICATION ERRORS 565

If the member is originally too short, then becomes negative and

these forces will reverse.

In global coordinates, these forces are

(14–26)

Matrix Analysis. In the general case, with the truss subjected to

applied forces, temperature changes, and fabrication errors, the initial

force-displacement relationship for the truss then becomes

(14–27)

Here is a column matrix for the entire truss of the initial fixed-end

forces caused by the temperature changes and fabrication errors of the

members defined in Eqs. 14–25 and 14–26.We can partition this equation

in the following form

Carrying out the multiplication, we obtain

(14–28)

(14–29)

According to the superposition procedure described above, the unknown

displacements are determined from the first equation by subtracting

and from both sides and then solving for This yields

Once these nodal displacements are obtained, the member forces are

then determined by superposition, i.e.,

If this equation is expanded to determine the force at the far end of the

member, we obtain

(14–30)

This result is similar to Eq. 14–23, except here we have the additional

term which represents the initial fixed-end member force due to

temperature changes and/or fabrication error as defined previously. Re-

alize that if the computed result from this equation is negative, the mem-

ber will be in compression.

The following two examples illustrate application of this procedure.

[-l

-l

T+ 1q

q = k¿TD + q

= K

-1

- K

- 1Q

.1Q

= K

+ K

+ 1Q

= K

+ K

+ 1Q

d= c

d+ c

Q = KD + Q

AE¢L

-l

¢L

EXAMPLE 14.7

Determine the force in members 1 and 2 of the pin-connected assembly

of Fig. 14–15 if member 2 was made 0.01 m too short before it was fitted

into place. Take AE = 8110

2 kN.

566

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

SOLUTION

Since the member is short, then and therefore

applying Eq. 14–26 to member 2, with we have

The structure stiffness matrix for this assembly has been established

in Example 14–5. Applying Eq. 14–27, we have

AE1-0.012

-0.8

-0.6

0.8

0.6

T= AE D

0.0016

0.0012

-0.0016

-0.0012

=-0.6,l

=-0.8,

¢L =-0.01 m,

(1)H

X= AE H

0.378 0.096 0 0 -0.128 -0.096 -0.25 0

0.096 0.405 0 -0.333 -0.096 -0.072 0 0

00000000

0 -0.333 0 0.333 0 0 0 0

-0.128 -0.096 0 0 0.128 0.096 0 0

-0.096 -0.072 0 0 0.096 0.072 0 0

-0.25 0 0 0 0 0 0.25 0

00000000

X H

X+ AE H

0.0016

0.0012

-0.0016

-0.0012

Fig. 14–15

4 m

3 m

Partitioning the matrices as shown and carrying out the multiplication

to obtain the equations for the unknown displacements yields

d= AE c

0.378 0.096

0.096 0.405

d+ AE c

00 -0.128 -0.096 -0.25 0

0 -0.333 -0.096 -0.072 0 0

V+ AE c

0.0016

0.0012

which gives

Solving these equations simultaneously,

Although not needed, the reactions Q can be found from the expan-

sion of Eq. (1) following the format of Eq. 14–29.

In order to determine the force in members 1 and 2 we must apply

Eq. 14–30, in which case we have

Member 1. so that

Ans.

Member 2. soAE = 8110

2 kN,L = 5 m,l

=-0.6,l

=-0.8,

=-5.56 kN

8110

[0 -101]D

-0.003704

-0.002084

T+ [0]

AE = 8110

2 kN,L = 3 m,l

= 1,l

= 0,

=-0.002084 m

=-0.003704 m

0 = AE[0.096D

+ 0.405D

] + AE[0] + AE[0.0012]

0 = AE[0.378D

+ 0.096D

] + AE[0] + AE[0.0016]

Ans. q

= 9.26 kN

8110

[0.8 0.6 -0.8 -0.6]D

-0.003704

-0.002084

T-

8110

2 1-0.012

14.8 TRUSSES HAVING THERMAL CHANGES AND FABRICATION ERRORS 567

EXAMPLE 14.8

Member 2 of the truss shown in Fig. 14–16 is subjected to an increase

in temperature of 150°F. Determine the force developed in member 2.

Take Each member has a cross-

sectional area of A = 0.75 in

E = 29110

2 lb>in

.a = 6.5110

-6

2>°F,

568

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

SOLUTION

Since there is a temperature increase, Applying

Eq. 14–25 to member 2, where we have

The stiffness matrix for this truss has been developed in Example 14–2.

T= AE16.52110

-6

211502D

0.7071

-0.7071

T= AE D

0.000689325

-0.000689325

= 0.7071,l

= 0.7071,

¢T =+150°F.

10 ft

(1)H

X= AE H

0.135 0.035 0 0 0 -0.1 -0.035 -0.035

0.035 0.135 0 -0.1 0 0 -0.035 -0.035

0 0 0.135 -0.035 0.035 -0.035 -0.1 0

0 -0.1 -0.035 0.135 -0.035 0.035 0 0

0 0 0.035 -0.035 0.135 -0.035 0 -0.1

-0.1 0 -0.035 0.035 -0.035 0.135 0 0

-0.035 -0.035 -0.1 0 0 0 0.135 0.035

-0.035 -0.035 0 0 -0.1 0 0.035 0.135

X+ AE H

0.000689325

-0.000689325

Fig. 14–16

Expanding to determine the equations of the unknown displacements,

and solving these equations simultaneously, yields

Using Eq. 14–30 to determine the force in member 2, we have

=-0.002027 ft

=-0.009848 ft

=-0.002027 ft

=-0.01187 ft

=-0.002027 ft

Ans.=-6093 lb =-6.09 k

0.75[29110

1022

[-0.707 -0.707 0.707 0.707]D

-0.002027

-0.01187

T- 0.75[29110

2][6.5110

-6

2]11502

Note that the temperature increase of member 2 will not cause any

reactions on the truss since externally the truss is statically determinate.

To show this, consider the matrix expansion of Eq. (1) for determining

the reactions. Using the results for the displacements, we have

+ 0 - 0.11-0.0020272] + AE[-0.000689325] = 0

= AE[-0.0351-0.0020272- 0.0351-0.011872+ 0

- 0.11-0.0020272+ 0 + 0] + AE[-0.000689325] = 0

= AE[-0.0351-0.0020272- 0.0351-0.011872

+ 0.0351-0.0098482- 0.0351-0.0020272] + AE[0] = 0

= AE[-0.11-0.0020272+ 0 - 0.0351-0.0020272

14.8 TRUSSES HAVING THERMAL CHANGES AND FABRICATION ERRORS 569