Example 32 Two Dimensional Plane Truss

Consider the pin-jointed, plane-frame ABC loaded as shown in Figure 3.9.

10 kN

10 kN

Construct a table in terms of tension coefficients and an X/Y co-ordinate system as shown in Table 3.1.

The equilibrium equations are solved in terms of the 't' values and hence the member forces and support reactions are evaluated and entered in the table as shown in Table 3.1. Consider joint B:

There are only two unknowns and two equations, hence: Adding both equations substitute for in tlic first equation ^ /rc = -4.76

Force in member AB - x ¿ab - + (1.43 x 3.0)- +7.13 kN HE Force in member = foe * = - (4.76x J.241) = - 20.2 kN STRUT

Joints A and C can be considered in a similar manner until all unknown values, including reactions, have been determined.

The reader should complete this solution to obtain the following values: FAC=+14.28

Joint

Equilibrium kqifeil in ns

Mcruhtr

/

Ltnglh

Fnrcc

<kNl

X

-l'Ait + + A, = û

AR

+ LJ,1

5,0

+ 7JÎ

Y

3^1. + A, ■ 0

AC

7

7,0

BC

- 4.7Ü

4J43

- 2oS8

1)

X

=0

Support RmclkittMkNl

V

-3JAIÎ - 3fiir - IO "0

C(iiii|iiiiini(

X

y

c

X

- % - Û

Support A

Y

+3/nr +1\ = 0

Support C

itro

In the case of a space frame, each joint has three co-ordinates and the forces have components in the three orthogonal X, Y and Z directions. This leads to (3xNumber. of joints) equations which can be solved as above to determine the 't' values and subsequently the member forces and support reactions.

+1 0

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