Analysis of Pin-jointed Trusses - Structural and Stress Analysis - page 107

so that

t EB − t EC + t EA = 0 (x)

from Eq. (viii)

t EB ( − 2 − 0) + t EC ( − 2 − 0) + t EA ( − 2 − 0) − 60 = 0

whence

t EB + t EC + t EA + =

30 0 (xi)

and from Eq. (ix)

t EB (2 + 2) + t EC ( − 4 + 2) + t EA ( − 4 + 2) = 0

which gives

t EB −

0 . 5 t EC − 0 . 5 t EA =

0 (xii)

Subtracting Eq. (xi) from Eq. (x) we have

− 2 t EC − 30 = 0

so that

t EC =−

15

Now subtracting Eq. (xii) from Eq. (xi) (or Eq. (x)) yields

1 . 5 t EC + 1 . 5 t EA + 30 = 0

which gives

t EA =− 5

Finally, from any of Eqs (x)-(xii),

t EB =− 10

The length of each of the members is now calculated, except that of EF which is given
(

= 2 m). Using Pythagoras' theorem

( x B −

x F ) 2 y F ) 2 z F ) 2

L FB = + ( y B − + ( z B −

whence

(

L FB = − 2 − 0) 2 + ( − 2 − 0) 2 + (2 − 0) 2 = 3 . 46m

Similarly

L FD = L EC = L EA = 3 . 46m L EB = 4 . 90m