Influence Lines - Structural and Stress Analysis - page 678

but in this case, R B in Eq. (x) is given by Eq. (viii). Substituting for R B from Eq. (viii)
in Eq. (x) we obtain

1
256 ( x 1 −

24 x 1 + R A =− S D =− 176 x 1 − 384) (4 . 0m ≤ x 1 ≤ 8 . 0m) (xiv)

Therefore the S D influence line consists of three segments, a 1 g, fb 1 and b 1 c 1 .

M D influence line

With the unit load between A and D

M D = R A × 2 − 1(2 − x 1 ) (xv)

Substituting for R A from Eq. (xi) in Eq. (xv) and simplifying, we obtain

1
128 ( x 1 +

M D = 48 x 1 ) (0 ≤ x 1 ≤ 2 . 0m) (xvi)

When x 1 = 0, M D = 0 and when x 1 = 2 . 0m, M D = 0 . 81, the ordinate d 2 hinthe M D

influence line in Fig. 20.21(d).

Now with the unit load between D and B

M D = R A × 2 (xvii)

Therefore, substituting for R A from Eq. (xi) we have

1
128 ( x 1 −

M D = 80 x 1 + 256) (2 . 0m ≤ x 1 ≤ 4 . 0m) (xviii)

From Eq. (xviii) we see that when x 1 = 2 . 0m, M D = 0 . 81, again the ordinate d 2 hin

Fig. 20.21(d). Also, when x 1 = 4 . 0m, M D = 0.

Finally, with the unit load between B and C, M D is again given by Eq. (xvii) but in
which R A is given by Eq. (xiv). Hence

1
128 ( x 1 −

24 x 1 + M D =− 176 x 1 − 384) (4 . 0m ≤ x 1 ≤ 8 . 0m) (xix)

The maximum ordinates in the S D and M D influence lines for the span BC may be
found by differentiating Eqs (xiv) and (xix) with respect to x 1 , equating to zero and
then substituting the resulting values of x 1 back in the equations. Thus, for example,
from Eq. (xiv)

d S D
d x 1 =

1
256 (3 x 1 −

48 x 1 +

=

176) 0

from which x 1 = 5 . 7m. Hence

S D (max) = 0 . 1

Similarly M D (max) =− 0 . 2at x 1 = 5 . 7m.