Influence Lines - Structural and Stress Analysis - page 655

Maximum positive shear force at K

It is clear from inspection that S K will be a maximum with the 5 kN load just to the
left of K, in which case the 3 kN load is off the beam and the ordinate under the 4 kN
load in the S K influence line is, from similar triangles, 0.1. Then

S K (max) = 5 × 0 . 3 + 4 × 0 . 1 = 1 . 9kN

Maximum negative shear force at K

There are two possible load positions which could give the maximum negative value of
shear force at K; neither can be eliminated by inspection. First we shall place the 3 kN
load just to the right of K. The ordinates under the 4 and 5 kN loads are calculated
from similar triangles and are − 0 . 5 and − 0 . 3, respectively. Then

S K = × − + × − + × − =−

3 ( 0 . 7) 4 ( 0 . 5) 5 ( 0 . 3) 5 . 6kN

Now with the 4 kN load just to the right of K, the ordinates under the 3 and 5 kN loads
are 0 . 1 and − 0.5, respectively. Then

S K = 3 × (0 . 1) + 4 × ( − 0 . 7) + 5 × ( − 0 . 5) =− 5 . 0kN

Therefore the maximum negative value of S K is − 5 . 6 kN and occurs with the 3 kN load

immediately to the right of K.

Maximum bending moment at K

We position the loads in accordance with the criterion of Eq. (20.14). The load per unit
length of the complete beam is (3 0 . 6 kN/m. Therefore if we position the

4 kN load at K and allocate 0 . 6 kN of the load to AK the load per unit length on AK is
(3

+ 4 + 5) / 20 =

0 . 6 kN/m.
The maximum bending moment at K therefore occurs with the 4 kN load at K; in this
example the critical load position could have been deduced by inspection.

+ 0 . 6) / 6 = 0 . 6 kN/m and the load per unit length on KB is (3 . 4 + 5) / 14 =

With the loads in this position the ordinates under the 3 and 5 kN loads in the M K
influence line are 1 . 4 and 3 . 0, respectively. Then

M K (max) = 3 × 1 . 4 + 4 × 4 . 2 + 5 × 3 . 0 = 36 . 0 kNm

DISTRIBUTED LOADS

Figure 20.7(a) shows a simply supported beam AB on which a uniformly distributed
load of intensity w and length l is crossing from left to right. Suppose we wish to
obtain values of shear force and bending moment at the section K of the beam. Again
we construct the S K and M K influence lines using either of the methods described in
Sections 20.1 and 20.2.