Calculation Procedure:
- Start the graphical construction
Draw a line through A and C, intersecting the vertical line through B at E. Draw a line
through B and C, intersecting the vertical line through A and F. Draw the vertical line GH
through D.
Let 9 denote the angle between AE and the horizontal. Lines through B and D perpen-
dicular to AE (omitted for clarity) make an angle 6 with the vertical. - Resolve the reaction into components
Resolve the reaction at A into the components ^ 1 and R 2 acting along AE and AB, respec-
tively (Fig. 52). - Determine the value of the first reaction
Let jc denote the horizontal distance from the right-hand support to the unit load, where x
has any value between O and L. Evaluate R 1 by equating the bending moment at B to zero.
Thus MB = R{b cos 6 - x = O; or = R 1 = x/(b cos B). - Evaluate the second reaction
Place the unit load within the interval CB. Evaluate R 2 by equating the bending moment at
C to zero. Thus Mc = R 2 d = O; /. ^ 2 = O. - Calculate the bending moment at D when the unit load lies
within the interval CB
Thus, MD = -R 1 V cos 6 = -[(v cos 6)/(b cos 0)]jt, or MD = -VxIb 9 Eq. a. When JC = TH, MD =
—vmlb. - Place the unit load in a new position, and determine
the bending moment
Place the unit load within the interval AD. Working from the right-hand support, proceed
in an analogous manner to arrive at the following result: MD - v'(L — x)/a, Eq. b. When x
= L — n, MD = v'nla. - Place the unit load within another interval, and evaluate
the second reaction
Place the unit load within the interval DC, and evaluate R 2. Thus Mc - R 2 d — (x — m) = O,
or R 2 = (X- m)ld.
Since both R 1 and R 2 vary linearly with respect to x, it follows that MD is also a linear
function of Jt. - Complete the influence line
In Fig. 526, draw lines BR and AS to represent Eqs. a and b, respectively. Draw the
straight line SR, thus completing the influence line. The point T at which this line inter-
sects the base is termed the neutral point. - Locate the neutral point
To locate T, draw a line through A and D in Fig. 52a intersecting BF at J. The neutral
point in the influence line lies vertically below J; that is, MD is zero when the action line
of the unit load passes through J.
The proof is as follows: Since MD = O and there are no applied loads in the interval
AD, it follows that the total reaction at A is directed along AD. Similarly, since Mc = O
and there are no applied loads in the interval CB, it follows that the total reaction at
B is directed along BC. Because the unit load and the two reactions constitute a bal-
anced system of forces, they are collinear. Therefore, J lies on the action line of the unit
load.
Alternatively, the location of the neutral point may be established by applying the geo-
metric properties of the influence line.