Page 93 - Handbook of Civil Engineering Calculations, Second Edition

1.76 STRUCTURAL STEEL ENGINEERING AND DESIGN

Calculation Procedure:

1. 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 denote the angle between AE and the horizontal. Lines through B and D perpen-
dicular to AE (omitted for clarity) make an angle with the vertical.
2. Resolve the reaction into components
Resolve the reaction at A into the components R 1 and R 2 acting along AE and AB, respec-
tively (Fig. 52).
3. Determine the value of the first reaction
Let x denote the horizontal distance from the right-hand support to the unit load, where x
has any value between 0 and L. Evaluate R 1 by equating the bending moment at B to zero.
Thus M B R 1 b cos x 0; or R 1 x/(b cos ).
4. 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 M C R 2 d 0; R 2 0.
5. Calculate the bending moment at D when the unit load lies
within the interval CB
Thus, M D R 1 v cos [(v cos )/(b cos )]x, or M D vx/b, Eq. a. When x m,
M D vm/b.
6. 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: M D v
(L x)/a, Eq. b. When
x L n, M D v
n/a.
7. 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 M C R 2 d (x m)
0, or R 2 (x m)/d.
Since both R 1 and R 2 vary linearly with respect to x, it follows that M D is also a linear
function of x.
8. Complete the influence line
In Fig. 52b, 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.
9. 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, M D is zero when the action line
of the unit load passes through J.
The proof is as follows: Since M D 0 and there are no applied loads in the interval
AD, it follows that the total reaction at A is directed along AD. Similarly, since M C 0
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.