Use of a Standard Parabola for Drawing Diagrams of Bending Moment. 663 
w — ~ and \w 
n " n^ 
or X = wn^ ju, = • ^. 
Since w occurs as a linear factor both in shear and in bending moment 
it may be neglected at present ; results got for 1 lb. per foot run can be used 
for any number {vi) of lbs. per foot run if we multiply the results for 
shear and for bending moment by the same factor w. 
We therefore take "h — n, — • 
It is open now to choose h so as to have a parabola of suitable shape. If 
the latus rectum is too great, the part near the vertex is difficult to draw 
and cut out accurately ; but again the parabola must be wide enough to 
allow easy spanning between two points of support. 
A convenient shape was a parabola drawn on squared paper with semi- 
latus rectum 20 divisions. Thus the equation *? — is satisfied by 
I = 20, „ = 20 ; hence A: = ^V- 
With this parabola we have -k — n, — \^n^. 
The parabola has been drawn out for a range | =: — 40 to | = +40. 
If we use tz, = 1, the greatest breadth of span for which this would serve 
would be 80', and then only if the bending moment for that span were 
symmetrical. With ordinary loading and supports the results for a span of 
60' could generally be got ; for broader spans a change in the value of n 
would be necessary. 
The parabola was drawn on 2 mm. squared paper pasted on cardboard 
and cut out ; the part used was the remainder of the paper after the area 
of the parabola had been removed ; the use of this part makes it easy to 
take the abscissa readings. 
When the parabola is placed in position, attention must be paid to 
the sign of the shear which is positive when the slope of the parabola 
is down. 
§ 4. Two examples are shown in the figure ; they have been constructed 
to show lack of symmetry. 
(1) A bar 80' long, 100 lbs. per foot run is supported at the ends, at 
the middle, and at one point midway between the middle and the ends, all 
the supports being at the same level. 
As the largest span is 40' we are safe in using the parabola with n — \. 
In Fig. I, A, B, c, D are the points of support. 
The bending moments at b and c for a bar of 1 lb. per foot run are 
obtained from the equations : 
120ao + 20G-3 = - K64000 4- 8000) 
20a2 + 80(^3 =r - i(8000 + 8000) 
whence G-.. = - 148, - - 13, 
