Then, 236,210 Ft

lb. - 112,250 ft.-lb. = 123,960 ft.-lb., the maximum bending moment. The bending moment in inch-pounds is 123,960 ft.-lb. X 12 = 1,487,520 in.-lb

If W= the load and L = the span, the maximum bending moment in a simple beam uniformly loaded is (W / 2)x(L/2)

- (W/2)x(L/4)= WL/8.(L/4 is the distance from the center of beam to center of gravity of each half of the uniform load.)

The maximum bending moment in a beam with a load con eentrated at the center is WL/4. Thus, a beam uniformly loaded will carry safely twice as much as if the. load were concentrated at the center.

While, by the preceding principles, the bending moment in any beam supported at one or two supports may be deter mined, it is more convenient to use concise formulas. The following table gives formulas for the maximum bending moment, maximum safe loads, and greatest deflections (or sag), for beams loaded and supported in different ways:

Table XXVI

Method of Loading.

Maximum

Bending

Moment.

if.

Maximum Load. W.

Deflection. D.

Length in Load in Feet. Pounds.

Ft.-Lb.

In.-Lb.

Lb.

In.

Table XXVI 246

WL/8

3WL/2

2QS/3L

5Wl3/384EI

Table XXVI 247

WL/6

2WL

QS/2L

Wl3/60EI

Table XXVI 248

WL/2

6WL

QS/6L

Wl3/8EI

Table XXVI 249

WL

12WL

QS/12L

Wl3/3EI

Table XXVI 250

WL 4

3WL

QS/3L

Wl3/48EI

Table XXVI 251

WA/2

6WA

QS/6A

(Wa/48EI)

X (3l2-4a2)

Between Supports.

WA/2

6WA

Q S/6A

(Wa/16EI) X

(I -2 a)2

L = length in feet; I = length in inches; W= total load in pounds; E = modulus of elasticity; I = moment of inertia; Q = section modulus; S = safe stress on the extreme fibers of the beam section (= modulus of rupture / factor of safety). In figuring deflections, all lengthsmust be expressed in inches; and small letters I, a, and b are used as reminders.