39. Maximum Shear. It is sometimes desirable to know the greatest or maximum value of the shear in a given case. This value can always be found with certainty by constructing the shear diagram, from which the maximum value of the shear is evident at a glance. In any case it can most readily be computed if one knows the section for which the shear is a maximum. The student should examine all the shear diagrams in the preceding articles and those that he has drawn, and see that

1. In cantilevers fixed in a wall, the maximum shear occurs at the wall.

2. In simple beams, the maximum shear occurs at a section next to one of the supports.

By the use of these propositions one can determine the value of the maximum shear without constructing the whole shear diagram. Thus, it is easily seen (referring to the diagrams, page 55) that for a

 Cantilever, end load P, maximum shear P " , uniform load W, " " W Simple beam, middle load P, " " ½P " " , uniform " W, "» " ½W

40. Bending Moment. By bending moment at (or for) a section of a loaded beam, is meant the algebraic sum of the moments of all the loads (including weight of beam) and reactions to the left or right of the section with respect to any point in the section.

41. Rule of Signs. We follow the rule of signs previously stated (Art. 29) that the moment of a force which tends to produce clockwise rotation is plus, and that of a force which tends to produce counter-clockwise rotation is minus; but in order to get the same sign for the bending moment whether computed from the right or left, we change the sign of the sum of the moments when computed from the loads and reactions on the right. Thus for section a. Fig. 8, the algebraic sums of the moments of the forces are: when computed from the left,

-1,000 X 5 + 3,000 X l= -2,000 foot-pounds; and when computed from the right,

1,000 X 19-3,000 X15 + 2,000 X13 + 2,000 X1= + 2,000 footpounds. The bending moment at section a is -2,000 foot-pounds.

Again, for section b, the algebraic sums of the moments of the forces are: when computed from the left,

-1,000 X 22 + 3,000 X 18-2,000 X 16-2,000 X 4+3,000 X 2= -2,000 foot-pounds; and when computed from the right,

1,000X2=+ 2,000 foot-pounds. The bending moment at the section is -2.000 foot-pounds.

BUILDINGS FOR THE PITTSBURG PLATE GLASS COMPANY, CRYSTAL CITY, MISSOURI.

Grinder and Polisher building, 408 ft. by 532 ft.: Laying Yard building, 100 ft. by 578 ft.; Lehr building, 68 ft. by 475 ft. and 114 ft. by 171 ft.; Stripper building, 90 ft. by 408 ft.; Warehouse, 153 ft. by 304 ft. and 90 ft. by 173 ft. Total weight of steel work, 6,280,000 lbs.

Courtesy of American Bridge Company.

It is usually convenient to compute the bending moment for a section from the forces to the right or left according as there are fewer forces (loads and reactions) on the right or left side of the section.

42. Units. It is customary to express bending moments in inch-pounds, but often the foot-pound unit is more convenient. To reduce foot-pounds to inch-pounds, multiply by twelve.

43. Notation. We shall use M to denote bending moment at any section, and the bending moment at a particular section will be denoted by that letter subscripted; thus M1 M2, etc., denote values of the bending moment for sections one, two, etc., feet from the left end of the beam.

Examples. 1. Compute the bending moments for sections one foot apart in the beam represented in Fig. 9, neglecting the weight of the beam. (The right and left reactions are 3,700 and 2,300 pounds respectively. See example 1, Art. 33.)

Since there are no forces acting on the beam to the left of the left support, M1=0. To the left of the section one foot from the left end there is but one force, the left reaction, and its arm is one foot; hence M1= + 2,300x1=2,300 foot-pounds. To the left of a section two feet from the left end there are two forces, 2,300 and 1,000 pounds, and their arms are 2 feet and 1 foot respectively; hence M,=+ 2,300 X 2-1,000x1=3,600 foot-pounds. At the left of all sections between B and C there are only two forces, 2,300 and 1,000 pounds; hence

 M3 = + 2,300 X 3 - 1,000 X 0 = + 4,900 foot-pounds, M4 = + 2,300 X 4 - 1,000 X 3 = + 6,200 " M5 = + 2,300 X 5 - 1,000 X 4 = + 7,500 " M6 = + 2,300 X 6 - 1,000 X 5 = + 8,800 "

To the right of a section seven feet from the left end there are two forces, the 3,000-pound load and the right reaction (3,700 pounds), and their arms with respect to an origin in that section are respectively one foot and three feet; hence

M7= - (-3,700 X 3 + 3,000 X1)= + 8,100 foot-pounds.

To the right of any section between E and D there is only one force, the right reaction; hence

M8= -(-3,700 X 2)=7,400 foot-pounds, M9= - (-3,700Xl)=3,700 "

Clearly M10=0.

2. A simple beam 10 feet long and supported at its ends weighs 400 pounds, and bears a uniformly distributed load of 1,600 pounds. Compute the bending moments for sections two feet apart.

Each reaction equals one-half the whole load, that is, ½| of (1,600 + 400) =1,000 pounds, and the load per foot including weight of the beam is 200 pounds. The forces acting on the beam to the left of the first section, two feet from the left end, are the left reaction (1,000 pounds) and the load (including weight) on the part of the beam to the left of the section (400 pounds). The arm of the reaction is 2 feet and that of the 400-pound force is 1 foot (the distance from the middle of the 400-pound load to the section). Hence

M2= + 1,000 X 2-400 X1= +1,600 foot-pounds.

The forces to the left of the next section, 4 feet from the left end, are the left reaction and all the load (including weight of beam) to the left (800 pounds). The arm of the reaction is 4 feet, and that of the 800-pound force is 2 feet; hence

M4= +1,000 X 4-800 X 2= + 2,400 foot-pounds.

Without further explanation the student should see that

M6= + 1,000 X 6-1,200 X 3= + 2,400 foot-pounds, M8= +1,000 X 8-1,600 X 4= +1,600

Evidently M0=M10=0.