This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

On almost every cross-section of a loaded beam there are three kinds of stress, namely tension, compression and shear. The first two are often called fibre stresses because they act along the real fibres of a wooden beam or the imaginary ones of which we may suppose iron and steel beams composed. Before taking up the subject of these stresses in beams it is desirable to study certain quantities relating to the loads, and on which the stresses in a beam depend. These quantities are called external shear and bending moment, and will now be discussed.

34. External Shear. By external shear at (or for) any section of a loaded beam is meant the algebraic sum of all the loads (including weight of beam) and reactions on either side of the section. This sum is called external shear because, as is shown later, it equals the shearing stress (internal) at the section. For brevity, we shall often say simply "shear" when external shear is meant.

35. Rule of Signs. In computing external shears, it is customary to give the plus sign to the reactions and the minus sign to the loads. But in order to get the same sign for the external shear whether computed from the right or left, we change the sign of the sum when computed from the loads and reactions to the right. Thus for section a of the beam in Fig. 8 the algebraic sum is, when computed from the left,

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

-1,000 + 3,000-2,000-2,000 = -2,000 pounds. The external shear at section a is +2,000 pounds.

Again, for section b the algebraic sum is, when computed from the left,

-1,000 + 3,000-2,000-2,000 + 3,000 = + 1,000 pounds; and when computed from the right, -1,000 pounds.

The external shear at the section is +1,000 pounds.

It is usually convenient to compute the shear at a section from the forces to the right or left according as there are fewer forces (loads and reactions) on the right or left sides of the section.

36. Units for Shears. It is customary to express external shears in pounds, but any other unit for expressing force and weight (as the ton) may be used.

37. Notation. We shall use V to stand for external shear at any section, and the shear at a particular section will be denoted by that letter subscripted; thus V1, V2, etc., stand for the shears at sections one, two, etc., feet from the left end of a beam.

The shear has different values just to the left and right of a support or concentrated load. We shall denote such values by V' and V"; thus V5" and V5" denote the values of the shear at sections a little less and a little more than 5 feet from the left end respectively.

Examples. 1. Compute the shears 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.)

All the following values of the shear are computed from the left. The shear just to the right of the left support is denoted by

V0", and V0" = 2,300 pounds. The shear just to the left of B is denoted by V1, and since the only force to the left of the section is the left reaction, V1'= 2,300 pounds. The shear just to the right of B is denoted by V1', and since the only forces to the left of this section are the left reaction and the 1,000-pound load,

V1' = 2,300 -1,000 = 1,300 pounds. To the left of all sections between B and C, there are but two forces, the left reaction and the 1,000-pound load; hence the shear at any of those sections equals 2,300-1,000 = 1,300 pounds, or

V2 = V3 = V4= V5 = V6'= 1,300 pounds.

The shear just to the right of C is denoted by V6"; and since the forces to the left of that section are the left reaction and the

1,000 - and 2,000-pound loads,

V6" = 2,300-1,000-2,000 = - 700 pounds.

Without further explanation, the student should understand that

V7 = + 2,300 - 1,000 - 2,000 = - 700 pounds,

V8' = -700,

V8" = + 2,300 - 1,000 - 2,000 - 3,000 = - 3,700,

V9 =V10'= - 3,700,

V10" = + 2.300 - 1,000 - 2,000 - 3,000 + 3,700 = 0

2. A simple beam 10 feet long, and supported at each end, weighs 400 pounds, and bears a uniformly distributed load of 1,G00 pounds. Compute the shears for sections two feet apart.

Evidently each reaction equals one-half the sum of the load and weight of the beam, that is, ½ (1,600 + 400) =1,000 pounds. To the left of a section 2 feet from the left end, the forces acting on the beam consist of the left reaction, the load on that part of the beam, and the weight of that part; then since the load and weight of the beam per foot equal 200 pounds,

V2= 1,000-200 X 2 = 600 pounds. To the left of a section four feet from the left end, the forces are the left reaction, the load on that part of the beam, and the weight ; hence

V = 1,000-200 X 4 = 200 pounds. Without further explanation, the student should see that

v6 | 1,000-200 X 6 = | -200 pounds, |

v8 = | 1,000-200 X 8 = | -000 pounds, |

v10 = | 1,000-200x10 = | -1,000 pounds, |

v10"= | 1,000-200x10 + | 1,000 = 0. |

3. Compute the values of the shear in example 1, taking into account the weight of the beam (400 pounds). (The right and left reactions are then 3,900 and 2,500 pounds respectively; see example 3, Art. 33.)

We proceed just as in example 1, except that in each computation we include the weight of the beam to the left of the section (or to the right when computing from forces to the right). The weight of the beam being 40 pounds per foot, then (computing from the left)

Vo" | = | + | 2,500 | pounds, | ||||||||||

v1' | = | + | 2,500 | - | 40 | = | + | 2,460 | ||||||

V1 " | = | + | 2,500 | - | 40- | -1,000 | = | + | -1,460, | |||||

v2 | = | + | 2,500 | - | 1,000 | - | 40 | X | 2 | + | 1,420, | |||

v3 | = | + | 2,500 | - | 1,000 | _ | 40 | X | 3 | = | + | 1,380, | ||

V4 | = | + | 2,500 | - | 1,000 | - | 40 | X | 4 | = | + | 1,340, | ||

v5 | = | + | 2,500 | - | 1,000 | - | 40 | X | 5 | = | + | 1,300, | ||

v6 | = | + | 2,500 | - | 1,000 | ___ | 40 | X | 6 | = | + | 1,260, | ||

v6" | = | + | 2,500 | - | 1,000 | - | 40 | X | 6- | -2,000 = | -740, | |||

v7 | = | + | 2,500 | - | 1,000 | 2,000- | 40 | X' | 7 = | -780, |

V 8 | = | + | 2,500 | - | 1,000 | - | 2,000 | - | 40x8: | = | -820, | |||||||

V8 | = | + | 2,500 | - | 1,000 | - | 2,000 | - | 40X8- | -3,000 | = | - | 3,820, | |||||

v9 | - | + | 2,500 | - | 1,000 | - | 2,000 | - | 3,000 | - | 40 | X | 9 = | = | - | 3,860, | ||

v10 | = | + | 2,500 | - | 1,000 | - | 2,000 | - | 3,000 | - | 40 | X | 10 | = | - | 3,900, | ||

V10 | = | + | 2,500 | - | 1,000 | - | 2,000 | - | 3,000 | __ | 40 | X | 10 | + | 3,900 | = | 0. |

Computing from the right, we find, as before, that

v7 | = | - | (3,000 | - | 3,000 | - | 40X3): | = | - | 780 pounds, | |||||

v8 | = | - | (3,900 | - | 3,000 | - | 40X2): | = | - | 820, | |||||

V8 | = | - | (3,900 | - | 40 | X | 2 | = | - | 3,820, |

1. Compute the values of the shear for sections of the beam represented in Fig. 10, neglecting the weight of the beam. (The right and left reactions are 3,300 and 4,000 pounds respectively; see example 2, Art. 33.)

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