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EXTERNAL SHEAR AND BENDING MOMENT.

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 sec tion 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 cus tomary 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 V V 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 V,' and V," denote the values of the shear at sec tions 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 support is denoted by V,", and 2,300 pounds. The shear just to the left of B is denoted by V,', and since the only force to the left of the section is the left reaction, V,'= 2,300 pounds. The shear just to the right of B is denoted by V,", and since the only forces to the left of this section are the left reaction and the 1,000-pound load, V," = 2,300-1,000= 1,800 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 V,= V,-= V8=1,300 pounds.

The shear just to the right of C is denoted by V,"; and since the forces to the left of that section are the left reaction and the 1,000- and 2,000-pound loads, V," = 2,300 -1,000- 2,000 =- 700 pounds.

AVithout further explanation, the student should understand that IT = +2,300- 1,000 - 2,000 = - 700 pounds, V, 700, V." = +2,300 - 1,000- 2,000 -3,000 3,700, V, =- 3,700, V"__- +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 1,600 pounds. Compute the shears for sections two feet apart.

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