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.

(1) The flexural compressive unit-stress on the upper fibre is greater than the direct unit-stress; that is, S1 is greater than S0. The resultant stress on the upper fibre is

Sc = S1 = S0 (compressive); and that on the lower fibre is

St = S2 + S0 (tensile).

The combined stress is as represented in Fig. 43, c, part tensile and part compressive.

(2) The flexural compressive unit-stress is less than the direct unit-stress; that is, S1 is less than S0. Then the combined unit-stress on the upper fibre is

Sc = S0 - S1 (tensile); and that on the lower fibre is

St = S2 + S0 (tensile).

The combined stress is represented by Fig. 43, d, and is all tensile.

Example. A steel bar 2x6 inches, and 12 feet long, is subjected to end pulls of 45,000 pounds. It is supported at each end, and sustains, as a beam, a uniform load of 6,000 pounds. It is required to compute the combined unit-fibre stresses.

Evidently the dangerous section is at the middle, and M = 1/8 Wl; that is,

Fig. 43.

M = 1/8 X 6,000 X 12 = 9,000 foot-pounds, or 9,000 X 12 = 108,000 inch-pounds.

The bar being placed with the six-inch side vertical, c1 = c2 = 3 inches, and

I = 1/12 X 2 X 63 = 36 inches4. (See Art. 52.)

Hence S1 = S2 = 108,000 X 3 / 36 = 9,000 pounds per square inch.

Since A = 2 X 6 = 12 square inches,

S0 = 45,000/12 = 3,750 pounds per square inch.

The greatest value of the combined compressive stress is

S1 - So = 9,000 - 3,750 = 5,250 pounds per square inch, and it occurs on the upper fibres of the middle section. The greatest value of the combined tensile stress is

S2 + So = 9,000 + 3,750 = 12,750 pounds per square inch, and it occurs on the lowest fibres of the middle section.

Change the load in the preceding illustration to one of 6,000 pounds placed in the middle, and then solve.

Ans. | Sc | = | 14.250 pounds per square inch. |

St | = | 21,750 " " " |

75. Flexure and Compression. Imagine the arrowheads on F reversed; then Fig. 43, a, will represent a beam under combined flexural and compressive stresses. The flexural unit-stresses are computed as in the preceding article. The direct stress is a compression equal to P, and the unit-stress clue to P is computed as in the preceding article. Evidently the effect of P is to increase the compressive stress and decrease the tensile stress due to the flexure. In combining, we have two cases as before:

(1) The flexural tensile unit-stress is greater than the direct unit-stress; that is, S2 is greater than S0. Then the combined unit-stress on the lower fibre is

St = S2 - S0 (tensile); and that on the upper fibre is

Sc = S1 + S0 (compressive). The combined fibre stress is represented by Fig. 44, a, and is part tensile and part compressive.

(2) The flexnral unit-stress on the lower fibre is less than the direct unit-stress; that is, S2 is less than So Then the combined unit-stress on the lower fibre is

St = S0 - S2 (compressive); and that on the upper fibre is

Sc = S0 + S1 (compressive). The combined fibre stress is represented by Fig. 44, b, and is all compressive.

Example. A piece of timber 6x6 inches, and 10 feet long, is subjected to end pushes of 9,000 pounds. It is supported in a horizontal position at its ends, and sustains a middle load of 400 pounds. Compute the combined fibre stresses.

Evidently the dangerous section is at the middle, and M = ¼ pl; that is,

Fig. 44.

M =1/4 x 400 X 10 = 1,000 foot-pounds, or 1,000 X 12 = 12,000 inch-pounds.

Since C1 = c2 = 3 inches, and

I = 1/12 Baz = 1/12 X 6 X 63 = 108 inches4, s1 = s2 = - 12,000 X 3 /108 = 333 1/3 pounds per square inch.

Since A = 6 X 6 = 36 square inches,

So = 9,000/36 = 250 Pounds per square inch.

Hence the greatest value of the combined compressive stress is So + S1 = 333 1/2 + 250 = 583 1/3 pounds per square inch.

It occurs on the upper fibres of the middle section. The greatest value of the combined tensile stress is

S2 - S0 = 333⅓ - 250 = 83⅓ pounds per square inch. It occurs on the lowest fibres of the middle section.

Change the load of the preceding illustration to a uniform load and solve.

Ans. | sc | = | 417 pounds per square inch. |

St | = | 83 " " " " (compression). |

76. Combined Flexural and Direct Stress by More Exact Formulas. The results in the preceding articles are only approximately correct. Imagine the beam represented in Fig. 45, a, to be first loaded with the transverse loads alone. They cause the beam to bend more or less, and produce certain flexural stresses at each section of the beam. Now, if end pulls are applied they tend to straighten the beam and hence diminish the flexural stresses. This effect of the end pulls was omitted in the discussion of Art. 74, and the results there given are therefore only approximate, the value of the greatest combined fibre unit-stress (St) being too large. On the other hand, if the end forces are pushes, they increase the bending, and therefore increase the flexural fibre stresses already caused by the transverse forces (see Fig. 45, b). The results indicated in Art. 75 must therefore in this case also be regarded as only approximate, the value of the greatest unit-fibre stress (Sc) being too small.

For beams loaded in the middle or with a uniform load, the following formulas, which take into account the flexural effect of the end forces, may be used :

M denotes bending moment at the middle section of the beam; I denotes the moment of inertia of the middle section with respect to the neutral axis;

Fig. 45.

S1, S2, cl and c2 have the same meanings as in Arts. 74 and 75, but refer always to the middle section; l denotes length of the beam;

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