Strength of Materials |
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Common terms and phrases
12 feet 20 feet long allowable stress allowable unit stress angle butt joint cantilever beam carries a load carries a uniformly cent compressive stress Compute the deflection Compute the maximum Compute the unit concentrated load concrete cross-section deformation described in Problem elastic curve elastic limit ends and carries equal equation Euler's curve Euler's formula expression factor of safety fixed end flange flexural stress foot-pounds formula free body free end horizontal I-beam inch in diameter inch thick inch-pounds inches deep inches wide lap joint left end linear foot long is supported maximum deflection maximum unit middle modulus of elasticity moment of inertia neutral axis pounds per linear pounds per square R₁ reaction right end rivets shaft shear diagram shown in Fig simple beam slope steel column straight-line tangent tensile stress ultimate strength uniformly distributed load unit shearing stress y₁ zero ΕΙ
Popular passages
Page 117 - Since, in general, the deflection due to the shearing forces is extremely small compared to that due to the flexural forces, it will be neglected in this discussion, and all beams will be treated as cases of simple flexure. The following assumptions are made in the case of simple flexure : 1. Plane transverse sections remain plane after bending. 2. The modulus of elasticity in tension is the same as that in compression. 3. The proportional elastic limit is not exceeded.
Page 82 - This is the equation of a straight line passing through the origin and sloping downward to the right. In Fig.
Page 269 - Problem 1. A wooden beam 8 inches wide, 16 inches deep, and 15 feet long is fixed at one end, supported on the same level at the other, and loaded with a uniformly distributed load of 1,200 pounds per linear foot. If E is 1,500,000 pounds per square inch, what is the maximum deflection? Ans. 0.138 inch. 150. Beam Fixed at One End, Supported at the Other, with Several Loads. — If a beam which is fixed at one end and supported at the other has several concentrated loads and partial or complete uniformly...
Page 31 - X 32,500 = 24,375 pounds per square inch, which corresponds to a factor of safety of 2.67.
Page 147 - ... theorem of three moments, however, gives the relation between the moments at three successive supports, from which the reactions and stresses may be obtained very easily. The theorem of three moments will not determine the deflections, but these are seldom required. Figure 123 represents two adjoining spans of a continuous beam which extends indefinitely in each direction.
Page 277 - If the spans are unequal or if the loading is unsymmetrical, the following method may be used. sum of the vertical forces equals zero and the sum of the moments with respect to any point equals zero, give two of the equations. The relation of the deflections at the ends from the tangent at the middle will give the other equation. EXAMPLE 1 Solve for the reactions Ri, R2, and R3 for the beam shown in Fig. 204. Solution'.
Page 105 - ... feet square placed in any position, whichever will produce the greater stresses. (c) Driveways shall be designed to support a uniformly distributed load of 100 pounds per square foot for vehicles weighing less than 3 tons with load, 150 pounds per square foot for vehicles weighing 3 to 10 tons with load, 200 pounds per square foot for vehicles weighing over 10 tons with load, or a concentrated load equal to the maximum expected wheel load on an area...
Page 183 - I-beam 24 feet long is supported at the ends and carries a concentrated load at the middle to cause a maximum stress of 16,000 pounds per square inch.
Page 183 - S,' = 5,150 pounds per square inch; maximum S/ = 10,010 pounds per square inch. Problem 3. A steel shaft 3 inches in diameter and 20 feet long is supported at the ends. At a point 1 foot from the left end is a pulley 2 feet in diameter weighing 50 pounds, upon which are downward vertical belt pulls of 500 pounds and 100 pounds. At a point 2 feet from the right end is a pulley 30 inches in diameter weighing 80 pounds upon which are downward vertical belt pulls of 440...
Page 68 - S, = 730 pounds per square inch; S, = 17,550 pounds per square inch. Problem 3. A reinforced concrete beam 16 inches wide, 14 inches deep, and 30 feet long is supported at the ends and at the middle. It is reinforced with eight steel rods, each J^ inch in diameter, placed 1 inch from the surface on the tension side. Using the value of E. as 30,000,000 pounds per square inch, Ec as 2,500,000 pounds per square inch, the allowable stress in the steel as 16,000 pounds per square inch...