Download solution manuals for students and more Summaries Design in PDF only on Docsity!
Design for Flexure and Shear
To determine the load capacity or the size of beam section, it must satisfy the allowable stresses in both flexure (bending) and shear. Shearing stress usually governs in the design of short beams that are heavily loaded, while flexure is usually the governing stress for long beams. In material comparison, timber is low in shear strength than that of steel. For any cross-sectional shape, flexure and shear are given in the following formulas: Flexure Formula Horizontal Shear Stress For rectangular beam, the following defines for flexure and shear: Flexure formula for rectangular beam Horizontal shear stress for rectangular beam Where fb = flexure stress fv = bending stress M = maximum moment applied to the beam V = maximum shear applied to the beam I = moment of inertia about the neutral axis Q = moment of area b = breadth d = depth Solution to Problem 580 | Design for Flexure and Shear Problem 580 A rectangular beam of width b and height h carries a central concentrated load P on a simply
supported span of length L. Express the maximum fv in terms of maximum fb. Solution 580 From the figure: From flexure formula: From shear stress formula: answer Problem 581 A laminated beam is composed of five planks, each 6 in. by 2 in., glued together to form a section 6 in. wide by 10 in. high. The allowable shear stress in the glue is 90 psi, the allowable shear stress in the wood is 120 psi, and the allowable flexural stress in the wood is 1200 psi. Determine the maximum uniformly distributed load that can be carried by the beam on a 6-ft simple span. Solution 581 Maximum moment for simple beam Maximum shear for simple beam
Solution 582 Based on bending stress (square b = d): Based on shear stress (square b = d): Use 145 mm × 145 mm square beam answer Problem 584 A wide-flange section having the dimensions shown in Fig. P-584 supports a distributed load of wo lb/ft on a simple span of length L ft. Determine the ratio of the maximum flexural stress to the maximum shear stress.
Solution 584 Bending stress: where: Thus, Shear stress: where: (see computation above) Thus, Ratio (flexural stress : shear stress) answer
answer Problem 586 The distributed load shown in Fig. P-586 is supported by a box beam having the same cross-section as that in Prob. 585. Determine the maximum value of wo that will not exceed a flexural stress of 10 MPa or a shearing stress of 1.0 MPa. Solution 586 From shear diagram Based on allowable bending stress Where (From Solution 585): c = 125 mm I = 334 375 000 mm^4 Thus,
Based on allowable shear stress Where (From Solution 585): Q = 1 781 250 mm^3 I = 334 375 000 mm^4 b = 100 mm Thus, For safe value of wo, use wo = 11.26 kN/m answer Problem 587 A beam carries two concentrated loads P and triangular load of 3P as shown in Fig. P-587. The beam section is the same as that in Fig. P-577 on this page. Determine the safe value of P if fb ≤ 1200 psi and fv ≤ 200 psi. Solution 587
- MD = MC + Area in shear diagram MD = -4P + 4P = 0
- The moment diagram at AB and CD are straight lines (1st degree curves) while at BC is 3rd degree curve. Based on allowable bending stress Where (From Solution 577) c = 6 in I = 350.67 in^4 Thus, Based on allowable shear stress Where (From Solution 577) Q = 35.5 in^3 I = 350.67 in^4 b = 0.75 in Thus, For safe value of P, use P = 740.85 lb. answer Problem 588 The distributed load shown in Fig. P-588 is supported by a wide-flange section of the given dimensions. Determine the maximum value of wo that will not exceed a flexural stress of 10 MPa or a shearing stress of 1.0 MPa.
Solution 588 From the shear diagram Maximum moment = sum of area in Shear Diagram at the left of point of zero shear Based on allowable flexural stress Where c = 150 mm I = 200(300^3 )/12 - 175(250^3 )/ I = 222 135 416.67 mm^4 Thus,
( okay! ) By transfer formula for moment of inertia
( okay! ) For M = -2W lb·ft Top fiber in tension Bottom fiber in compression For M = W lb·ft Top fiber in compression Bottom fiber in tension Based on allowable shear stress:
Solution 590 Check MC from the overhang segment ( okay! ) Based on allowable bending stress Where M = 2.5P + 1700 lb·ft c = 12/2 = 6 in I = 10(12^3 )/12 - 8(10^3 )/12 = 773.33 in^4 Thus, Based on allowable shear stress Where V = 0.5P + 1160 lb
Q = 10(1)(5.5) + 2 [ 5(1)(2.5) ] = 80 in^3 b = 2 in Thus, For safe value of P, use P = 3480 lb answer
Spacing of Rivets or Bolts in Built-Up Beams
When two or more thin layers of beams are fastened together with a bolt or a rivet so that they act as a unit to gain more strength, it is necessary to design the to size or spacing of these bolts or rivets so that it can carry the shearing force acting between each adjacent layers. Consider the beam shown in the figure.
Solution 592 Where Thus, Spacing of bolts answer Problem 593 A box beam, built up as shown in Fig. P-593, is secured by screws spaced 5 in. apart. The beam supports a concentrated load P at the third point of a simply supported span 12 ft long. Determine
the maximum value of P that will not exceed fv = 120 psi in the beam or a shearing force of 300 lb in the screws. What is the maximum flexural stress in the beam? Solution 593 Based on allowable shearing force of beam
