433_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

433_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 5.6 Design of Beams 427 Solution 5.6-20 N UMERICAL DATA (a) I GNORE BEAM SELF WEIGHT - FIND Equate M max1 to M max2 & solve for b min b min ± 11.91 mm ; b min ± a 9 8 qL 2 s a b 1 3 and M max2 ± s a S S ± 2 3 b 3 M max1 ± 1.5 q L 2 2 at B b min s a ± 60 MPa g ± 77 kN m 3 L ± 150 mm q ± 4 kN m (b) N OW MODIFY - INCLUDE BEAM WEIGHT Equate M max1 to M max2 & solve for b min Insert numerical values, then solve for b b min ± 11.92 mm ; a 2 3 s a b b 3 ² 1 g L 2 2 b 2 ² 3 4 qL 2 ± 0 and M max ± s a a 2 3 b 3 b M max ± (1.5 q + w ) L 2 2 w ± g A w ± g 1 2 b 2 2 Problem 5.6-21 A retaining wall 5 ft high is constructed of horizontal wood planks 3 in. thick (actual dimension) that are supported by vertical wood piles of 12 in. diameter (actual dimension), as shown in the figure. The lateral earth pressure is p 1 ± 100 lb/ft 2 at the top of the wall and p 2 ± 400 lb/ft 2 at the bottom. Assuming that the allowable stress in the wood is 1200 psi,
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Unformatted text preview: calculate the maximum permissible spacing s of the piles. ( Hint: Observe that the spacing of the piles may be governed by the load-carrying capacity of either the planks or the piles. Consider the piles to act as cantilever beams subjected to a trapezoidal distribution of load, and consider the planks to act as simple beams between the piles. To be on the safe side, assume that the pressure on the bottom plank is uniform and equal to the maximum pressure.) 3 in. s 5 ft Top view Side view 3 in. 12 in. diam. 12 in. diam. p 1 = 100 lb/ft 2 p 2 = 400 lb/ft 2 05Ch05.qxd 9/24/08 4:59 AM Page 427...
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