10-03 Chap Gere - SECTION 10.4 Method of Superposition 655...

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Problem 10.4-15 A temporary wood flume serving as a channel for irrigation water is shown in the figure. The vertical boards forming the sides of the flume are sunk in the ground, which provides a fixed support. The top of the flume is held by tie rods that are tightened so that there is no deflection of the boards at that point. Thus, the vertical boards may be modeled as a beam AB , supported and loaded as shown in the last part of the figure. Assuming that the thickness t of the boards is 1.5 in., the depth d of the water is 40 in., and the height h to the tie rods is 50 in., what is the maximum bending stress s in the boards? ( Hint: The numerically largest bending moment occurs at the fixed support.) Solution 10.4-15 Side wall of a wood flume SECTION 10.4 Method of Superposition 655 h = 50 in. t = 1.5 in. d = 40 in. B A Select R B as redundant. Equilibrium: R ELEASED STRUCTURE AND FORCE - DISPL . EQS . From Table G-1, Case B: ( d B ) 2 5 R B L 3 3 EI ( d B ) 1 5 q 0 a 4 30 EI 1 q 0 a 3 24 EI ( L 2 a ) 5 q 0 a 3 120 EI (5 L 2 a ) M A 5 q 0 a 2 6 2 R B L C OMPATIBILITY d B 5 ( d B ) 1 2 ( d B ) 2 5 0 [ M AXIMUM BENDING MOMENT N UMERICAL VALUES a 5 40 in. L 5 50 in. t 5 1.5 in. b 5 width of beam g 5 62.4 lb/ft 3 5 0.03611 lb / in. 3 Pressure p 5 g aq 0 5 pb 5 g ab s 5 M max S 5 509 psi S 5 bt 2 6 5 0.3750 b M max 5 g a 3 b 120 L 2 (20 L 2 2 15 aL 1 3 a 2 ) 5 191.05 b s 5 M max S S 5 bt 2 6 5 q 0 a 2 120 L 2 L 2 2 15 aL 1 3 a 2 ) M max 5 M A 5 1 6 q 0 a 2 2 R B L R B 5 q 0 a 3 L 2 a ) 40 L 3 B A L a R A R B M A q 0 B A q 0 ( d B ) 1 B A ( d B ) 2 5 R B L 3 3 EI R B t N.A. b
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Problem 10.4-16 Two identical, simply supported beams AB and CD are placed so that they cross each other at their midpoints (see figure). Before the uniform load is applied, the beams just touch each other at the crossing point. Determine the maximum bending moments ( M AB ) max and ( M CD ) max in beams AB and CD , respectively, due to the uniform load if the intensity of the load is q 5 6.4 kN/m and the length of each beam is L 5 4 m. Solution 10.4-16 Two beams that cross 656 CHAPTER 10 Statically Indeterminate Beams A B C D q F 5 interaction force between the beams U PPER BEAM ( d B ) 1 5 downward deflection due to q ( d B ) 2 5 upward deflection due to F d AB 5 ( d B ) 1 2 ( d B ) 2 L OWER BEAM C OMPATIBILITY d AB 5 d CD [ F 5 5 qL 16 5 qL 4 384 EI 2 FL 3 48 EI 5 FL 3 48 EI d CD 5 FL 3 48 EI 5 5 qL 4 384 EI 2 FL 3 48 EI 5 FL 3 48 EI 5 5 qL 4 384 EI U PPER BEAM L OWER BEAM N UMERICAL VALUES q 5 6.4 kN/m ( M AB ) max 5 6.05 kN ? m L 5 4 m ( M CD ) max 5 8.0 kN ? m ( M CD ) max 5 5 qL 2 64 M max 5 FL 4 5 5 qL 2 64 ( M AB ) max 5 121 qL 2 2048 M 1 5 3 qL 2 64 M max 5 121 qL 2 2048 x 1 5 11 L 32 R A 5 11 qL 32 A L 2 B q F L 2 C L 2 D F L 2 A 5 qL 16 B q R A R B 5 R A M max M 1 x 1 x 1 M O M O
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Problem 10.4-17 The cantilever beam AB shown in the figure is an S 6 3 12.5 steel I-beam with E 5 30 3 10 6 psi. The simple beam DE is a wood beam 4 in. 3 12 in. (nominal dimensions) in cross section with E 5 1.5 3 10 6 psi. A steel rod AC of diameter 0.25 in., length 10 ft, and E 5 30 3 10 6 psi serves as a hanger joining the two beams. The hanger fits snugly between the beams before the uniform load is applied to beam DE .
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10-03 Chap Gere - SECTION 10.4 Method of Superposition 655...

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