Mechanics_of_Materials_Chap_09-06 - SECTION 9.11 615...

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SECTION 9.11 Representation of Loads on Beams by Discontinuity Functions 615 Representation of Loads on Beams by Discontinuity Functions Problem 9.11-1 through 9.11-12 A beam and its loading are shown in the figure. Using discontinuity functions, write the expression for the intensity q ( x ) of the equivalent distributed load acting on the beam (include the reactions in the expression for the equivalent load). Solution 9.11-1 Cantilever beam P a L b x B D A y F ROM EQUILIBRIUM : R A 5 PM A 5 Pa U SE T ABLE 9-2. 52 P K x L 2 1 1 Pa K x L 2 2 1 P K x 2 a L 2 1 q ( x ) R A K x L 2 1 1 M A K x L 2 2 1 P K x 2 a L 2 1 P a L b x B D A y R A M A Problem 9.11-2 Solution 9.11-2 Cantilever beam q a L b x B D A y F ROM EQUILIBRIUM : R A 5 qb U SE T ABLE 9-2. 1 q K x 2 a L 0 2 q K x 2 L L 0 qb K x L 2 1 1 qb 2 (2 a 1 b ) K x L 2 2 q ( x ) R A K x L 2 1 1 M A K x L 2 2 1 q K x 2 a L 0 2 q K x 2 L L 0 M A 5 qb 2 a 1 b ) q a L b x B D A y M A R A
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Problem 9.11-3 Solution 9.11-3 Cantilever beam 616 CHAPTER 9 Deflections of Beams q = 2 k/ft P = 4 k 3 ft 6 ft x B AD y a 5 6 ft 5 72 in. b 5 3 ft 5 36 in. L 5 9 ft 5 108 in. q 5 2 k / ft 5 1 6 k / in. F ROM EQUILIBRIUM : R A 5 16 k M A 5 864 k-in. U SE T ABLE 9-2. Units: kips, inches (Units: x 5 in., q 5 k / in.) 1 4 K x 2 108 L 2 1 52 16 K x L 2 1 1 864 K x L 2 2 1 1 6 K x L 0 2 1 6 K x 2 72 L 0 1 P K x 2 L L 2 1 q ( x ) R A K x L 2 1 1 M A K x L 2 2 1 q K x L 0 2 q K x 2 a L 0 q = 2 k/ft P = 4 k b 5 3 ft a 5 6 ft x B y M A R A Problem 9.11-4 Solution 9.11-4 Simple beam b L a x B D A y P F ROM EQUILIBRIUM : U SE T ABLE 9-2. 2 Pa L K x 2 L L 2 1 Pb L K x L 2 1 1 P K x 2 a L 2 1 q ( x ) R A K x L 2 1 1 P K x 2 a L 2 1 2 R B K x 2 L L 2 1 R B 5 Pa L R A 5 pb L b L a x B D A y P R A R B
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Problem 9.11-5 Solution 9.11-5 Simple beam SECTION 9.11 Representation of Loads on Beams by Discontinuity Functions 617 b L a x B D A y M 0 F ROM EQUILIBRIUM : U SE T ABLE 9-2. 1 M 0 L K x 2 L L 2 1 52 M 0 L K x L 2 1 1 M 0 K x 2 a L 2 2 q ( x ) R A K x L 2 1 1 M 0 K x 2 a L 2 2 1 R B K x 2 L L 2 1 R B 5 M 0 L (downward) R A 5 M 0 L b L a x B D A y M 0 R A R B Problem 9.11-6 Solution 9.11-6 Simple beam a L x B D a A y P E P F ROM EQUILIBRIUM : R A 5 R B 5 P U SE T ABLE 9-2. 2 P K x 2 L L 2 1 P K x L 2 1 1 P K x 2 a L 2 1 1 P K x 2 L 1 a L 2 1 2 R B K x 2 L L 2 1 q ( x ) R A K x L 2 1 1 P K x 2 a L 2 1 1 P K x 2 L 1 a L 2 1 a L x B D a A y P E P R A R B
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618 CHAPTER 9 Deflections of Beams Problem 9.11-7 Solution 9.11-7 Simple beam 10 ft 16 ft x B D A y M 0 = 20 k-ft P = 18 k M 0 5 20 k-ft 5 240 k-in. P 5 18 k a 5 16 ft 5 192 in. b 5 10 ft 5 120 in. L 5 26 ft 5 312 in. F ROM EQUILIBRIUM : R A 5 7.692 k R B 5 10.308 k U SE T ABLE 9-2. Units: kips, inches (Units: x 5 in., q 5 k / in.) 2 10.308 K x 2 312 L 2 1 52 7.692 K x L 2 1 1 240 K x L 2 2 1 18 K x 2 192 L 2 1 2 R B K x 2 L L 2 1 q ( x ) R A K x L 2 1 1 M 0 K x L 2 2 1 P K x 2 a L 2 1 b = 10 ft a = 16 ft x B D A y M 0 = 20 k-ft P = 18 k R A R B Problem 9.11-8 Solution 9.11-8 Simple beam a L q x B D A y F ROM EQUILIBRIUM : U SE T ABLE 9-2.
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Mechanics_of_Materials_Chap_09-06 - SECTION 9.11 615...

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