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09-06 Chap Gere

# 09-06 Chap Gere - SECTION 9.11 Representation of Loads on...

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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 P M A Pa U SE T ABLE 9-2. P x 1 Pa x 2 P x a 1 q ( x ) R A x 1 M A x 2 P x a 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 qb U SE T ABLE 9-2. q x a 0 q x L 0 qb x 1 qb 2 (2 a b ) x 2 q ( x ) R A x 1 M A x 2 q x a 0 q x L 0 M A qb 2 (2 a 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 A D y a 6 ft 72 in. b 3 ft 36 in. L 9 ft 108 in. q 2 k ft 1 6 k in. F ROM EQUILIBRIUM : R A 16 k M A 864 k-in. U SE T ABLE 9-2. Units: kips, inches (Units: x in., q k in.) 4 x 108 1 16 x 1 864 x 2 1 6 x 0 1 6 x 72 0 P x L 1 q ( x ) R A x 1 M A x 2 q x 0 q x a 0 q = 2 k/ft P = 4 k b 3 ft a 6 ft x B A D 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. Pa L x L 1 Pb L x 1 P x a 1 q ( x ) R A x 1 P x a 1 R B x L 1 R B Pa L R A Pb L b L a x B D A y P R A R B
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. M 0 L x L 1 M 0 L x 1 M 0 x a 2 q ( x ) R A x 1 M 0 x a 2 R B x L 1 R B M 0 L (downward) R A 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 R B P U SE T ABLE 9-2. P x L 1 P x 1 P x a 1 P x L a 1 R B x L 1 q ( x ) R A x 1 P x a 1 P x L a 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 20 k-ft 240 k-in. P 18 k a 16 ft 192 in. b 10 ft 120 in. L 26 ft 312 in. F ROM EQUILIBRIUM : R A 7.692 k R B 10.308 k U SE T ABLE 9-2. Units: kips, inches (Units: x in., q k in.) 10.308 x 312 1 7.692 x 1 240 x 2 18 x 192 1 R B x L 1 q ( x ) R A x 1 M 0 x 2 P x a 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. q x a 0 ( qa 2 2 L ) x L 1 ( qa 2 L )(2 L a ) x 1 q x 0 q ( x ) R A x 1 q x 0 q x a 0 R B x L 1 R B qa 2 2 L R A qa 2 L (2 L a ) a L q x B D A y R A R B
SECTION 9.11 Representation of Loads on Beams by Discontinuity Functions 619 Problem 9.11-9 Solution 9.11-9 Simple beam L /3 L /3 L /3 q 0 x B E D A y F ROM EQUILIBRIUM : U SE T ABLE 9-2. (5 q 0 L 54) x L 1 (3 q 0 L ) x 2 L 3 1 q 0 x 2 L 3 0 (2 q 0 L 27) x 1 (3 q 0 L ) x L 3 1 q 0 x 2 L 3 0 R B x L 1 q ( x ) R A x 1 3 q 0 L x L 3 1 3 q 0 L x 2 L 3 1 R B 5 q 0 L 54 R A 2 q 0 L 27 L /3 L /3 L /3 q 0 x B E D A y R A R B Problem 9.11-10 Solution 9.11-10 Simple beam P = 120 kN q = 20 kN/m x B D C A y 10 m 5 m 5 m F ROM EQUILIBRIUM : R A 180 kN R B 140 kN U SE T ABLE 9-2. Units: kilonewtons, meters (Units: x meters, q kN m) 120 x 15 1 140 x 20 1 180 x 1 20 x 0 20 x 10 0 P x 3 L 4 1 R B x L 1 q ( x ) R A x 1 q x 0 q x L 2 0 P = 120 kN q

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