**preview**has

**blurred**sections. Sign up to view the full version! View Full Document

**Unformatted text preview: **Shear Forces and Bending Moments Problem 4.3-1 Calculate the shear force V and bending moment M at a cross section just to the left of the 1600-lb load acting on the simple beam AB shown in the figure. Solution 4.3-1 Simple beam 4 Shear Forces and Bending Moments 259 A B 1600 lb 800 lb 120 in. 30 in. 60 in. 30 in. S M A 5 0: R B 5 1400 lb S M B 5 0: R A 5 1000 lb Free-body diagram of segment DB 5 42,000 lb-in. © M D 5 0: Ê M 5 (1400 lb)(30 in.) 5 200 lb © F VERT 5 0: Ê V 5 1600 lb 2 1400 lb A B 1600 lb 800 lb 30 in. 60 in. 30 in. D R A R B B 1600 lb 30 in. D R B V M Problem 4.3-2 Determine the shear force V and bending moment M at the midpoint C of the simple beam AB shown in the figure. Solution 4.3-2 Simple beam A C B 2.0 kN/m 6.0 kN 1.0 m 1.0 m 4.0 m 2.0 m A C B 2.0 kN/m 6.0 kN 1.0 m 1.0 m 2.0 m R A R B S M A 5 0: R B 5 4.5 kN S M B 5 0: R A 5 5.5 kN Free-body diagram of segment AC © M C 5 0: Ê M 5 5.0 kN ? m © F VERT 5 0: Ê V 5 2 1.5 kN A C 6.0 kN 1.0 m 1.0 m R A V M Problem 4.3-3 Determine the shear force V and bending moment M at the midpoint of the beam with overhangs (see figure). Note that one load acts downward and the other upward. Solution 4.3-3 Beam with overhangs 260 CHAPTER 4 Shear Forces and Bending Moments P P b b L 5 P ¢ 1 1 2 b L ≤ Ê (upward) R A 5 1 L [ P ( L 1 b 1 b ) ] © M B 5 Free-body diagram ( C is the midpoint ) M 5 PL 2 1 Pb 2 Pb 2 PL 2 5 M 5 P ¢ 1 1 2 b L ≤ ¢ L 2 ≤ 2 P ¢ b 1 L 2 ≤ © M C 5 0: 5 2 bP L V 5 R A 2 P 5 P ¢ 1 1 2 b L ≤ 2 P © F VERT 5 0: © M A 5 0: Ê R B 5 P ¢ 1 1 2 b L ≤ Ê (downward) P P b b L A B R A R B P b L/2 A C R A V M Problem 4.3-4 Calculate the shear force V and bending moment M at a cross section located 0.5 m from the fixed support of the cantilever beam AB shown in the figure. Solution 4.3-4 Cantilever beam A B 1.5 kN/m 4.0 kN 1.0 m 1.0 m 2.0 m Free-body diagram of segment DB Point D is 0.5 m from support A . 5 2 9.5 kN ? m 5 2 2.0 kN ? m 2 7.5 kN ? m 2 (1.5 kN / m)(2.0 m)(2.5 m) © M D 5 0: Ê M 5 2 (4.0 kN)(0.5 m) 5 4.0 kN 1 3.0 kN 5 7.0 kN V 5 4.0 kN 1 (1.5 kN / m)(2.0 m) © F VERT 5 0: A B 1.5 kN/m 4.0 kN 1.0 m 1.0 m 2.0 m D B 1.5 kN/m 4.0 kN 1.0 m 0.5 m 2.0 m V M Problem 4.3-5 Determine the shear force V and bending moment M at a cross section located 16 ft from the left-hand end A of the beam with an overhang shown in the figure. Solution 4.3-5 Beam with an overhang SECTION 4.3 Shear Forces and Bending Moments 261 A C B 400 lb/ft 200 lb/ft 6 ft 6 ft 10 ft 10 ft S M B 5 0: R A 5 2460 lb S M A 5 0: R B 5 2740 lb Free-body diagram of segment AD Point D is 16 ft from support A . 5 2 4640 lb-ft 2 (400 lb / ft)(10 ft)(11 ft) © M D 5 0: Ê M 5 (2460 lb)(16 ft) 5 2 1540 lb V 5 2460 lb 2 (400 lb / ft)(10 ft) © F VERT 5 0: A C B 400 lb/ft 200 lb/ft 6 ft 6 ft 10 ft 10 ft R A R B A D 400 lb/ft 6 ft 10 ft R A V M Problem 4.3-6 The beam ABC shown in the figure is simply supported at A and B and has an overhang from B to C . The loads consist of a horizontal force P 1 5 4.0 kN acting at the 4....

View Full Document