Homework#9-Solutions

# Homework#9-Solutions - 05Ch05.qxd 9/25/08 444 2:29 PM Page...

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444 CHAPTER 5 Stresses in Beams (Basic Topics) Solution 5.8-4 Shear stresses in a cantilever beam Eq. (5-39): t ± 50 * 10 ² 6 (10,000 ² y 1 2 ) ( y 1 ± mm; t ± MPa) t ± 8,000 2(80 * 10 6 ) c (200) 2 4 ² y 1 2 d ( t ± N/mm 2 ± MPa) h ± 200 mm ( y 1 ± mm) I ± bh 3 12 ± 80 * 10 6 mm 4 V ± P ± 8.0 kN ± 8,000 N t ± V 2 I a h 2 4 ² y 1 2 b Distance from the y 1 top surface (mm) (mm) 0 100 0 0 25 75 0.219 219 50 50 0.375 375 75 25 0.469 469 100 (N.A.) 0 0.500 500 G RAPH OF SHEAR STRESS t (kPa) (MPa) t t Solution 5.8-3 Maximum load M max q ± 5.833 1b ft q ± g A weight of beam per unit distance g ± 35 lb ft 3 A ± b # h h ± 6 in. t allow ± 200 psi L ± 8 ft b ± 4 in. ; M max ± 25.4 k-ft M max ± 2 AL 3 t allow ² qL 2 2 M ± 2 AL 3 t max ² qL 2 2 t max ± 3 V 2 A ± 3 2 A a M L + qL 2 b V ± M L + qL 2 Problem 5.8-4 A cantilever beam of length supports a load (see figure). The beam is made of wood with cross-sectional dimensions . Calculate the shear stresses due to the load at points located , , and from the top surface of the beam. from these results, plot a graph showing the distribution of shear stresses from top to bottom of the beam. 100 mm 75 mm 25 mm P 120 mm * 200 mm P ± 8.0 kN L ± 2 m 200 mm 120 mm L = 2 m P = 8.0 kN 05Ch05.qxd 9/25/08 2:29 PM Page 444

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SECTION 5.8 Shear Stresses in Rectangular Beams 449 Solution 5.8-10 Simply supported wood beam L ± 1.2 m q ± g bh ± 181.44 N/m g ± 5.4 kN/m 3 S ± bh 2 6 ± 1344 * 10 3 mm 3 A ± bh ± 33,600 mm 2 b ± 140 mm h ± 240 mm (a) A LLOWABLE BASED UPON BENDING STRESS Equate values of and solve for : or P ± 38.0 kN ; 0.3 P + 32.66 ± 11,424 P ± 37,970 N P M max ± 11,424 N # m M max ± S s allow ± (1344 * 10 3 mm 3 )(8.5 MPa) ( P ± newtons; M ± N # m) ± 0.3 P + 32.66 N # m + (181.44 N/m)(1.2 m) 2 8 M max ± PL 4 + qL 2 8 ± P (1.2 m) 4 s allow ± 8.5 MPa s ± M max S P From Appendix F: Select in. beam (nominal dimensions) (b) R EPEAT ( A ) CONSIDERING THE WEIGHT OF THE BEAM R B ± 7.83 * 10 3 1b R B ± 7.725 * 10 3 1b + q beam L 2 q beam ± 20.964 1b ft g ± 35 1b ft 3 q beam ± g A A ± 86.25 in. 2 S ± 165.3 in. 3 ; 8 * 12 s max ± M S S req ± M max s allow S req ± 120.6 in. 3 beam is still satisfactory for shear. beam is still satisfactory for moment. ; Use 8 * 12 in. beam 8 * 12 S req ± M max s allow S req ± 122.3 in. 3 < S M max ± 2.293 * 10 4 1b-ft M max ± R B d ² qd 2 2 q total ± q + q beam q total ± 145.964 1b ft 8 * 12 A req ± 73.405 in. 2
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## This note was uploaded on 11/26/2011 for the course EM 319 taught by Professor Kennethm.liechti during the Spring '08 term at University of Texas at Austin.

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Homework#9-Solutions - 05Ch05.qxd 9/25/08 444 2:29 PM Page...

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