01-01ChapGere.0005

# 01-01ChapGere.0005 - C B A a b Solution 1.2-7 Two steel...

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SECTION 1.2 Normal Stress and Strain 5 Solution 1.2-6 Car on inclined track F REE - BODY DIAGRAM OF CAR W 5 Weight of car T 5 Tensile force in cable a 5 Angle of incline A 5 Effective area of cable R 1 , R 2 5 Wheel reactions (no friction force between wheels and rails) E QUILIBRIUM IN THE INCLINED DIRECTION T 5 W sin a © F T 5 0 Ê Q 1 b 2 T 2 W sin a 5 0 T ENSILE STRESS IN THE CABLE S UBSTITUTE NUMERICAL VALUES : W 5 130 kN a 5 30 8 A 5 490 mm 2 5 133 MPa s t 5 (130 kN)(sin 30 8 ) 490 mm 2 s t 5 T A 5 W sin a A a W R 1 R 2 Problem 1.2-7 Two steel wires, AB and BC , support a lamp weighing 18 lb (see figure). Wire AB is at an angle a 5 34° to the horizontal and wire BC is at an angle b 5 48°. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.) Determine the tensile stresses s AB and s BC in the two wires.
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Unformatted text preview: C B A a b Solution 1.2-7 Two steel wires supporting a lamp F REE-BODY DIAGRAM OF POINT B S UBSTITUTE NUMERICAL VALUES : 2 T AB (0.82904) 1 T BC (0.66913) 5 T AB (0.55919) 1 T BC (0.74314) 2 18 5 S OLVE THE E QUATIONS : T AB 5 12.163 lb T BC 5 15.069 lb T ENSILE STRESSES IN THE WIRES s BC 5 T BC A 5 21,300 psi s AB 5 T AB A 5 17,200 psi T AB T BC W = 18 lb y x a b E QUATIONS OF E QUILIBRIUM S F x 5 2 T AB cos a 1 T BC cos b 5 S F y 5 T AB sin a 1 T BC sin b 2 W 5 a 5 34 8 b 5 48 8 d 5 30 mils 5 0.030 in. A 5 p d 2 4 5 706.9 3 10 2 6 in. 2 A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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## This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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