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Unformatted text preview: Normal Stress: 1 Bars CE and DE in the truss have a crosssectional
area A=25cm'. Find the normal stress in each of these bars. Solution Using the method ofjoints. and the fact that angle DEC=53.13°‘ we
have ch Sin(53.13°)=20kN. then Fc5=25.00kN. Then. the stress in bar CE 15 A C l'. similarly. FDE=Fag Cos(53.13°)=15.00kN. and 0' DF : I‘m“ = ——152 = —6, 000kPa (compression)
' Am: 25*1/100
Normal Stress: 3 Beam AB is vertical and the 20KN load 20“ is applied in the ydirection. Find the stress in the two
identical cables, of diameter d=2cm. Solution Due to symmetry, the two cables will be subjected to
identical tensile stress Coordinates of point B (x,y,z) in meters: (0,0,5)
Coordinates of point C: (l,4,0) Vector BC: (l,—4,5) Magnitude of vector BC: (—1)2 + (—4)2 + 52 = 6.48m _ _ . _ —l —4 —5
Unit vector in direction BC: —,—,— 6.48 6.48 6.48
Let Fnc denote the magnitude of the force in cable BC. Then the component of this force in the y
4 direction is —mFBC . Then for equilibrium in the ydirection (since the component of the force in BD
is the same as the force in BC) 4
2—F =20kN—>F =16.2kN.
6.48 3° “C Then, aBC =Fﬁ=ﬂ=srs923kpa A n0.022/4 Normal Stress: 6* A taw truck is using a cable to pull the classic car up a 15° hill. If the classic car
weighs 4000 lbs and the cable has a diameter of 3/4 inch, ﬁnd the stress in the cable when the truck comes
to a stop while on the hill. Ignore friction between the car and the pavement. ...
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 Summer '08
 FRANTZISKONIS

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