deflection and member deformation
Post on 19-Jan-2015
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- 1. Problem Number (1)
A 3-mm thick hollow polystyrene cylinder E = 3GPa and a rigid circular plate (only part of which is shown) are used to support a 250-mm long steel rod AB (E = 200 GPa) of 6-mm diameter. If a 3.2KB load P is applied at B, determine (a) the elongation of rod AB, (b) the deflection of point B, (c) the average normal stress in rod AB.
Solution:
L= FLA E= 32000.253.14910-6200109=1.4 10-4 m
Deflection of B =
3200 0.033.14252- 22210-6 3 109+ 1.4 10-4=0.214 mm
= FA= 32003.14 9 10-6=113.2 MPa
Problem Number (2)
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is made of steel E = 200GPa and rod BC of brass E = 105GPa. Determine (a) the total deformation of the composite rod ABC, (b) the deflection of point B.
Solution:
Assume that the force 40KN is directed to downward at point B
L= 30 103 0.253.14 15 15 10-6 200 109+ 70 103 0.33.14 25 25 10-6 105 109=0.393 mm
Deflection of Point B = 70 103 0.33.14 25 25 10-6 105 109 = 0.102 mm
Problem Number (3)
Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4KN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B.
Solution:
LBC=LAB
(Q-4000) 0.53.14 0.03 0.03 70 109= 4000 0.43.14 0.01 0.01 70 109
Then,Q = 32800 N
Then,Deflection of B = (32800-4000) 0.53.14 0.03 0.03 70 109= 0.0728 mm
Problem Number (4)
The rod ABC is made of an aluminum for which E = 70GPa. Knowing that P = 6KN and Q = 42 KN, determine the deflection of (a) point A, (b) point B.
Solution:
Deflection of A = LAB- LBC
= 6000 0.43.14 0.01 0.01 70 109- (42000-6000) 0.53.14 0.03 0.03 70 109=0.01819 mm
Deflection of B = 42000-60000.53.14 0.03 0.03 70 109=0.091 mm
Problem Number (5)
Each of the links AB and CD is made of steel (E = 200GPa) and has a uniform rectangular cross section of 6 * 24 mm. Determine the largest load which can be suspended from point E if the deflection of E is not to exceed 0.25 mm.
Solution:
MB = P(375 + 250) FDC (250) = 0
FDC=2.5P (Tension)
Fy = FDC FBA P = 0
FBA=1.5P (Tension)
CD= FDC LDCEDC ADC= 2.5P (200)(10)-3200(10)9(6)(24)(10)-6=1.736P (10)-8 m (Downward)
BA= FBALABEAB AAB= 1.5P (200)(10)-3200(10)9(6)(24)(10)-6=1.0416P (10)-8 m (Upward)
From geometry of the deflected structure:
E=250+375250 C- (375250)B
E=2.5-1.736P(10)-8- (1.5)(1.0416P)(10)-8= -2.7776P(10)-8 m
For maximum deflection E=0.25mm
2.7776P(10)-8=0.25(10)-3
P)max = 9.57 KN
Problem Number (6)
The length of the 2-mm diameter steel wire CD has been adjusted so that with no load applied, a gap of 1.5mm exists between the end B of the rigid beam ACB and a contact point E. knowing that E = 200 GPa, determine where a 20-kg block should be placed on the beam in order to cause contact between B and E.
Solution: