# 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: