deflection and member deformation
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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.
∆ L= F× LA×E
Deflection of B =
3.14 ((25 )2−(22 )2 )×10−6×3×109+1.4×10−4=0.214mm
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.
Assume that the force 40KN is directed to downward at point B
∆ L= 30×103×0.253.14×15×15×10−6×200×109 +
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.
∆ LBC=∆ LAB
Then, Q = 32800 N
Then, Deflection of B = (32800−4000)×0.5
3.14×0.03×0.03×70×109 = 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.
Deflection of A = ∆ LAB−∆ LBC
Deflection of B = (42000−6000 )×0.5
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.
∑MB = P(375 + 250) – FDC (250) = 0
∑Fy = FDC – FBA – P = 0
∴ ΔCD=FDCLDCEDC ADC
∴ ΔBA=FBA LABEAB A AB
From geometry of the deflected structure:
∴ ΔE=(250+375250 )ΔC−( 375
∴ ΔE=(2.5 ) (−1.736 P ) ¿
For maximum deflection |ΔE|=0.25mm
∴2.7776 P ¿
∴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.