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= F× LA×E

= 3200×0.25

3.14×9×10−6×200×109=1.4×10−4m

Deflection of B =

3200×0.03

3.14 ((25 )2−(22 )2 )×10−6×3×109+1.4×10−4=0.214mm

τ= FA

= 3200

3.14×9×10−6=113.2MPa

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 =0.393mm

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)×0.5

3.14×0.03×0.03×70×109 =4000×0.4

3.14×0.01×0.01×70×109

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.

Solution:

Deflection of A = ∆ LAB−∆ LBC

= 6000×0.4

3.14×0.01×0.01×70×109 −(42000−6000)×0.5

3.14×0.03×0.03×70×109 =0.01819mm

Deflection of B = (42000−6000 )×0.5

3.14×0.03×0.03×70×109=0.091mm

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)

∴ ΔCD=FDCLDCEDC ADC

=2.5P (200)¿¿

∴ ΔBA=FBA LABEAB A AB

=1.5P (200)¿¿

From geometry of the deflected structure:

∴ ΔE=(250+375250 )ΔC−( 375

250)ΔB

∴ 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.

Solution: