Answer:

Problem1.1 A rod 150 cm long and of diameter 2.0 cm is SUBJECTED to an axial pull of 20 kN.

If the modulus of elasticity of the material of the rod is 2 x 105 N/mm2

; DETERMINE:

The stress,

The strain, and.

The elongation of the rod

Sol. Given: length of the rod, L=150cm

Diameter of the rod D=2.0cm = 20mm

Area A=pi

4

(20)

2 = 100π mm2

Axial pull, P 20kN= 20,000N

Modulo of elasticity, E 2.0 x 105 N/mm2

The stress (α) is given by equation (1.1) as

α=

P

A

=

2000

100π

= 63.662 /2

Using equation (1.5), the strain is obtained as

E=α

 Strain, e=

=

63.662

2 105 = 63.662/2 Ans.

Elongation is obtained by using equation (1.2) as

=

Elongation DL= e x L

= 0.000318 x 150 = 0.0477 cm. Ans.

Problem 1.2. find the minimum diameter of a steel wire, which is used or raise a load of

4000N if the stress I n the rod is not exceed 95MN/m2

Sol. Given: load, P = 4000N

Stress,  = 95 MN/m2 = 95 x 106N/m2

( M= Mega = 106

)

= 95 N/mm2 ( 106 N/m2 = 1 N/mm2

)

Let D = diameter in a wire in MM

 Area, =

4

2

Now Stress =

=

4000

4

2 = 4000 4 /

2 or

2 =

40004

95

= 53.61

D = 7.32 mm. take help of this