Question

The multi-material rod below has force P=3 kN applied to it. The center portion is steel, with E=207 GPa and yield strength Sy=250 MPa. The outer shell is aluminum with E=69 GPa and yield strength Sy=276 MPa. The center portion has cross-sectional area of 7.07 mm2 while the outer shell has cross-sectional area of 5.50 mm² . The load P is applied through rigid plates (not pictured) so that the force is shared between the two materials. 

Note: This scenario is used for this and the next two problems. 

What is the stress in the steel portion, in units of MPa, under this loading? 

What is the stress in the aluminum portion, in units of MPa, under this loading? 

What is the deformation of the system, in units of mm, under this loading if the bar is 1 min length?

Answer 1

Answer 2

SOLUTION :

Deformation (D)  is equal for both the materials' portions.. 

Load (P) is shared by the two portions..

So,

Ps + Pa = Load applied = 3 kN = 3000 N

Ds = Da

=> Ps L / (As * Es) = Pa L / (Aa * Ea)

=> Ps / Pa = (Es * As) / (Ea * Aa)

=> Ps / Pa = (207*10^9 * 7.07 * 10^(-6)) / (69*10^9 * 5.50 *10^(-6)) =  3.856 (pure ratio) 

But,

Ps + Pa = 3000

=> Pa (Ps/Pa + 1) = 3000

=> Pa (3.856 + 1) = 3000

=> Pa = 3000/4.856

=> Pa = 617.79 N

=> Ps = 3000 - 617.79 = 2382.21 N 

So,

Stress in steel portion, (fs),  

= Ps / As 

= 2382.21 / (7.07*10^(-6)) 

= 336.95 * 10^6  

= 336.95 MPa (ANSWER)

Stress in aluminium portion, (fa)

= Pa / Aa 

= 617.79 / (5.50*10^(-6))

= 112.33 * 10^6 

= 112.33 MPa (ANSWER).

So,

Deformation may be calculated either based on steel portion or aluminium portion.

Both will give the same result.

Deformation of the system (based on steel portion) per meter length of system

= fs  / Es

= 336.95 *10^6 / (207*10^9)

= 0.00163 m 

= 1.63 mm (ANSWER).