1. Problem Number (1)
A 250-mm bar of 15 * 30-mm rectangular cross section
consists of two aluminum layers, 5-mm think, brazed to a center
brass layer of the same thickness. If it is subjected to centric
forces of magnitude P = 30KN, and knowing that Ea = 70GPa
and Eb = 105GPa, determine the normal stress (a) in the
aluminum layers, (b) in the brass layers.
Solution:
Fa : is the force in the aluminum layers.
Fs : is the force in the steel layer.
Fa + Fs = 30KN (1)
∆ 𝐴 =
𝐹𝑎 𝐿 𝑎
𝐸 𝑎 𝐴 𝑎
=
𝐹𝑎 (250)(10)−3
(70)(10)9(10)(30)(10)−6
= 1.19(10)−8
𝐹𝑎 𝑚𝑚
Δ 𝑠 =
𝐹𝑠 𝐿 𝑠
𝐸𝑠 𝐴 𝑠
=
𝐹𝑠(250)(10)−3
(105)(10)9(5)(30)(10)−6
= 1.587(10)−8
𝐹𝑠 𝑚𝑚
Compatibility condition:
∆a = ∆s
∴ 𝐹𝑎 = 1.334 𝐹𝑠
From (1) :

2. ∴ 𝐹𝑠 = 12.85 𝐾𝑁 ∴ 𝜎𝑠 = 85.67 𝑀𝑃𝑎
∴ 𝐹𝑎 = 17.15 𝐾𝑁 ∴ 𝜎𝑎 = 57.17 MPa
Problem Number (2)
Compressive centric forces of 160KN are applied at both
ends of the assembly shown by means of rigid plates. Knowing
that Es = 200 GPa and Ea = 70GPa, determine (a) the normal
stresses in the steel core and the aluminum shell, (b) the
deformation of the assembly.
Solution:
Let the force Ps be the force that affect on the steel core and the
force Pa be the force that affect on the Aluminum shell.
Then, Ps + Pa = P
∵ ΔL =
P L
A E
, ΔLs = ΔLa
∴
Ps L
As Es
=
Pa L
Aa Ea
As = 3.14 * 12.5 * 12.5 * 10-6 = 4.9 * 10-4 m
Aa = 3.14 * ((31)2 – (12.5)2) * 10-6 = 2.527 * 10-3 m

3. Ps = P - Pa
∴
(P − Pa) L
As Es
=
Pa L
Aa Ea
∴ Pa =
AaEa P
AaEa + AsEs
= 102848.9 N
∴ σa = 40.7 MPa
∴Ps = 160000 – 102848.9 = 57151.1 N
∴ σs = 116.3 MPa
The deformation of the assembly
=
102848.9 × 0.25
2.527 × 10−3 × 70 × 109
= 0.145 mm
Problem Number (3)
Three steel rods E = 200 GPa support a 36-KN load P.
Each of the rods AB and CD has a 200-mm2
cross sectional area
and rod EF has a 625-mm2
cross sectional area. Neglecting the
deformation of rod BED, determine (a) the change in length of
rod EF, (b) the stress in each rod.

4. Let the force P1 affect on the rod EF of L1 = 0.4m and of cross
section area 625 * 10-6 m2 and the force P2 affect on the rod CD
of L2 = 0.5 m and of cross section area 200 * 10-6 m2 and the
force P3 affect on the rod AB of L3 = 0.5 m and of cross section
area 200 * 10-6 m2 .
Then, P = P1 + P2 + P3
Since, P2 = P3 , then P = P1 + 2P2 , then P2 =
P− P1
2
ΔL1 = ΔL2
𝑃1 𝐿1
𝐴1 𝐸
=
𝑃2 𝐿2
𝐴2 𝐸
P1 L1
A1
=
(P − P1) L2
2A2
P1 =
P L2 A1
L2 A1 + 2A2 L1
=
36000 ∗ 0.5 ∗ 625 ∗ 10−6
(0.5 ∗ 625 ∗ 10−6) + (2 ∗ 200 ∗ 0.4 ∗ 10−6)
= 23.81 KN
Δ𝐿1 =
𝑃1 𝐿1
𝐴1 𝐸
=
23810 ∗ 0.4
625 ∗ 200 ∗ 103
= 0.0762 mm
𝜎1 =
𝑃1
𝐴1
=
23810
625 ∗ 10−6
= 38.1 MPa
𝜎2 = 𝜎3 =
𝑃2
𝐴2
=
(36 − 23.81) ∗ 103
2 ∗ 200 ∗ 10−6
= 30.475 MPa

5. Problem Number (4)
Two cylindrical rods, one of steel and the other of brass,
are joined to C and restrained by rigid supports at A and E. for
the loading shown and knowing that Es = 200 GPa and Eb = 105
GPa, determine (a) the reactions at A and E, (b) the deflection of
point C.
By dividing the rod to four parts 1) DE = 0.1m , 2) CD = 0.1m ,
3) BC = 0.12m , 4) AB = 0.18m, and by taking the force at point
B to direction of the force at D.
Δ1 = 0
Δ2 =
40 ∗ 1000 ∗ 0.1
15 ∗ 15 ∗ 3.14 ∗ 105 ∗ 103
= 5.4 ∗ 10−5
𝑚
Δ3 =
40 ∗ 1000 ∗ 0.12
20 ∗ 20 ∗ 3.14 ∗ 200 ∗ 103
= 1.91 ∗ 10−5
𝑚
Δ4 =
100 ∗ 1000 ∗ 0.18
20 ∗ 20 ∗ 3.14 ∗ 200 ∗ 103
= 7.166 ∗ 10−5
𝑚
∆ = Δ1 + Δ2 + Δ3 + Δ4 = 0.14476 mm
RA + RE = 100KN
By dividing the rod to two parts 1) CE and 2) AC
Δ1 + Δ2 = 0.14476 𝑚𝑚

6. Δ1 =
𝑅 𝐸 ∗ 0.2
15 ∗ 15 ∗ 3.14 ∗ 105 ∗ 103
→ (1)
Δ2 =
𝑅 𝐸 ∗ 0.3
20 ∗ 20 ∗ 3.14 ∗ 200 ∗ 103
→ (2)
𝑡ℎ𝑒𝑛, 𝑅 𝐸 (
0.2
15 ∗ 15 ∗ 3.14 ∗ 105 ∗ 103
+
0.3
20 ∗ 20 ∗ 3.14 ∗ 200 ∗ 103
) = 0.14476
Then, RE = 37.21 KN , RA = 62.79 KN
The deflection of the point C is :
( 1.91 * 10-5 ) + ( 7.166 * 10-5 ) – ( 4.443 * 10-5 ) = 46.3 𝜇𝑚