1. Prob. 2.15: A 3-mm thick hollow polystrene cylinder (𝑬 = 𝟑 𝑮𝑷𝒂)
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. 𝟐 𝑲𝑵 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.
𝑺𝒕𝒓𝒆𝒔𝒔 & 𝑆𝑡𝑟𝑎𝑖𝑛 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠
𝑷𝒂𝒓𝒕 𝟐

2. Solution:
ΔL =
𝑭𝑳
𝑬𝑨
For the rod AB:
ΔL1 =
𝟑𝟐𝟎𝟎×𝟎.𝟐𝟓
𝟐𝟎𝟎∗𝟏𝟎𝟗∗𝜋(0.003)2 = 1.414*𝟏𝟎−𝟒
m
The elongation of rod AB = 0.1414 mm
For the cylinder:
The part of which thickness is 3 mm will be affected by
the stress resulting from the force F=3.2 KN
The area of this part (A) = 𝝅((𝟎. 𝟎𝟐𝟓)𝟐
− (𝟎. 𝟎𝟐𝟐)𝟐
A = 4.429*𝟏𝟎−𝟒
𝒎𝟐
ΔL2 =
−𝟑𝟐𝟎𝟎∗𝟎.𝟎𝟑
𝟑∗𝟏𝟎𝟗∗𝟒.𝟒𝟐𝟗∗𝟏𝟎−𝟒 = -7.225*𝟏𝟎−𝟓
m
UB = ΔL1 – ΔL2 = 1.414*𝟏𝟎−𝟒
- (-7.225*𝟏𝟎−𝟓
)
UB = 2.14 mm
Normal stress for rod AB:
σ =
𝑭
𝑨
=
𝟑𝟐𝟎𝟎
𝝅(𝟎.𝟎𝟎𝟑)𝟐 = 𝟏𝟏𝟑𝟏𝟕𝟔𝟖𝟒𝟖. 𝟒
𝑵
𝒎𝟐

3. Prob. 2.17: Two solid cylindrical rods are joined at B and loaded as
shown. Rod AB is made of steel (𝐄 = 𝟐𝟎𝟎 𝐆𝐏𝐚) and rod BC of brass
(𝐄 = 𝟏𝟎𝟓 𝐆𝐏𝐚). Determine (a) the total deformation of the composite
rod ABC, (b) the deflection of point B.
Solution:
UB = ΔL1 =
−(𝟑𝟎𝟎𝟎𝟎+𝟒𝟎𝟎𝟎𝟎)𝑳
𝑬𝑨
=
− 𝟑𝟎𝟎𝟎𝟎+𝟒𝟎𝟎𝟎𝟎 (𝟎.𝟑)
𝟐𝟎𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟏𝟓)𝟐
UB = -1.485*𝟏𝟎−𝟒
m
ΔL2 =
−(𝟑𝟎𝟎𝟎𝟎)𝑳
𝑬𝑨
=
− 𝟑𝟎𝟎𝟎𝟎 (𝟎.𝟐𝟓)
𝟏𝟎𝟓∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟐𝟓)𝟐
= −𝟑. 𝟔𝟑𝟕 ∗ 𝟏𝟎−𝟓
𝒎
Total deflection of rod ABC = ΔL1 + ΔL2
= 1.8487*𝟏𝟎−𝟒
𝒎

4. Prob. 2.19: Both portions of the rod ABC are made of an aluminum
for which 𝐄 = 𝟕𝟎 𝐏𝐚 . Knowing that the magnitude of P is 4 KN,
determine (a) the value of 𝐐 so that the deflection at A is zero, (b) the
corresponding deflection of B.
Solution:
We will call the portion BC: part (1)
And the portion AB: part (2)
ΔL =
𝑭𝑳
𝑬𝑨
For part (1):
F = Q-P = Q-4000
ΔL1 =
− 𝑸−𝟒𝟎𝟎𝟎 (𝟎.𝟓)
𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟑)𝟐

5. For part (2):
ΔL2 =
𝟒𝟎𝟎𝟎 (𝟎.𝟒)
𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟏)𝟐
Since the deflection at A = 0, then
ΔL1 = ΔL2
− 𝑸−𝟒𝟎𝟎𝟎 (𝟎.𝟓)
𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟑)𝟐 =
𝟒𝟎𝟎𝟎 (𝟎.𝟒)
𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟏)𝟐
(-Q+4000) (0.5) = (4000) (0.4) (9)
-0.5Q+2000 = 14400
0.5Q = - 16400
Q = - 32800 N
Q = 32.8 KN
Prob. 2.20: The rod ABC is made of an aluminum for which
𝐄 = 𝟕𝟎 𝐆𝐏𝐚. Knowing that 𝐏 = 𝟔 𝐊𝐍 and 𝐐 = 𝟒𝟐 𝐊𝐍, determine the
deflection of (a) point A, (b) point B. (𝑺𝒆𝒆 𝒕𝒉𝒆 𝒑𝒓𝒆𝒗𝒊𝒐𝒖𝒔 𝒇𝒊𝒈𝒖𝒓𝒆)
Solution:
We will call the portion BC: part (1)
And the portion AB: part (2)
ΔL =
𝑭𝑳
𝑬𝑨
For part (1):
F = Q-P = -42+6 -36 KN

6. ΔL1 =
− 𝟑𝟔𝟎𝟎𝟎 (𝟎.𝟓)
𝟕𝟎∗𝟏𝟎𝟗∗𝝅(𝟎.𝟎𝟑)𝟐 = UB = 9.09𝟏𝟎−𝟓
𝒎
ΔL2 =
𝟔𝟎𝟎𝟎 𝟎.𝟒
𝟕𝟎∗𝟏𝟎𝟗∗𝝅 𝟎.𝟎𝟏 𝟐 = 1.09*𝟏𝟎−𝟒 𝒎
UA = ΔL1 + ΔL2 = 9.09𝟏𝟎−𝟓 + 1.09*𝟏𝟎−𝟒 =
1.81*𝟏𝟎−𝟓
𝒎
Prob. 2.28: The length of the 2-mm-diameter steel wire CD has
been adjusted so that with no load applied, a gap of 1.5 mm 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.

7. Solution: