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×𝐿
3200 × 0.25
=
𝐴 × 𝐸
3.14 × 9 × 10−6 × 200 × 109
= 1.4 × 10−4 𝑚
Deflection of B =
3200 × 0.03
+ 1.4 × 10−4
2
2
−6
9
3.14((25) − (22) ) × 10 × 3 × 10
= 0.214 𝑚𝑚
𝜏=
𝐹
𝐴
=
3200
3.14 ×9 × 10−6
= 113.2 𝑀𝑃𝑎
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 ×0.25
3.14 ×15 ×15 × 10−6 ×200 × 109
70 × 103 ×0.3
3.14 ×25 ×25 × 10−6 ×105 × 109
Deflection of Point B =
0.102 mm
Problem Number (3)
+
= 0.393 𝑚𝑚
70 × 103 ×0.3
3.14 ×25 ×25 × 10−6 ×105 × 109
=
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 =
0.0728 mm
Problem Number (4)
(32800−4000) ×0.5
3.14 ×0.03 ×0.03 ×70 × 109
=
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
=
0.01819 𝑚𝑚
Deflection of B =
(42000−6000)×0.5
3.14 ×0.03 ×0.03 ×70 × 109
Problem Number (5)
= 0.091 𝑚𝑚
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
𝐹𝐷𝐶 𝐿𝐷𝐶
2.5𝑃 (200)(10)−3
=
=
(200)(10)9 (6)(24)(10)−6
𝐸𝐷𝐶 𝐴𝐷𝐶
= 1.736𝑃 (10)−8 𝑚 (𝐷𝑜𝑤𝑛𝑤𝑎𝑟𝑑)
∴ ΔBA
𝐹𝐵𝐴 𝐿𝐴𝐵
1.5𝑃 (200)(10)−3
=
=
(200)(10)9 (6)(24)(10)−6
𝐸𝐴𝐵 𝐴𝐴𝐵
= 1.0416𝑃 (10)−8 𝑚 (𝑈𝑝𝑤𝑎𝑟𝑑)
From geometry of the deflected structure:
∴ Δ𝐸 = (
250 + 375
375
) ΔC − (
)Δ
250
250 B
∴ Δ𝐸 = (2.5)(−1.736𝑃)(10)−8 − (1.5)(1.0416𝑃)(10)−8
= −2.7776𝑃(10)−8 𝑚
For maximum deflection |Δ𝐸 | = 0.25𝑚𝑚
∴ 2.7776𝑃(10)−8 = 0.25(10)−3
∴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: