Deflection and member deformation

1. 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:
∆�=
��
?????? × �
=
3200×0.25
3.14×9×10
−6
×200×10
9
=1.4 × 10
−4
�
Deflection of B =
3200 ×0.03
3.14((25)
2
− (22)
2)×10
−6
× 3 × 10
9
+ 1.4 × 10
−4
=0.214 ��

??????=
�
�
=
3200
3.14 ×9 × 10
−6
=113.2 ��??????

2. 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
∆�=
30 × 10
3
×0.25
3.14 ×15 ×15 × 10
−6
×200 × 10
9
+

70 × 10
3
×0.3
3.14 ×25 ×25 × 10
−6
×105 × 10
9
=0.393 ��
Deflection of Point B =
70 × 10
3
×0.3
3.14 ×25 ×25 × 10
−6
×105 × 10
9
=
0.102 mm



Problem Number (3)

3. 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:
∆�
��=∆�
��
(�−4000) ×0.5
3.14 ×0.03 ×0.03 ×70 × 10
9
=
4000 ×0.4
3.14 × 0.01 ×0.01 ×70 × 10
9

Then, Q = 32800 N
Then, Deflection of B =
(32800−4000) ×0.5
3.14 ×0.03 ×0.03 ×70 × 10
9
=
0.0728 mm



Problem Number (4)

4. 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 = ∆�
��− ∆�
��
=
6000 ×0.4
3.14 × 0.01 ×0.01 ×70 × 10
9
−
(42000−6000) ×0.5
3.14 ×0.03 ×0.03 ×70 × 10
9
=
0.01819 ��
Deflection of B =
(42000−6000)×0.5
3.14 ×0.03 ×0.03 ×70 × 10
9
=0.091 ��






Problem Number (5)

5. 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
∴ �
��=2.5� (����??????��)
∑Fy = FDC

– FBA – P = 0
∴�
��=1.5� (����??????��)
∴Δ
CD=
�
�� �
��
�
�� ??????
��
=
2.5� (200)(10)
−3
(200)(10)
9
(6)(24)(10)
−6
=1.736� (10)
−8
� (��??????�????????????��)
∴ Δ
BA=
�
���
��
�
�� ??????
��
=
1.5� (200)(10)
−3
(200)(10)
9
(6)(24)(10)
−6
=1.0416� (10)
−8
� (��????????????��)
From geometry of the deflected structure:
∴Δ
�=(
250+375
250
)Δ
C− (
375
250
)Δ
B

6. ∴Δ
�=(2.5)(−1.736�)(10)
−8
− (1.5)(1.0416�)(10)
−8
= −2.7776�(10)
−8
�
For maximum deflection |Δ
�|=0.25��
∴2.7776�(10)
−8
=0.25(10)
−3

∴P)max = 9.57 KN

7. 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: