Patrick Clarence DG. De Guzman
BSME
PROBLEM SET#5
Problem 5.1
When the two forces are placed on the beam, the diameter of the A-36 steel
rod BC decreases from 40 mm to 39.99 mm. Determine the magnitude of each force
P.
Given:
a) D = 40 mm
b) D = 39.99 mm
Required:
a) Determine the magnitude of each force P
Solution:
∑MA = 0 ] ↻+
4
−FBC ( )(3 m) + P + P(1) = 0
5
FBC = 1. 25 P
For Normal Stress and Strain:
𝑑−𝑑𝑜 39.99−40
𝑚𝑚
εlat =
=
= −0.25(10−3)
𝑚𝑚
𝑑𝑜
40
0.78125(10−3 )𝑚𝑚
εlat = −vεa ; −0.25(10−3) = −(0.32)εa ; εa =
σBC= E εa ∴ σBC = 200(109)(0.78125(10−3))
σBC= 156.25 MPa
𝐹
1.25 𝑃
σBC= 𝐵𝐶 ; 156.25(106) = 𝜋
𝐴𝐵𝐶
4
(0.04)2
P=157.08 × 103N ≈ 157 kN
𝑚𝑚
Problem 5.2
If P 150 kN, determine the elastic elongation of rod BC and the decrease in its
diameter. Rod BC is made of A-36 steel and has a diameter of 40 mm.
Given:
a) P = 150 kN
b) D = 40 mm
Required:
a) Determine the elastic elongation of rod BC
b) The decrease in its diameter
Solution:
∑MA = 0 ] ↻ +
4
−FBC (5) (3) + 150(2) + 150(1) = 0
FBC= 187.5 kN
𝐹
187.5(103 )
𝐴𝐵𝐶
(0.04)2
4
σBC= 𝐵𝐶 = 𝜋
= = 149.21 MPa
σBC= E εBC
= 149.21(106) = 200(109)εBC ; εBC= 0.7460 × 10−3mm/mm
LBC=√7502 + 10002 = 1250𝑚𝑚
δBC=εBC LBC = 0.07460 × 10−3(1250)
δBC= 0.933 mm
εlat= −vεa
εlat= −(0.32)(0.7460 × 10−3) = −0.2387 × 10−3
δd= εlatdBC
= (−0.2387 × 10−3)(40)
δd= −9. 55 × 10−3mm
Problem 5.3
The friction pad A is used to support the member, which is subjected to an
axial force of P=2kN. The pad is made from a material having a modulus of elasticity
of E=4 MPa and Poisson’s ratio v=0.4. If slipping does not occur, determine the
normal and shear strains in the pad. The width is 50 mm. Assume that the material is
linearly elastic. Also, neglect the effect of the moment acting on the pad.
Given:
a)
b)
c)
d)
P = 2 kN
E = 4 MPa
v = 0.4
w = 50 mm
Required:
a) Determine the normal and shear strains in the pad
Solution:
∑Fx = 0 ] → + ; V − 2cos (60°) = 0 ; V = 1 kN
∑Fy = 0 ] ↓− ; N − 2sin(60°) = 0 ; N = 1.732 kN
𝑉
1𝑋103
𝐴
𝑁
0.1(0.05)
1.732𝑋103
τ= =
σ=
G=
𝐴
=
= 200 kPa
0.1(0.05)
𝐸
2(1 + 𝑣)
=
= 346.41 kPa
4
2(1 + 0.4)
= 1.429 MPa
σ = Eε
346.41×103= 4(106)ε
ε = 0. 08660 mm/mm
τ =Gγ
200×103= 1.429(106)γ
γ= 0.140 rad
Problem 5.4
The lap joint is connected together using a 1.25 in. diameter bolt. If the bolt is
made from a material having a shear stress–strain diagram that is approximated as
shown, determine the shear strain developed in the shear plane of the bolt when P =
75 kip.
Given:
a) D = 1.25 in
b) P = 75 kip
Required:
a) Determine the shear strain developed in the shear
plane of the bolt
Solution:
∑Fx = 0 ] → +
75 − 2(V) = 0 ; V = 37.5 kip
𝑉
τ= = 𝜋
𝐴
4
37.5
(1.252 )
τ = 30.56 ksi
30.56
𝛾
=
50
0.005
γ= 3.06 × 10−3 rad
Problem 5.5
The lap joint is connected together using a 1.25 in. diameter bolt. If the bolt is
made from a material having a shear stress–strain diagram that is approximated as
shown, determine the permanent shear strain in the shear plane of the bolt when
the applied force P = 150 kip is
removed.
Given:
a) D = 1.25 in
b) P = 150 kip
Required:
a) Determine the permanent shear strain in
the shear plane of the bolt
Solution:
∑Fx = 0 ] → +
150 − 2(V) = 0 ; V = 75 kip
75
𝑉
τ= = 𝜋
= 61.12 ksi
2)
𝐴
(1.25
4
61.12−50
=
75−50
𝛾−0.005
0.05−0.005
γ = 0.02501 rad
G=
50
0.005
= 10 × 103ksi
τ = Gγr
61.12 = 10 × 103(γr)
γr=6.112 × 10−3rad
γP = γ − γr
= 0.02501 − 6.12 × 10−3
γP = 0. 0189 rad
Problem 5.6
A shear spring is made by bonding the rubber annulus to a rigid fixed ring and
a plug. When an axial load P is placed on the plug, show that the slope at point y in
the rubber is dy/dr = − tan P /(2πhGr) or small angles we can write Integrate this
dy/dr = −P/(2πhGr) expression and evaluate the constant of integration using the
condition that y=0 at r = r0. From the result compute the deflection y = δ of the plug.
Given:
a) y = 0
b) r = r0
c) y = δ
Required:
a) Compute the deflection y = δ of the plug
Solution:
𝜏
𝑃
γ= 𝐴 =
𝐺
2𝜋𝑟ℎ
;
𝑑𝑦
𝑑𝑟
= - tanγ = -tan(
𝑃
2𝜋𝐺𝑟
)
then tanγ = γ. Therefore,
𝑃
𝑃
𝑑𝑟
𝑃
𝑑𝑦
=; y=;
y=ln r + C
∫
𝑑𝑥
2𝜋ℎ𝐺𝑟
2𝜋ℎ𝐺
𝑟
2𝜋ℎ𝐺
At r = r0 , y = 0
0=-
𝑃
2𝜋ℎ𝐺
Then, y = -
ln r0 + C ; C = 𝑃
2𝜋ℎ𝐺
At r = ri , y = δ
δ=
𝑷
𝟐𝝅𝒓𝒉
𝒓
𝐥𝐧 𝟎
𝒓𝒊
𝑟
ln( 0)
𝑟
𝑃
2𝜋ℎ𝐺
ln r0