Mekanika Bahan
S
TRESS
!Stress
!
Average Normal Stress in an Axially Loaded
Bar
!
Average Shear Stress
!
Allowable Stress
(2)
Equilibrium of a Deformable Body
A
D
w(s) C
FR
.
w
s
• Body Force
• Support Reaction
DyAy Ax
(3)
Type of Connection
Idealized
Symbol Reaction Number of Unknowns
Table 1 Supports for Coplanar Structures
(1) θ
Light cable
One unknown. The reaction is a force that acts in the direction of the cable or link.
θ θ
One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact.
F
Rollers and rockers (2)
One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact. (3)
F
(4)
Type of Connection
Idealized
Symbol Reaction Number of Unknowns
Three unknowns. The reactions are the moment and the two force components.
M Fy
Fx
(6)
Fixed support
Two unknowns. The reactions are two force components.
Fy
Fx
(4)
Smooth pin or hinge
Two unknowns. The reactions are a force and moment.
(5)
Internal pin
Fy Fx
(5)
• Equation of Equilibrium
Σ
Σ
Σ
Σ
F = 0
Σ
Σ
Σ
Σ
M
O= 0
Σ
Σ
Σ
Σ
F
x= 0
Σ
Σ
Σ
Σ
F
y= 0
Σ
Σ
Σ
Σ
F
z= 0
Σ
Σ
Σ
(6)
• Internal Resultant Loadings
F
2F
1F
3F
4section
F
2F
1x
y
F
3F
4V
N
M
OV
N
(7)
Example 1
Determine the resultant internal loadings acting on the cross section at C of the beam shown.
270 N/m
3 m 6 m
A B
(8)
• Support Reactions
(1/2)(9)(270) = 1215 N
3 m 6 m
A
B C
270 N/m
540 N
4 m
B 2 m
180 N/m
C
540 N =
V
CN
CM
C0 = -540(2) = -1080 N•m =
540 N 135 N
90 N/m
MA = 3645 N•m
3 m
A C
270 N/m
180 N/m
540 N
1080 N•m
0
• Internal Loading
MA = 3645 N•m
(9)
b
h tw
(1/2)(9)(270) = 1215 N
3 m 6 m
A
B C
270 N/m
• Shear and bending moment diagram
3645 N•m
1215 N
V (N) 1215
M (N•m)
+
3645
-540
1080
1080 N•m
C flanges web
(10)
Example 2
Determine the resultant internal loadings acting on the cross section at C of the machine shaft shown. The shaft is supported by bearings at A and B, which exert only vertical forces on the shaft.
225 N 800 N/m
200 mm
50 mm
100 mm 100 mm
50 mm A
C
B
(11)
(800 N/m)(0.15 m) = 120 N/m 225 N
0.275 m 0.125 0.1 m
A B
D • Support Reactions
0.2 m A
C
800(0.05) = 40 N
VC MC
NC • Internal Loading
+ Σ M
A = 0: -(120)(0.275) + By(0.4) - (225)(0.5) = 0, By = 363.75 N
+ ΣM
A = 0:
ΣFx = 0: NC = 0
+
ΣFy = 0:
-18.75 - 40 - VC = 0,
VC = -58.75 N
+
Σ Fy = 0: Ay - 120 +363.75 - 225 = 0 Ay = -18.75 N,
+
By Ay
Ax
Σ Fx = 0: Ax = 0 Ax = 0,
(12)
58.75 N 5.69 N•m
0
0.2 m
0.025 m A
C
800(0.05) = 40 N
18.75 N
VC = -58.75 N
MC = -5.69 N•m
NC = 0
0.2 m
0.025 m A
C
800(0.05) = 40 N
18.75 N
(13)
Example 3
The hoist consists of the beam AB and attached pulleys, the cable, and the motor. Determine the resultant internal loadings acting on the cross section at C if the motor is lifting the 2 kN load W with constant velocity. Neglect the weight of the pulleys and beam.
1 m 2 m 0.5 m
W 0.15 m
0.15 m C
A D
(14)
1 m 2 m 0.5 m
W 0.15 m
0.15 m C
A D
B
W = 2 kN
1 m 0.15 m
C A
W = 2 kN
VC MC NC TC = 2 kN
+ Σ M
C = 0:
MC - 2(0.15) + 2(1.15) = 0 Σ Fx = 0:
2 + NC = 0
NC = -2 kN
+
Σ Fy = 0:
-2 - VC = 0,
VC = -2 kN
(15)
Example 4
Determine the resultant internal loadings acting on the cross section at G of the wooden beam shown . Assume the joints at A, B, C, D, and E are pin-connected.
6.5 kN
0.6 m 0.6 m 2 m
1 m
A
B C
D
E G
(16)
6.5 kN
0.6 m 0.6 m (2/3)(2) = 4/3 m
1 m
A
B C
D
E G
4 kN/m (1/2)(2)(4) = 4 kN 2/3 m
FBC
Ex Ey • Support Reactions
+ Σ M
E = 0: 4(4/3) + 6.5(3.2) - FBC(1) = 0, FBC = 26.1 kN
Σ Fy = 0: -6.5 - 4 + Ey = 0 Ey = 10.5 kN
+
Two - force
Σ Fx = 0: 26.13 + Ex = 0 Ex = -26.1 kN,
(17)
• Internal loadings acting on the cross section at G
+ ΣM
G = 0: MG -34.02 sin39.81o(0.6) + 6.5(0.6) = 0 MG = 9.17 kN•m ΣFx = 0: 34.02 cos39.81o + N
G = 0 NG = -26.1 kN +
VC
MC NC 0.6 m
6.5 kN
A
G
B
FBC = 26.13 kN
50.19o
39.81o
FBD FBA
FAB
Joint B
ΣFx = 0:
-FBA sin50.19o + 26.13 = 0
FBA = 34.0 kN, (T)
+
ΣFy = 0:
-FBA cos50.19o -F
BD = 0
FBD = -21.8 kN, (C)
+
= -21.78 kN 34.02 kN =
= 34.02 kN
(18)
Example 5
Determine the resultant internal loadings acting on the cross section at B of the pipe shown. The pipe has a mass of 2 kg/m and is subjected to both a vertical force of 50 N and a couple moment of 70 N•m at its end A. It is fixed to the wall at C.
50 N
1.25 m 0.75 m
0.5 m
70 N•m A
B C
D x
y z
(19)
(MB)y
50 N
70 N•m A
B
D
x y
z
(FB)y (FB)x
(FB)z
0.25 m
WBD= 9.81 N W
BD = (2 kg/m)(0.5 m)(9.81 N/kg) = 9.81 N
0.625 m
0.625 m WAD
WAD = (2 kg/m)(1.25 m)(9.81 N/kg) = 24.52 N
• Free-Body Diagram
• Equilibrium of Equilibrium
Σ
Fy = 0: (FB)y = 0Σ
Fz = 0: (FB)z - 9.81 -24.525 -50 = 0, (FB)z = 84.3 NΣ
(MB)x = 0: (MB)x + 70 - 50(0.5) - 24.525(0.5) - 9.81(0.25) = 0, (MB)x = -30.3 N•mΣ
(MB)y = 0: (MB)y + 24.525(0.625) + 50(1.25) = 0, (MB)y = -77.8 N•m(MB)x
(MB)z
(MB)y
(MB)x
(MB)z
Σ
Fx = 0: (FB)x = 0(20)
• Free-Body Diagram
50 N
70 N•m A
B
D
x y
z
0.25 m 9.81 N
0.625 m
0.625 m 24.5 N
70 N•m
(FB)x = 0
(FB)y = 0
(FB)z = 84.3 N
(MB)x = -30.3 N•m
(MB)y = -77.8 N•m
(MB)z = 0
84.3 N
30.3 N•m
(21)
+
z
Face
+
x
Face
x
y
z
Stress
(22)
x
y
z
-
x
Face
τ
yzτ
yxσ
yτ
zyτ
zxσ
zσ'
xτ'
xzτ'
xy-
y
Face
σ'
yτ'
yzτ'
yx-
z
Face
τ'
zyτ'
zxσ'
zy
Top View
τ
xy∆
y
∆
z
σ
∆
y
∆
z
τ
yx∆
x
∆
z
σ
y∆
x
∆
z
σ
'
x∆
y
∆
z
τ
'
xy∆
y
∆
z
σ
'
y∆
x
∆
z
τ
'
yx∆
x
∆
z
σ
x=
σ'
x
σ
y=
σ'
y
σ
z=
σ’
z
τ
xy=
τ'
xy
τ
yx=
τ'
yx
τ
zx=
τ’
xz
By compatibility,
τ
xzτ
xyσ
x(23)
ΣFy = 0: - σ
'
y∆
x
∆
z +
σ
y∆
x
∆
z
= 0
+
σ
'
y=
σ
yΣFx = 0:
σ
x∆
y
∆
z -
σ
'
x∆
y
∆
z
= 0
+
σ
x=
σ
'
x+ ΣM
O´ = 0: (τxy
∆
y
∆
z
)(
∆
x
) - (
τ
yx∆
x
∆
z
)(
∆
y
) = 0
x
y
τ
xy∆
y
∆
z
σ
x∆
y
∆
z
τ
yx∆
x
∆
z
σ
y∆
x
∆
z
σ
'
x∆
y
∆
z
τ
'
xy∆
y
∆
z
σ
'
y∆
x
∆
z
τ
'
yx∆
x
∆
z
O
σ
x∆
y
∆
z
σ
'
x∆
y
∆
z
τ
xy∆
y
∆
z
τ
'
xy∆
y
∆
z
σ
y∆
x
∆
z
σ
'
y∆
x
∆
z
τ
yx∆
x
∆
z
τ
'
yx∆
x
∆
z
(24)
z
x
y
∆
x
∆
y
τ
yz∆
x
∆
z
σ
y∆
x
∆
z
τ
yx∆
x
∆
z
σ
z∆
x
∆
y
τ
zy
∆
x
∆
y
τ
zx∆
x
∆
y
∆
z
τ
xz∆
y
∆
z
τ
xy∆
y
∆
z
σ
x∆
y
∆
z
σ
'
z∆
x
∆
y
σ
'
x∆
y
∆
z
σ
'
y∆
x
∆
z
z
x
y
σ
y∆
x
∆
z
σ
z∆
x
∆
y
σ
x∆
y
∆
z
z
y
τ
zx∆
x
∆
y
τ
xz∆
y
∆
z
τ
'
xz∆
y
∆
z
τ
'
∆
x
∆
y
z
y
τ
yz∆
x
∆
z
τ
zy∆
x
∆
y
τ
'
yz∆
x
∆
z
τ
∆
∆
z
x
y
τ
∆
y
∆
z
τ
yx∆
x
∆
z
τ
'
yx∆
x
∆
z
(25)
Average Normal Stress in an Axially Loaded Bar
• Assumptions
The material must be
- Homogeneous material
- Isotropic material
P
P
P
External force
Internal force
(26)
P
P x
y z
P
x y
σ
∆F = σ∆A
P
σ σ
∫
=∫
Σ = ↑
+
A z
Rz F dF dA
F ;
σ
A P =
σ
A P =
σ
(27)
Example 6
The bar shown has a constant width of 35 mm and a thickness of 10 mm.
Determine the maximum average normal stress in the bar when it is subjected to the loading shown.
9 kN
9 kN
4 kN
4 kN
A B C D
35 mm
22 kN 12 kN
(28)
9 kN
9 kN
4 kN
4 kN
A B C D
35 mm
22 kN 12 kN
12
22 30
P (kN)
x 10 mm
35 mm
30 kN σ
MPa 7
. 85 )
m 01 . 0 )( m 035 . 0 (
kN 30
max = = = =
A P σ
σ BC
(29)
9 kN
9 kN
4 kN
4 kN
A B C D
35 mm
22 kN 12 kN
34.3
62.9 85.7
σ
(MPa)x 10 mm
35 mm
30 kN σ
MPa 7
. 85 kN
30
max = = = =
P σ
σ BC
(30)
Example 7
The 80 kg lamp is supported b two rods AB and BC as shown . If AB has a diameter of 10 mm and BC has a diameter of 8 mm, determine which rod is subjected to the greater average normal stress.
60o A
B
C
3 4 5
(31)
B
x y
• Internal Loading
0 60 cos ) 5 4 ( ;
0 − =
= Σ →
+ o
BA BC
x F F
F 0 8 . 784 60 sin ) 5 3 ( ;
0 + − =
= Σ ↑ + o BA BC
y F F
F
F = 395.2 N, F = 632.4 N
• Average Normal Stress
60o A B C 3 4 5 d = 10 mm
d = 8 mm
MPa 86 . 7 ) m 004 . 0 ( N 2 . 395 2 = = = π A F σ BC BC BC MPa 05 . 8 ) m 005 . 0 ( N 4 . 632 2 = = = π A F σ BA BA BA
The average normal stress distribution acting over a cross section of rod AB.
8.05 MPa 8.05 MPa 60o FBA 3 4 5 FBC
(32)
Example 8
Member AC shown is subjected to a vertical force of 3 kN. Determine the
position x of this force so that the average compressive stress at C is equal to the average tensile stress in the tie rod AB. The rod has a cross-sectional area of 400 mm2 and the contact area at C is 650 mm2.
200 mm 3 kN
A B
C x
(33)
200 mm 3 kN
A
C x
FAB
FC ) 1 ( 0
3 ;
0 + − = −−−−
= Σ ↑
+ Fy FAB FC
+ Σ M
A = 0: -3(x) + FC(200) = 0 ---(2) • Internal Loading
• Average Normal Stress
650 400
C AB F
F = =
σ
) 3 ( 625
.
1 −−−−
= AB
C F
F
Substituting (3) into (1), solving for FAB, the solving for FC, we obtain F = 1.143 kN F = 1.857 kN
(34)
V
V
P
Connections
P
A
B
C
D
ττττ
avgA V
avg =
τ
P
A
A
V
(35)
• Single Shear
P
P
P
V
=
P
P
P
Top View
P
P
Side View
P
P
P
V
=
P
P
P
Top View
P
P
(36)
• Double Shear
P
P
Top View
P
/2
P
P
/2
P
P
Side View
P
V = P
/2
P
P
Top View
P
/2
P
P
/2
P
P
Side View
P
(37)
Normal Stress: Compression and Tension Load
P
P
P
P´ b
d t a
a b
b
bt P
a a− = @
σ
b
t
A
=
bt
P
Section a-a
t d b
P
b b
) (
@
− =
−
σ
b
t
d
A = (b-d)t
P
(38)
t
b
d
P
P
b´
b´
t
d Abearing = dt
Bearing Stress
dt P
bearing =
σ
P
/2
P
/2
P
(39)
dθ d b t t d b P normal ) ( − =
σ
ΣFx = 0:
+ ) cos 0
2 ( 90 90 = −
∫
° °−
σ
tθ
dθ
d P b P d t d b = ° ° − 90 90 sin ) 2
(
σ
θ
θ
o bearing td P 90 sin =
σ
θ dt P bearing =σ
(40)
P
d
P
b
t
Tension
Compression
• Axial Force Diagram for Compression Load on Plate
-P
-P
dt P Stress
Bearing bearing =
−
σ
n compressio t
bd P Stress
Normal ,
) (
: = −
(41)
b
t
Tension
Compression
• Axial Force Diagram for Tension Load on Plate
tension t
d b
P Stress
Normal ,
) (
max :
− + =
−
σ
P dt
P Stress
Bearing bearing =
−
σ
P
P
d
P
a
P
/2
P
P
τ
τ
(42)
• Shearing Stress on pin
P
P
P
P
P
P
P
P
P
P
P
P
A P =
τ
(43)
Double
ShearP/
2
P
P/
2
P
/2
P
P
/2
P/
2
P/
2
P
P/
2
P/
2
P/
2
P
P/
2
P/
2
P/
2
P
A P
τ = / 2
A P
(44)
z
x
y
∆
y
∆
x
∆
z
Section plane
Pure shear
τ
yzτ
zyτ
yzτ
zyPure
Shear(45)
Example 9
The bar shown a square cross section for which the depth and thickness are 40 mm. If an axial force of 800 N is applied along the centroidal axis of the bar’s cross-sectional area, determine the average normal stress and average shear stress acting on the material along (a) section plane a-a and (b) section plane b-b.
800 N
a
a
b
b
30o
20 mm 20 mm
(46)
N = 800 N (a) section plane a-a
800 N
a
a
kPa 500 )
m 04 . 0 )( m 04 . 0 (
N 800
= =
= A N σ
0 =
avg
τ
800 N
a
a
b
b
30o
20 mm 20 mm 40 mm
(47)
(b) section plane b-b 800 N b b 30o 800 N 30o
N = 800sin30o = 400 N
V = 800 cos30o = 692.82 N
kPa 125 ) m 08 . 0 )( m 04 . 0 ( N 400 = = = A N σ kPa 51 . 216 N 82 . 692 = = =V 30o x 40 mm mm 80 ; 40 30
sin = x =
x o τavg 800 N a a b b 30o 20 mm 20 mm 40 mm σ 216.51 kPa b
(48)
Example 10
The wooden strut shown is suspended form a 10 mm diameter steel rod, which is fastened to the wall. If the strut supports a vertical load of 5 kN, compute the average shear stress in the rod at the wall and along the two shaded planes of the strut, one of which is indicated as abcd.
5 kN a b
c
d
20 mm 40 mm
(49)
5 kN
63.7 MPa
V = 5 kN
• Average shear stress in the rod at the wall
5 kN
d = 10 mm
MPa 7 . 63 ) m 005 . 0 ( kN 5 2 = = = π A V τ avg
• Average shear stress along the two shaded plane
MPa 12 . 3 ) m 02 . 0 )( m 04 . 0 ( kN 5 . 2 = = = A V τ avg 5 kN 3.12 MPa 5 kN a b c d 20 mm 40 mm
V = 2.5 kN
a b
c
d
5 kN
(50)
Example 11
The inclined member show is subjected to a compressive force of 3 kN.
Determine the average compressive stress along the areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by EDB.
25 mm
40 mm 80 mm 50 mm
3 kN 4
5 3
(51)
FBC= 3(4/5) = 2.4 kN V = 1.8 kN
3 kN 4 5 3 3 kN 4 5 3 MPa 8 . 1 kPa 1800 ) m 040 . 0 )( m 025 . 0 ( kN 8 . 1 = = = = AB AB AB A F σ
• The average compressive stress along the areas of contact defined by AB and BC :
1.8 MPa
25 mm
40 mm 80 mm 50 mm
3 kN 4
5 3
(52)
V = 1.8 kN
FBC= 3(4/5) = 2.4 kN
3 kN 4 5
3
FAB = 3(3/5) = 1.8 kN
• The average shear stress along the horizontal plane defined by EDB :
25 mm
40 mm 80 mm 50 mm
3 kN 4
5 3
MPa 562
. 0
kPa 5 . 562 )
m 08 . 0 )( m 04 . 0 (
kN 8 . 1
=
= =
= A V τ
avg
(53)
Allowable Stress
allow fail
P P S
F. =
allow fail
S F
σ
σ
= .
allow fail
S F
τ
τ
= .
6. Design of Simple Connections
allow
P A
σ
=
allow
V A
τ
=
• Cross-Sectional Area of a Tension Member
P
P
a
a
P
τ
allowP A =
(54)
(σ
b)
allowP
V = P
P
• Cross-Sectional Area of a Connector Subjected to Shear
allow
P A
σ
=
P
P
ττττ
allow• Required Area to Resist Bearing
allow b
P A
) (
σ
=(55)
• Required Area to Resist Shear by Axial Load
P
d P l
allow
π
τ
=
P
d
(56)
Example 12a
The control arm is subjected to the loading shown. (a) Determine the shear stress for the steel at all pin (b) Determine normal stress in rod AB, plate D and E. The thickness of plate D and E is 10 mm.
pin C 13 kN 25 kN
3 4
5
75 mm 50 mm 200 mm
A B
C D E
φ
= 10 mmφ
= 20 mmD E
φ
= 25 mm13 kN 50 mm
25 kN 50 mm
(57)
25 kN 13 kN
3 4
5
75 mm 50 mm 200 mm
• Reaction C
0 ) 125 . 0 ( 87 . 36 sin 25 ) 075 . 0 ( 13 ) 2 . 0 ( + − = AB F + ΣM
C = 0:
FAB = 4.5 kN , ←
ΣFy = 0:
+ C
y + 13 - 25sin36.87o = 0 = 2 kN, ↑ ΣFx = 0:
+
-4.5 - Cx + 25cos36.87o = 0
Cx = 15.5 kN, ←
25 kN 13 kN
75 mm 50 mm 200 mm FAB Cy Cx A B C C 36.87o
(58)
25 kN 13 kN
75 mm 50 mm 200 mm
4.5 kN (a) Shear stress
15.63 kN
15.63 kN
Pin C (Double shear)
7.815 kN 7.815 kN
Pin D and E (Single shear)
pin C 13 kN 25 kN
3 4
5
75 mm 50 mm 200 mm
A B
C D E
φ
= 35 mmφ
= 20 mmφ
= 25 mmφ
= 25 mm⇐ =
=
= VC 7.815 kN
⇐ =
=
= 26.48 MPa
) 025 . 0 )( 4 / ( kN 13 2 π A V τ D D D ⇐ = =
= 50.93 MPa
) 025 . 0 )( 4 / ( kN 25 2 π A V τ E E E
(59)
25 kN 13 kN
75 mm 50 mm 200 mm
4.5 kN (b) Normal stress
15.63 kN
Cale AB Plate D
φ
rod = 10 mm AC D E σ ⇐ = =
= 56.7 MPa
) 010 . 0 )( 4 / ( kN 5 . 4 2
π
σ
AB A P D Eφ
= 25 mm13 kN 50 mm
25 kN 50 mm
φ = 25 mm
13 kN
σD
13 kN
13 kN = ⇐
= = MPa 26 ) 01 . 0 )( 05 . 0 ( kN 13 A P σ D
(60)
25 kN 13 kN
75 mm 50 mm 200 mm
4.5 kN
15.63 kN
φ
rod = 35 mm AC
D E
Plate E
⇐ = − = = MPa 100 ) 01 . 0 )( 025 . 0 05 . 0 ( kN 25 A P E
σ
D Eφ
= 25 mm13 kN 50 mm
25 kN 50 mm
φ
= 25 mm50 mm 25 kN 25 kN 50 mm σE 0.05 m 0.02 5 m t = 0
.01 m
(61)
Example 12b
The control arm is subjected to the loading shown. Determine the required diameter of the steel pin at C if the allowable shear stress for the steel is τallow = 55 MPa. Note in the figure that the pin is subjected to double shear.
22 kN 13 kN
3 4
5
75 mm 50 mm 200 mm
A B
(62)
• Internal Shear Force 0 ) 125 . 0 ( 87 . 36 sin 22 ) 075 . 0 ( 13 ) 2 . 0 ( − − = AB F + ΣM
C = 0:
FAB = 13.125 kN
ΣFy = 0:
+ C
y - 13 - 22sin36.87o = 0
Cy = 26.2 kN ΣFx = 0:
+
-13.125 - Cx + 22cos36.87o = 0 Cx = 4.47 kN
22 kN 13 kN
75 mm 50 mm 200 mm FAB Cy Cx 22 kN 13 kN 3 4 5
75 mm 50 mm 200 mm
A B
C
(63)
• Required Area 2 2 6 6 3 mm 242 m 10 6 . 241 10 55 10 29 . 13 = × = × × = = − allow τ V A 2 2 mm 242 ) 2
(d = π
d = 17.55 mm 22 kN
13 kN 3
4 5
75 mm 50 mm 200 mm A B C 22 kN 13 kN
75 mm 50 mm 200 mm 13.125 kN 26.58 kN 26.58 kN 13.29 kN 13.29 kN
(64)
Example 13a
The two members are pinned together at B as shown below. Top views of the pin connections at A and B are also given. If the pins have an allowable shear stress of
τ
allow = 86 MPa , the allowable tensile stress of rod CB is (σ
t)allow = 112 MPa and the allowable bearing stress is (σ
b)allow = 150 MPa, determine to the smallest diameter of pins A and B ,the diameter of rod CB and the thickness of ABnecessary to support the load.
13 kN
A
B
C
1 m 0.6 m
3 4 5 Top view
A
(65)
• Internal Force
13 kN
A B
FBC
1 m 0.6 m
36.87o Ay
Ax
+ Σ M
A = 0: −13(1)+ FBC sin36.87(1.6) =0 FBC = 13.54 kN Σ Fy = 0:
+ Ay −13+13.54sin36.87 =0
Ay = 4.88 kN Σ Fx = 0:
+ − +13.54cos36.87 =0
x
A
(66)
• Diameter of the Pins
13 kN
A B
FBC = 13.54 kN
1 m 0.6 m
36.87o Ay = 4.88 kN
Ax = 10.83 kN
Top view A B 11.88 kN Pin A 11.88 kN
VA = 5.94 kN
VA = 5.94 kN
2 2
3 69.07 mm
m / kN 10 86 kN 94 . 5 = × = = allow A A τ V A 2 2 mm 07 . 69 ) (
4 d A = π
Pin B
FBC = 13.54 kN
VB= 13.54 kN
2 2
3 157.4 mm
m / kN 10 86 kN 54 . 13 = × = = allow B B V A
τ
2 2 mm 4 . 157 ) (4 dB =
(67)
• Diameter of Rod
13.54 kN
13.54 kN
2 2
3 120.9 mm
m / kN 10
112
kN 54 . 13 )
( = × =
=
allow t
BC
σ P A
2 2
mm 9 . 120 )
(
4 dBC = π
dBC = 12.4 mm, Use dBC = 15 mm 13 kN
A B
FBC = 13.54 kN
1 m 0.6 m
36.87o Ay = 4.88 kN
Ax = 10.83 kN 11.88 kN
(68)
tAB
φ
B = 15 mm 20 kNA B
13.54 kN 11.88 kN
tAB
φ
A = 10 mmA B
(σb )A (σb )B
m 00792 . 0 ) 010 . 0 ( 10 88 . 11 10 150 ) ( 3 6 = × = × = AB AB allow b t t A P σ
• Thickness
13 kN
A B
FBC = 13.54 kN
1 m 0.6 m
36.87o Ay = 4.88 kN
Ax = 10.83 kN 11.88 kN m 00602 . 0 ) 015 . 0 ( 10 54 . 13 10 150 ) ( 3 6 = × = × = AB AB allow b t t A P σ mm 8 = AB t use comparison By
(69)
Example 13b
The two members are pinned together at B as shown below. Top views of the pin connections at A and B are also given. If the pins have an allowable shear stress of
τ
allow = 86 MPa, the allowable tensile stress of rod CB is (σ
t)allow = 112 MPa and the allowable bearing stress is (σ
b)allow = 150 MPa, determine to themaximum load P that the beam can supported.
P
A
B
C
1 m 0.6 m
3 4 5 Top view
A
B
φ
= 15 mmφ
= 10 mmt = 8 mm
φ
= 15 mm t = 8 mm(70)
• Internal Force
P
A B
FBC
1 m 0.6 m
36.87o Ay
Ax
+ Σ M
A = 0: −P(1)+ FBC sin36.87(1.6) = 0
FBC = 1.042P (T) 0
87 . 36 sin 042
. 1 ;
0 − + =
= Σ ↑
+ Fy Ay P P
Ay = 0.375P 0
87 . 36 cos 042
. 1 ;
0 − + =
= Σ → +
P A
Fx x
(71)
Pin A (double shear) 0.914P
VA = 0.457P VA = 0.457P
457 . 0 kPa
10
86 × 3 = 2
=
P A
V τ
A A allow
Pin B (single shear)
FBC = 1.042P
VB= 1.042P
042 . 1 kPa
10
86 × 3 = 2
=
P A
V τ
B B allow
P
A B
FBC = 1.042P
1 m 0.6 m
36.87o Ay = 0.375P
0.834P
Top view A
B
0.914P
φ
= 10 mm(72)
1.042P
1.042P
Rod AB
P
A B
FBC = 1.042P
1 m 0.6 m
36.87o Ay = 0.375P
0.834P
0.914P
kN 19
) 015 . 0 )( 4 / (
042 . 1 kPa
10
112 3 2
= = ×
=
P
π
P A
P σ
BC allow
(73)
tAB = 10 mm
φ
B = 15 mm 20 kNA B
FBC = 1.042P
0.914P
tAB = 10 mm
φ
A = 10 mmA B
(σb )A (σb )B
kN 41 . 16 ) 010 . 0 )( 010 . 0 ( 914 . 0 10 150 ) ( 6 = = × = P P A P σ allow b kN 60 . 21 ) 01 . 0 )( 015 . 0 ( 042 . 1 10 150 ) ( 6 = = × = P P A P σ allow b P A B
FBC = 1.042P
1 m 0.6 m
36.87o Ay = 0.375P
0.834P
Top view A
B
0.914P
φ
= 10 mmφ
= 15 mmt = 10 mm
(74)
Example 14
The suspender rod is supported at its end by a fixed-connected circular disk as shown. If the rod passes through a 40-mm diameter hole, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 20 kN load. The allowable normal stress for the rod is σallow = 60 MPa, and the allowable shear stress for the disk is τallow = 35 MPa.
d
40 mm t
20 kN 20 kN
40 mm
(75)
d
40 mm t
20 kN 20 kN
40 mm
τallow
• Diameter of Rod
2 3 m / kN 10 60 kN 20 × = = allow σ P A 2 3 2 m / kN 10 60 kN 20
4 = ×
= π d A
d = 0.0206 m = 20.6 mm
• Thickness of Disk
2 3 m / kN 10 35 kN 20 × = = allow τ V A 2 3 m / kN 10 35 kN 20 ) m 02 . 0 ( 2 × = t π
(76)
Example 15
An axial load on the shaft shown is resisted by the collar at C, which is attached to the shaft and located on the right side of the bearing at B. Determine the largest value of P for the two axial forces at E and F so that the stress in the collar does not exceed an allowable bearing stress at C of (σb)allow = 75 MPa and allowable shearing stress of the adhesive at C of τallow = 100 MPa , and the average normal stress in the shaft does not exceed an allowable tensile stress of (σt)allow = 55 MPa.
A B
P
C E
F
2P
60 mm 20 mm
(77)
A B P C E F 2P
60 mm 20 mm
80 mm 3P Position Load 3P 2P
• Axial Stress
3P P 2P A P allow t) =
(
σ
2 2 3 ) m 03 . 0 ( 3 m / kN 10 55 π P = ו Bearing Stress
A P
allow b) =
(
σ
] ) m 03 . 0 ( ) m 04 . 0 ( [ 3 m / kN 1075 3 2 2 2
π π P − = × B C 60 mm 80 mm 3P
(78)
A B P C E F 2P
60 mm 20 mm
80 mm 3P Position Load 3P 2P shear allow A P =
τ
)] 020 )(. m 04 . 0 ( 2 [ 3 m / kN 10100 3 2
π
P =
×
P3 = 55 kN • Shearing Stress
B C 60 mm 80 mm 3P 20 mm
• Axial Stress
• Shearing Stress
• Bearing Stress
P3 = 55 kN P1 = 51.8 kN
(79)
Example 16
The rigid bar AB shown supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross-sectional area of 1800 mm2 . The18-mm-diameter pins at A and C are subjected to single shear. If the failure stress for the steel and aluminum is (σst)fail = 680 Mpa and (σal)fail = 70 MPa, respectively, and the failure shear stress for each pin is τfail = 900 MPa, determine the largest load P that can be applied to the bar. Apply a factor of safety of F.S = 2.0.
P
Steel 0.75 m
2 m
Aluminum
A B
(80)
+ ΣM
B = 0: P(1.25 m) - FAC(2 m) = 0 ---> FAC = 0.625P 1.25 m P A B FAC FB 0.75 m + ΣM
A = 0: P(0.75 m) - FB(2 m) = 0 ---> FB = 0.375P
P Steel 0.75 m 2 m Aluminum A B C
• Rod AC
AC AC fail st allow st A F S F = = . ) ( )
(
σ
σ
2 3 m) π(0.01 0.625 2 kPa 10 680 P = ×
• Block B
B B fail al allow al A F S F = = . ) ( )
(
σ
σ
2 6 3 m 10 1800 0.375 2 kPa 10 70 − × = × P
(81)
P
Steel 0.75 m
2 m
Aluminum
A B
C
• Pin A or C
pin AC fail
allow
A F S
F =
= .
τ
τ
2 3
m)
π(0.009
0.625 2
kPa 10
900 P
= ×
P3 = 183 kN
Largest load P = 168 kN
1.25 m
P
A B
FAC= 0.625P
FB= 0.375P 0.75 m
• Pin A or C P3 = 183 kN
Summary
• Rod AC P1 = 171 kN • Block B P2 = 168 kN
(82)
STRAIN
!
Deformation
!
Strain
!
Normal Strain
!Shear Strain
(83)
Deformation
Deformed body
θ
A B C
Undeformed body
u(A)
A´
B´ C´
(84)
Strain
x y
z
x y
z
x y
1 εx 1
x y
1
εy
(85)
x y
x y
2
π x
y
xy
γ
xy
γ
εy
1
1 εx
(86)
Example 1
The slender rod shown subjected to an increase of temperature along its axis, which creates a normal strain in the rod of
ε
z = 40(10-3)z1/2, where z is given inmeters. Determine (a) the displacement of the end B of the rod due to the temperature increase, and (b) the average normal strain in the rod.
200 mm z
dz
A
(87)
200 mm z
dz
A
B
(a) the displacement of the end B of the rod due to the temperature increase
dz z
dz'=[40(10−3) 1/2]
∫
−=
∆ 0.2m
0 2 / 1 3 ] ) 10 ( 40
[ z dz
B
↓ =
= −
m z ) 0.00239 3 2 )( 10 ( 40 m 2 . 0 0 2 / 3 3
(b) the average normal strain in the rod.
mm / mm 0119 . 0 mm 200 mm 39 . 2 ' = = ∆ ∆ − ∆ = s s s avg
ε
(88)
Example 2
A force acting on the grip of the lever arm shown causes the arm to rotate clockwise through an angle of θ = 0.002 rad. Determine the average normal strain developed in the wire BC.
B
A
L 2L
(89)
B
A
L 2L
C B´
2L
B
A
L 2L
C
Lcos
θ
Lsin θθ
• Solution
001 . 0 2
002 . 0 2 2
'
= =
= =
−
=
θ
θ
ε
L L CB
CB CB
avg
Since θ is small, the stretch in wire CB, is using Eq.2-6. The average normal strain in the wire is then
, sinθ Lθ L ≈
(90)
Example 3
The plate is deformed into the dashed shape shown. If in this deformed shape horizontal lines on the plate remain horizontal and do not change their length, determine (a) the average normal strain along the side AB, and (b) the average shear strain in the plate relative to the x and y axes.
250 mm
300 mm
3 mm
2 mm B
C x
y
(91)
250 mm 3 mm 2 mm B y A B´ mm 018 . 248 ) 3 ( ) 2 250 (
'= − 2 + 2 =
AB
(a) the average normal strain along the side AB
250 250 018 . 248 ' ) ( = − = − AB AB AB avg AB
ε
mm/mm ) 10 ( 93 .7 −3
− =
(b) the average shear strain in the plate relative to the x and y axes.
250 mm
3 mm 2 mmB
C x y A B´
γ
xy θ´ rad 0121 . 0 ) 2 250 3 (tan 1 =
− = − xy γ ' 2
θ
π
γ
xy = −(92)
Example 4
The plate shown is held in the rigid horizontal guides at its top and bottom, AD and BC. If its right side CD is given a uniform horizontal displacement of 2 mm, determine (a) the average normal strain along the diagonal AC, and (b) the shear strain at E relative to the x, y axes.
150 mm
150 mm A
B C
D
E
2 mm x y
(93)
A B 75 mm C´ D´ E´ 75 mm
76 mm 76 mm
(a) the average normal strain along the diagonal AC m 21213 . 0 ) 150 . 0 ( ) 150 . 0
( 2 + 2 =
= AC m 21355 . 0 ) 152 . 0 ( ) 150 . 0 (
'= 2 + 2 =
AC 150 mm 150 mm A B C D E 2 mm x y mm/mm 00669 . 0 21213 . 0 21213 . 0 21355 . 0 ' ) ( = − = − = AC AC AC avg AC
ε
θ
´(94)
A B 75 mm C´ D´ E´ 75 mm
76 mm 76 mm
θ
´(b) the shear strain at E relative to the x, y axes.
75 76 )
2 ' tan(
θ
= 150 mm 150 mm A B C D E 2 mm x y rad 58404 . 1 ) 759 . 90 ( 180 759 . 90'= o = π o o =
θ rad 0132 . 0 58404 . 1
2 − = −
= π
(95)
M
ECHANICAL
P
ROPERTIES OF
M
ATERIALS
!
Simple Tension Test
!
The Stress-Strain Diagram
!
Stress-Strain Behavior of Ductile and Brittle
Materials
!
Hooke’s Law
!Strain Energy
!Poisson’s Ratio
!
The Shear Stress-Strain Diagram
(96)
400 350 300 250 200 150 100 50
0
εεεε
(mm/mm)
σ
σ
σ
σ
(MPa)
0.10 0.20 0.30 0.40 Upper scale
0.00
0.0020 0.0030
0.0010 0.0040 Lower scale
0.0000
Stress Strain Relationship
L0 d0
P P
(97)
necking
strain
hardening
yielding
elastic
region
The Stress-Strain Diagram
εεεε
σ
σ
σ
σ
σ
plproportional limit
elastic limit
σ
fσ
uσ
f ´fracture stress
ultimate stress
elastic
plastic behavior
true fracture stress
yield stress
(98)
σY
0.002 (0.2% offset)
300 250 200 150 100 50
0
ε
(mm/mm)
σ
(MPa)
0.005 0.010
Offset yield strength for material with no yield points Offset Yield Stress
(99)
(σy)u= 230 MPa (σy)l= 220 MPa
0 A
P =
σ
0 L
δ
ε
=400 350 300 250 200 150 100 50
0
ε (mm/mm)
σ
(MPa)
0.10 0.20 0.30 0.40 Upper scale
0.00
0.0040 0.0000
E = = 200x103 MPa= 200 GPa 0.001
σ
fail = 295 MPaσpl= 200 MPa
σ
u= 390 MPa200
L0 d0
P P
(100)
400 350 300 250 200 150 100 50
0
ε
(mm/mm)
σ
(MPa)
0.10 0.20 0.30 0.40 Upper scale
0.00
0.0020 0.0030
0.0010 0.0040 Lower scale
0.0000
σ
pl= 180 MPaσ
y= 250 MPaσ
u= 360 MPaσ
fail= 310 MPaE = = 200x103 MPa = 200 GPa 0.0005
(1)
NOTE
2 3 2
2 2
) (
) 10
200 ( )
(
r KL
MPa r
KL E cr
× =
=
π
π
σ
KL/r 89 100 125 150 175 200 225
σcr (MPa) 250 197 126 88 64 49 39
Structural steel A 36
E = 200 GPa
400
200
σ , MPa
100
197 MPa 300
88 MPa
49 MPa
σy = 250
2 2
) (
r L K
E cr
π
σ
=42 . 60 2
. 66
) 10 8 )( 5 . 0 ( ) (
3 = ×
=
x r KL
61.9 MPa
6 . 178 5
. 24
) 10 5 )( 7 . 0 ( )
(
3 = ×
=
y r KL
(2)
26
Example 4The aluminum column is fixed at its bottom and is braced at its top by two rods so as to prevent movement at the top along the x axis, If it is assumed to be fixed at its base, determine the largest allowable load P that can be applied. Use a
factor of safety for buckling of F.S. = 3.0. Take Eal = 70 GPa, σy = 215 MPa, A = 7.5(10-3) m2, I
x = 61.3(10-6) m4, Iy = 23.2(10-6) m4.
x
z
y
5 m
P
(3)
5 m
y-y axis buckling x-x axis buckling
5 m
Eal = 70 GPa,
σ
y = 215 MPa, A = 7.5(10-3) m2rx = 90 mm Free (top)
Fixed (bottom) Kx = 2
Ix = 61.3(10-6) m4 I
y = 23.2(10-6) m4
Pinned (top) Fixed (bottom) Ky = 0.7
ry = 50 mm
• Bucking x-x axis
kN m
Pa A
PY =
σ
y = (215×106 )(7.5×10−3 2) =1612 • Yield Stress (σσσσy)kN m
kPa EI
P ) y (70 10 )(23.2 10 ) 1308
(
4 6 6
2 2
= ×
× =
=
−
π
π
• Bucking y-y axis
kN m
m kPa
KL EI P
x x x
cr 425
) 5 2 (
) 10
3 . 61 )( 10
70 ( )
( )
( 2
4 6 6
2 2
2
= ×
× ×
= =
−
π
π
kN kN
S F
P
Pallow cr 141
3 425
. = =
=
x
z
y
5 m
P
(4)
28
NOTEAluminum
E = 70 Gpa σσσσy = 215 MPa
2 3 2
2 2
) (
) 10
70 ( )
(
r KL
MPa r
KL E cr
× =
=
π
π
σ
KL/r
57
75 100 125 150 175 200
σcr (MPa)
215
122.8 69.1 44.2 30.7 22.6 17.3
1 . 111 90
10 5 2 )
(
3
= ×
× =
mm mm r
KL x
70 50
10 5 7 . 0 )
(
3
= ×
× =
mm mm r
KL y
<---occur buckling
56 MPa
111
2 2
) (
r KL
E cr
π
σ
=200
100
0 KL/r
σcr , MPa
100 200
50
50 150
150
57
69 MPa
30 MPa
17 MPa
(5)
Example 5
Determine the maximum load P the column can support before it either begins to buckle or the steel yields. Assume that member BC is pinned at its end for the x-x axis and fixed for y-y axis buckling. Take E = 200 GPa,
σ
y = 250 MPa.2 m
P
4 m
1 m 35 mm
4 5
3 x
y
35 mm 25 mm
A C
(6)
30
2 m P 4 m 1 m 35 mm 4 5 3 x y 35 mm 25 mm A C B F+ ΣM
A = 0: 5 )(4) (2) 0
3
( + =
− F P
* 6 5 − − − = P F ) 035 . 0 )( 025 . 0 ( ) 495 ( ) 10 200 ( 6 5 2 9 2 × =
π
P 495 ] ) 035 . 0 )( 025 . 0 ( ) 035 . 0 )( 025 . 0 ( 12 1 [ ) 5 ( 1 ) ( 2 / 1 3 = = • x r KL 346 ] ) 025 . 0 )( 035 . 0 ( ) 025 . 0 )( 035 . 0 ( 12 1 [ ) 5 ( 5 . 0 ) ( 2 / 1 3 = = • y r KL A r KL E A F cr 2 2 ) (π
σ
= = •P = 8.46 kN < 262.5 kN
Ax Ay kN P kN A P F Y Y Y Y 5 . 262 8 . 218 ) 035 . 025 )(. 10 250 ( 6 5 3 = = × × = = =