STRESS ! !
! ! !
Stress Average Normal Stress in an Axially Loaded Bar Average Shear Stress Allowable Stress Design of Simple Connections
1
Equilibrium of a Deformable Body • Body Force w
FR w(s)
Ax
A
.
C D
Ay
• Support Reaction
s
Dy
2
Table 1 Supports for Coplanar Structures Type of Connection (1)
Idealized Symbol
θ
θ
Reaction θ
Light cable (2)
F
Number of Unknowns One unknown. The reaction is a force that acts in the direction of the cable or link. One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact.
Rollers and rockers (3) Smooth surface
F
One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact.
3
Type of Connection
Idealized Symbol
Reaction
(4)
Fy Fx
Number of Unknowns
Two unknowns. The reactions are two force components.
Smooth pin or hinge (5)
Fy Two unknowns. The reactions are a force and moment.
Fx
Internal pin (6) Fx Fixed support
M
Fy
Three unknowns. The reactions are the moment and the two force components.
4
• Equation of Equilibrium
ΣF=0 Σ MO = 0
Σ Fx = 0
Σ Fy = 0
Σ Fz = 0
Σ Mx = 0
Σ My = 0
Σ Mz = 0
5
• Internal Resultant Loadings section
F2
F3
F1
F4 y
F2 N F1
V
MO x
MO
F3
V N F4
6
Example 1 Determine the resultant internal loadings acting on the cross section at C of the beam shown.
270 N/m
A
B 3m
C
6m
7
• Support Reactions 270 N/m
(1/2)(9)(270) = 1215 N
A MA = 3645 N•m
B 3m
C
6m
Ay = 1215 N
• Internal Loading
180 N/m
540 N
540 N = VC -540(2) = -1080 N•m = MC 135 N 540 N 270 N/m 90 N/m MA = 3645 N•m
0 = NC
B C
2m
4m
180 N/m
0 A 3m
Ay = 1215 N
C
1080 N•m
540 N
8
• Shear and bending moment diagram (1/2)(9)(270) = 1215 N 270 N/m 3645 N•m A B C
3m
1215 N
6m
V (N) 1215 540 +
M (N•m) -
1080 3645
b flanges
C 540 N Side View
1080 N•m
tw
web 540 N
Front View
h
9
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 A
B 200 mm
C 100 mm 50 mm
D 100 mm
50 mm
10
(800 N/m)(0.15 m) = 120 N/m 225 N
• Support Reactions A
Ax
B D 0.275 m
0.125
By
Ay + Σ MA = 0:
-(120)(0.275) + By(0.4) - (225)(0.5) = 0,
By = 363.75 N
Ax = 0
Ax = 0,
Ay - 120 +363.75 - 225 = 0
Ay = -18.75 N,
+ Σ Fx = 0: Σ Fy = 0:
+
• Internal Loading
+ ΣF = 0: x
800(0.05) = 40 N A
18.75 N
+
MC 0.2 m
0.1 m
C
ΣFy = 0:
NC VC
0.025 m
+ ΣMA = 0:
NC = 0
-18.75 - 40 - VC = 0, VC = -58.75 N 18.75(0.25) + 40(0.025) + MC = 0 MC = -5.69 N•m
11
• Internal Loading 800(0.05) = 40 N MC = -5.69 N•m
A 0.2 m 18.75 N
C
NC = 0 VC = -58.75 N
0.025 m 800(0.05) = 40 N 5.69 N•m
A
0 0.2 m
18.75 N
C
58.75 N
0.025 m
12
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.
1m
0.15 m
2m
0.5 m B
A
C
0.15 m
D
W
13
1m
0.15 m
2m
0.5 m B
A
0.15 m
C
D
W W = 2 kN 1m
0.15 m A
W = 2 kN
TC = 2 kN
+ Σ F = 0: x
NC
C VC
MC
+
Σ Fy = 0:
2 + NC = 0 NC = -2 kN -2 - VC = 0, VC = -2 kN
+ Σ MC = 0:
MC - 2(0.15) + 2(1.15) = 0 MC = -2 kN•m
14
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. B
C
6.5 kN 1m G
E
A D 4 kN/m 0.6 m 0.6 m
2m
15
• Support Reactions
Two - force
B
C
6.5 kN
FBC
Ey G
E
1m Ex
A D 4 kN/m 0.6 m 0.6 m
2/3 m
(1/2)(2)(4) = 4 kN (2/3)(2) = 4/3 m
+ Σ ME = 0:
4(4/3) + 6.5(3.2) - FBC(1) = 0,
FBC = 26.1 kN
+ Σ F = 0: x
26.13 + Ex = 0
Ex = -26.1 kN,
-6.5 - 4 + Ey = 0
Ey = 10.5 kN
+
Σ Fy = 0:
16
• Internal loadings acting on the cross section at G B FBC = 26.13 kN 34.02 kN = FBA 6.5 kN
50.19o
FBD = -21.78 kN FAB = 34.02 kN +
39.81o A G 0.6 m VC
Joint B ΣFx = 0:
MC NC +
ΣFy = 0:
-FBA sin50.19o + 26.13 = 0 FBA = 34.0 kN, (T) -FBA cos50.19o -FBD = 0 FBD = -21.8 kN, (C)
Member AG + ΣMG = 0: + ΣF = 0: x +
ΣFy = 0:
MG -34.02 sin39.81o(0.6) + 6.5(0.6) = 0
MG = 9.17 kN•m
34.02 cos39.81o + NG = 0
NG = -26.1 kN
-6.5 + 34.02 sin39.81o - VG = 0
VG = 15.3 kN
17
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. z
C x
0.75 m 50 N
70 N•m
A
B
0.5 m D
1.25 m
y
18
• Free-Body Diagram
z
(MB)z
(FB)z (FB)y WBD= 9.81 N B 0.25 m (FB)x (MB)x WAD D 50 N 0.625 m y
(MB)y
x 70 N•m
A
0.625 m
WBD = (2 kg/m)(0.5 m)(9.81 N/kg) = 9.81 N WAD = (2 kg/m)(1.25 m)(9.81 N/kg) = 24.52 N
• Equilibrium of Equilibrium
Σ Fx = 0:
(FB)x = 0
Σ Fy = 0:
(FB)y = 0
Σ Fz = 0:
(FB)z - 9.81 -24.525 -50 = 0,
(FB)z = 84.3 N
Σ (MB)x = 0:
(MB)x + 70 - 50(0.5) - 24.525(0.5) - 9.81(0.25) = 0,
(MB)x = -30.3 N•m
Σ (MB)y = 0:
(MB)y + 24.525(0.625) + 50(1.25) = 0,
(MB)y = -77.8 N•m
Σ (MB)x = 0:
(MB)z = 0
19
• Free-Body Diagram (FB)x = 0
(MB)x = -30.3 N•m
(FB)y = 0
(MB)y = -77.8 N•m
(FB)z = 84.3 N
(MB)z = 0 z
30.3 N•m x 70 N•m
77.8 N•m 84.3 N 9.81 N B 0.25 m 24.5 N
50 N A
D
0.625 m y 0.625 m
20
Stress
z
+z Face +y Face
y
+x Face x
21
z σz
By compatibility,
τzy
τzx
-x Face
σ'x
τ'xy τ'yx
σ'y
τxz
τ'yz
-y Face
σx
σz = σ’z τxy = τ'xy
σy
τyx
τyx = τ'yx y
τ'zx
τ'zy
-z Face x
σy = σ'y
τyz
τ'xz τxy
σx = σ'x
τ'xy∆y∆z
σ'z σ'y∆x∆z
x
τzx = τ’xz σ'x∆y∆z
τ'yx∆x∆z
y Top View
σx∆y∆z
τyx∆x∆z
σy∆x∆z
τxy∆y∆z
22
σ'y∆x∆z
τ'yx∆x∆z
O
τ'xy∆y∆z
σ'x∆y∆z y
O´ τyx∆x∆z
x + ΣF = 0: y
σx∆y∆z
σy∆x∆z
τxy∆y∆z
- σ'y∆x∆z + σy∆x∆z = 0
σ'y = σy + ΣF = 0: x
σx∆y∆z - σ'x∆y∆z = 0 σx = σ'x
+ ΣMO´ = 0:
(τxy∆y∆z)(∆x) - (τyx∆x∆z)(∆y) = 0
τxy = τyx
23
z
τzx∆x∆y
z
σz∆x∆y τ ∆x∆y zy τyz∆x∆z σy∆x∆z τyx∆x∆z
∆z τxz∆y∆z τxy∆y∆z σx∆y∆z
∆x
σz∆x∆y σ'x∆y∆z σy∆x∆z
σ'y∆x∆z
y
σx∆y∆z
∆y
y
σ'z∆x∆y
x
x z
τ'yz∆x∆z
z
τzy∆x∆y τyz∆x∆z
τ'yx∆x∆z
τ'zy∆x∆y
x
τxy∆y∆z
τzx∆x∆y
τyx∆x∆z τxz∆y∆z
y
y x
z
τ'xy∆y∆z
x
τ'zx∆x∆y
τ'xz∆y∆z y
24
Average Normal Stress in an Axially Loaded Bar • Assumptions The material must be - Homogeneous material - Isotropic material P
Internal force Cross-sectional area
External force P
P
25
z P
P
∆F = σ∆A
σ
x
σ
x
σ
σ
y
y P + ↑ FRz = ΣFz ;
P
∫ dF = ∫ σdA A
P = σA
σ=
P A
26
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.
A
B 9 kN
C 4 kN
12 kN
D 22 kN
9 kN 35 mm
4 kN
27
A
B 9 kN
C 4 kN
D
12 kN
22 kN 9 kN
4 kN
35 mm P (kN) 30
22
12
x 10 mm
30 kN 35 mm
σ max = σ BC =
σ
PBC 30 kN = = 85 . 7 MPa A ( 0 . 035 m )( 0 . 01 m )
28
A
B 9 kN
C 4 kN
D
12 kN
22 kN 9 kN
4 kN
35 mm σ (MPa) 85.7
62.9
34.3
x 10 mm
30 kN 35 mm
σ max = σ BC =
σ
PBC 30 kN = = 85 . 7 MPa A ( 0 . 035 m )( 0 . 01 m )
29
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. C A
5
3
4 60o
B
30
• Average Normal Stress C
• Internal Loading y
A d = 10 mm 5 3 4
FBC
FBA 5
3
B 4
60o
60o
B
x
σ BC =
FBC 395 . 2 N = = 7 . 86 MPa 2 ABC π ( 0 . 004 m )
σ BA =
FBA 632 . 4 N = = 8 . 05 MPa ABA π ( 0 . 005 m ) 2
80(9.81) = 784.8 N + → ΣFx = 0;
+ ↑ ΣFy = 0;
4 FBC ( ) − FBA cos 60o = 0 5 3 FBC ( ) + FBA sin 60o − 784.8 = 0 5
FBC = 395.2 N,
d = 8 mm
The average normal stress distribution acting over a cross section of rod AB. 8.05 MPa
8.05 MPa
FBA = 632.4 N 8.05 MPa
632.4 N
31
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.
B 3 kN x A C 200 mm
32
FAB
3 kN x
A C 200 mm • Internal Loading + ↑ ΣFy = 0;
+ Σ MA = 0:
FC FAB + FC − 3 = 0
− − − −(1)
-3(x) + FC(200) = 0
• Average Normal Stress
σ=
------(2)
FAB F = C 400 650
FC = 1.625 FAB
− − − −(3)
Substituting (3) into (1), solving for FAB, the solving for FC, we obtain FAB = 1.143 kN
FC = 1.857 kN
The position of the applied load is determined from (2);
x = 124 mm
33
Connections • Simple Shear
P A C D
B P
P
A τavg
A V
V
V
τ avg =
V A
34
P
• Single Shear
P
P
P
P
P
P
P
Top View P
Top View P
P
P Side View
Side View P
V=P
P
V=P
35
P
• Double Shear
P/2
P
P/2 P/2
P/2
P
P
P
P
Top View P
Top View P
P
P
Side View
Side View P
V = P/2 V = P/2
P
V = P/2 V = P/2
36
Normal Stress: Compression and Tension Load • Tension Load
P P
a b
P
b
P´ b
t
d
a
P
P
A = bt t b
σ @ a −a =
Section a-a
P bt
σ @ b −b =
d
A = (b-d)t
b
P (b − d )t
t
Section b-b
37
• Bearing Stress
t
b´
P
d
P
b
b´ t
P/2 P P/2
d
Abearing = dt
Bearing Stress
σ bearing =
P dt
38
σ normal =
P (b − d )t
t
dθ θ
σ bearing = + ΣF = 0: x
P−∫
90°
−90°
b
d
P dt d 2
σ b ( )t cosθdθ = 0
d ( )tσ b sin θdθ 2
90° −90°
=P
σ bearing =
P td sin 90o
σ bearing =
P td
− − − − −*
39
• Axial Force Diagram for Compression Load on Plate
P
b
P
d
t Tension
-P Compression
-P
− Normal
Stress :
− Bearing
Stress
−P , compression (bd )t P σ bearing = dt
σ=
40
• Axial Force Diagram for Tension Load on Plate a d
b
P
P
t Tension P
P
Compression − Normal − Bearing − Shearing
Stress : max
σ=
+P , (b − d )t P = dt
Stress
σ bearing
Stress
P τ= 2at
tension
τ P/2
P t
τ P/2 a
41
• Shearing Stress on pin P
Single Shear
P
P
P
P
P
P
P
τ=
P
P P
P A
P
42
P
Double Shear
P/2 P/2
P/2
P
P/2
P/2
P/2
P/2
P/2 P
τ =
P /2 A
τ =
P /2 A
P/2
P/2 P
P P/2
P/2
P/2
P/2
43
Pure Shear z
Section plane
τzy ∆z y ∆y x
∆x
τyz
τyz
τzy
Pure shear
τzy = τyz
44
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.
a
b
20 mm
800 N b a
30o
20 mm
45
a
40 mm
b
20 mm
800 N b a
30o
20 mm
(a) section plane a-a a 800 N
N = 800 N a
σ =
N 800 N = = 500 kPa A ( 0 . 04 m )( 0 . 04 m )
τ avg = 0
500 kPa
46
a
40 mm
20 mm
b
800 N b
30o
a
(b) section plane b-b
20 mm
V = 800 cos30o = 692.82 N 30o
800 N
b 30o x 40 mm 30o
sin 30 o =
τavg σ State of Stress @ b-b b
σ =
b 800 N
N = 800sin30o = 400 N 40 x
; x = 80 mm
N 400 N = = 125 kPa A ( 0 . 04 m )( 0 . 08 m )
b 216.51 kPa V 692 . 82 N τ = = = 216 . 51 kPa avg 125 kPa A ( 0 . 04 m )( 0 . 08 m )
47
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.
c b
20 mm d
a
40 mm
5 kN
48
• Average shear stress in the rod at the wall
c b
20 mm d
a
5 kN
V = 5 kN τ avg =
40 mm d = 10 mm
V 5 kN = = 63 . 7 MPa A π ( 0 . 005 m ) 2
5 kN 63.7 MPa
5 kN
• Average shear stress along the two shaded plane V = 2.5 kN
V = 2.5 kN c τ avg =
b d a
2 . 5 kN V = = 3 . 12 MPa A ( 0 . 04 m )( 0 . 02 m )
3.12 MPa 5 kN
5 kN
49
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. 5 4 3
3 kN
25 mm 40 mm
50 mm 80 mm
50
• The average compressive stress along the areas of contact defined by AB and BC : 3 kN 5 4 3 3 kN 5 4 3 25 mm 40 mm
3 kN
50 mm 80 mm
V = 1.8 kN
FAB = 3(3/5) = 1.8 kN
5 4 3
FBC= 3(4/5) = 2.4 kN
σ AB =
1 . 8 kN FAB = = 1800 kPa = 1 . 8 MPa 1.8 MPa AAB ( 0 . 025 m )( 0 . 040 m )
σ BC =
FBC 2 . 4 kN = = 1200 kPa = 1 . 2 MPa ABC ( 0 . 050 m )( 0 . 040 m)
1. 2 MPa
51
• The average shear stress along the horizontal plane defined by EDB : 3 kN 5 4 3 3 kN 5 4 3 25 mm 50 mm
40 mm
80 mm
V = 1.8 kN τ avg =
FAB = 3(3/5) = 1.8 kN FBC= 3(4/5) = 2.4 kN
V 1 . 8 kN = = 562 . 5 kPa A ( 0 . 04 m)( 0 . 08 m) = 0 . 562 MPa
1.8 kN 0.56 MPa
52
Allowable Stress F .S =
Pfail
F .S =
Pallow
σ fail σ allow
F .S =
τ fail τ allow
6. Design of Simple Connections A=
P
A=
σ allow
V
τ allow
• Cross-Sectional Area of a Tension Member P
P
τallow
a a
P
A=
P
σ allow
53
• Cross-Sectional Area of a Connector Subjected to Shear P
P τallow A=
V=P
P
P
σ allow
P
• Required Area to Resist Bearing
(σb)allow
A=
P (σ b ) allow
54
• Required Area to Resist Shear by Axial Load Uniform shear stress (τallow) d P
l=
P
P
τ allowπd
55
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.
B
φ = 10 mm φ = 25 mm D
A
φ = 25 mm
50 mm
200 mm
50 mm D
C pin C
E
φ = 20 mm
13 kN
E
13 kN 75 mm 50 mm
25 kN
3
5 4
25 kN
56
B
FAB
A
200 mm
200 mm C 3 13 kN 75 mm 50 mm
Cx
5 4
25 kN
36.87o 13 kN
Cy
75 mm 50 mm
25 kN
• Reaction C + ΣMC = 0:
FAB (0.2) + 13(0.075) − 25 sin 36.87(0.125) = 0
FAB = 4.5 kN , ← + ΣF = 0: x
-4.5 - Cx + 25cos36.87o = 0 Cx = 15.5 kN, ←
+ Cx Cy
C
ΣFy = 0:
Cy + 13 - 25sin36.87o = 0 Cy = 2 kN, ↑ C = (15 . 5 ) 2 + ( 2 ) 2 = 15 . 63 kN
57
B
φ = 35 mm A
(a) Shear stress
200 mm
200 mm
φ = 25 mm φ = 25 mm
D
C pin C
4.5 kN
E
3
φ = 20 mm
13 kN 75 mm 50 mm
Pin C (Double shear)
5 4
25 kN
13 kN
75 mm 50 mm
25 kN
Pin D and E (Single shear)
15.63 kN 7.815 kN
τD =
13 kN VD = = 26 . 48 MPa ⇐ 2 AD (π / 4 )( 0 . 025 )
τE =
VE 25 kN = = 50 . 93 MPa ⇐ 2 AE ( π / 4 )( 0 . 025 )
7.815 kN τC =
15.63 kN
VC 7 . 815 kN = = 24 . 88 MPa ⇐ AC ( π / 4 )( 0 . 02 ) 2
58
(b) Normal stress 4.5 kN
φrod = 10 mm φ = 25 mm
D
E
A
200 mm φ = 25 mm D
50 mm
C 50 mm
25 kN
15.63 kN
13 kN
Cale AB σ
σ AB
E
4.5 kN P = = = 56.7 MPa ⇐ AAB (π / 4)(0.010) 2
13 kN
75 mm 50 mm Plate D 13 kN
13 kN
σD
13 kN t = 0.01 m 0.05 m
σD =
25 kN
P A
13 kN ( 0 . 05 )( 0 . 01 ) = 26 MPa ⇐ =
59
4.5 kN
φrod = 35 mm φ = 25 mm
D
E
A
200 mm
φ = 25 mm D
50 mm
C 50 mm
25 kN
15.63 kN
13 kN
σE
50 mm
25 kN
0.0 1
P A
25 kN (0.05 − 0.025)(0.01) = 100 MPa ⇐ =
25
m
t=
σE =
25 kN
0. 0
25 kN
5m
50 mm
0. 0
25 kN
13 kN
75 mm 50 mm
m
Plate E
E
60
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.
B
A
200 mm C 3 13 kN 75 mm 50 mm
5 4
22 kN
61
B
FAB
A
200 mm
200 mm C 3 13 kN 75 mm 50 mm
Cx
5 4
22 kN
13 kN
Cy
75 mm 50 mm
22 kN
• Internal Shear Force + ΣMC = 0:
FAB (0.2) − 13(0.075) − 22 sin 36.87(0.125) = 0
FAB = 13.125 kN + ΣF = 0: x
-13.125 - Cx + 22cos36.87o = 0 Cx = 4.47 kN
+ Cx Cy
C
ΣFy = 0:
Cy - 13 - 22sin36.87o = 0 Cy = 26.2 kN
C = ( 4 . 47 ) 2 + ( 26 . 2 ) 2 = 26 . 58 kN
62
B
13.125 kN
A
200 mm
200 mm C 3 13 kN 75 mm 50 mm
5 4
22 kN
26.58 kN
13 kN
75 mm 50 mm
22 kN
• Required Area A=
26.58 kN
V τ allow
13 . 29 × 10 3 −6 2 2 = = 241 . 6 × 10 m = 242 mm 55 × 10 6 d π ( ) 2 = 242 mm 2 2
13.29 kN 13.29 kN
d = 17.55 mm Use d = 20 mm
63
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 τallow = 86 MPa , the allowable tensile stress of rod CB is (σt)allow = 112 MPa and the allowable bearing stress is (σb)allow = 150 MPa, determine to the smallest diameter of pins A and B ,the diameter of rod CB and the thickness of AB necessary to support the load. A C
Top view 13 kN
5
3 4
A B 1m
0.6 m
B
64
• Internal Force 13 kN
Ay
FBC B
A
Ax
1m
36.87o
0.6 m
+ Σ MA = 0: − 13(1) + FBC sin 36.87(1.6) = 0 FBC = 13.54 kN +
Σ Fy = 0: Ay − 13 + 13.54 sin 36.87 = 0 Ay = 4.88 kN
+ Σ F = 0: x
− Ax + 13.54 cos 36.87 = 0
Ax = 10.83 kN
65
• Diameter of the Pins Ay = 4.88 kN
11.88 kN
13 kN
FBC = 13.54 kN B
A Ax = 10.83 kN
1m
36.87o B
0.6 m
A Top view Pin B
Pin A
FBC = 13.54 kN
11.88 kN VA = 5.94 kN VA = 5.94 kN AA =
VA
5 . 94 kN 2 = = 69 . 07 mm 86 × 10 3 kN / m 2
τ allow π (d A ) 2 = 69 . 07 mm 2 4
dA = 9.38 mm, Use dA = 10 mm
VB= 13.54 kN AB =
π 4
VB
τ allow
=
13.54 kN = 157.4 mm 2 3 2 86 × 10 kN / m
( d B ) 2 = 157.4 mm 2
dB = 14.16 mm, Use dB = 15 mm
66
• Diameter of Rod 11.88 kN
Ay = 4.88 kN
13 kN
FBC = 13.54 kN B
A Ax = 10.83 kN
1m
36.87o
0.6 m
13.54 kN
13.54 kN
ABC =
P 13 . 54 kN = = 120 . 9 mm 2 3 2 (σ t ) allow 112 × 10 kN / m π (d BC ) 2 = 120 . 9 mm 2 4
dBC = 12.4 mm, Use dBC = 15 mm
67
• Thickness 13 kN
Ay = 4.88 kN
11.88 kN
FBC = 13.54 kN B
A Ax = 10.83 kN
φA = 10 mm P (σ b ) allow = A 11 . 88 × 10 3 6 150 × 10 = ( 0 . 010 )t AB
1m
0.6 m
20 kN
tAB A
36.87o
φB = 15 mm B tAB
11.88 kN
13.54 kN
A
B
(σb )A
(σb )B
t AB = 0 . 00792 m
P A 13 . 54 × 10 3 6 150 × 10 = ( 0 . 015 )t AB (σ b ) allow =
t AB = 0 . 00602 m
By comparison use t AB = 8 mm
68
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 τallow = 86 MPa, the allowable tensile stress of rod CB is (σt)allow = 112 MPa and the allowable bearing stress is (σb)allow = 150 MPa, determine to the maximum load P that the beam can supported.
φ = 10 mm
A
φ = 15 mm C
Top view t = 8 mm P
5
3 4
A B 1m
0.6 m
B
φ = 15 mm
t = 8 mm
69
• Internal Force P
Ay Ax
FBC B
A 1m
36.87o
0.6 m
+ Σ MA = 0: − P(1) + FBC sin 36.87(1.6) = 0 FBC = 1.042P (T) + ↑ ΣFy = 0;
Ay − P + 1.042 P sin 36.87 = 0
Ay = 0.375P + → ΣFx = 0; − Ax + 1.042 P cos 36.87 = 0
Ax = 0.834 kN
70
Ay = 0.375P
0.914P
P
FBC = 1.042P B
A 0.834P
1m φ = 10 mm
A
36.87o B
0.6 m
φ = 15 mm
Top view Pin A (double shear)
Pin B (single shear)
0.914P
FBC = 1.042P
VA = 0.457P VA = 0.457P τ allow = 86 × 10 3 kPa =
VA AA
0 . 457 P ( π / 4 )( 0 . 01 ) 2
P = 14 . 78 kN
VB= 1.042P τ allow = 86 × 10 3 kPa =
VB AB
1 . 042 P (π / 4 )( 0 . 015 ) 2
P = 14 . 58 kN
71
0.914P
Ay = 0.375P
P
FBC = 1.042P B
A 0.834P
Rod AB
1m
0.6 m
φrod = 15 mm
1.042P σ allow =
1.042P
36.87o
112 × 10 3 kPa =
P ABC 1 . 042 P ( π / 4 )( 0 . 015 ) 2
P = 19 kN
72
P
Ay = 0.375P
0.914P
FBC = 1.042P B
A 0.834P A
1m φ = 10 mm
36.87o B
0.6 m
φ = 15 mm
t = 10 mm
Top view t = 10 mm
φA = 10 mm (σ b ) allow
P = A
150 × 10 6 =
tAB = 10 mm A
20 kN
φB = 15 mm B tAB = 10 mm
0.914P
0 . 914 P ( 0 . 010 )( 0 . 010 )
FBC = 1.042P
A
B
(σb )A
(σb )B
P = 16 . 41 kN
By comparison all P = 14 . 58 kN ⇐
(σ b ) allow =
P A
150 × 10 6 =
1 . 042 P ( 0 . 015 )( 0 . 01 )
P = 21 . 60 kN
73
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. 40 mm
t 40 mm τallow
d
20 kN
20 kN
74
40 mm
t
40 mm τallow
d
20 kN
20 kN • Diameter of Rod A=
A=
P σ allow
π d 4
2
=
20 kN 60 × 10 3 kN / m 2
=
20 kN 60 × 10 3 kN / m 2
d = 0.0206 m = 20.6 mm
• Thickness of Disk A=
V τ allow
=
20 kN 35 × 10 3 kN / m 2
2π ( 0 . 02 m )t =
20 kN 35 × 10 3 kN / m 2
t = 4.55x10-3 m = 4.55 mm
75
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 P
2P F
E
60 mm
B
20 mm C
80 mm
76
A P
2P F
60 mm
B
20 mm 3P C
E
80 mm
Load 3P 2P Position • Bearing Stress
• Axial Stress P
2P
(σ t ) allow
P = A
3P 55 × 10 kN / m = π ( 0 . 03 m ) 2 3
2
P1 = 51.8 kN
B 3P
3P
60 mm C
(σ b ) allow =
75 × 10 3 kN / m 2 =
80 mm
P A
3P [π ( 0 . 04 m ) 2 − π ( 0 . 03 m ) 2 ]
P2 = 55 kN
77
A P
2P F
60 mm
B
20 mm 3P C
E
80 mm
Load 3P 2P Position • Shearing Stress B 3P
P1 = 51.8 kN
60 mm C
τ allow =
• Axial Stress
P
80 mm 20 mm
Ashear
100 × 10 3 kN / m 2 =
3P [ 2π ( 0 . 04 m )(. 020 )]
• Bearing Stress P2 = 55 kN • Shearing Stress P3 = 55 kN
P3 = 55 kN The largest load that can be applied to the shaft is P = 51.8 kN.
78
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-mmdiameter 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.
C P Steel 0.75 m B
A
Aluminum 2m
79
C P Steel 0.75 m
P
FAC B
A
B
A
Aluminum
0.75 m
1.25 m FB
2m + ΣMB = 0:
P(1.25 m) - FAC(2 m) = 0
---------> FAC = 0.625P
+ ΣMA = 0:
P(0.75 m) - FB(2 m) = 0
----------> FB = 0.375P
• Rod AC (σ st ) allow =
• Block B (σ st ) fail F .S
F = AC AAC
680 × 10 3 kPa 0.625 P = 2 π(0.01 m) 2
P1 = 171 kN
(σ al ) allow =
(σ al ) fail F .S
=
FB AB
70 × 10 3 kPa 0.375 P = 2 1800 × 10 −6 m 2
P2 = 168 kN
80
C P Steel 0.75 m
FAC= 0.625P P B
A
B
A
Aluminum
0.75 m
FB= 0.375P
2m Summary
• Pin A or C
τ allow
1.25 m
τ fail
F = = AC F .S Apin
900 × 10 kPa 0.625 P = 2 π(0.009 m) 2 3
P3 = 183 kN
• Rod AC
P1 = 171 kN
• Block B
P2 = 168 kN
• Pin A or C P3 = 183 kN Largest load P = 168 kN
81
