fundamentals of machine elements_ch05
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Fundamentals of Machine Elements_Ch05TRANSCRIPT
Hamrock • Fundamentals of Machine Elements
Chapter 5: Deformation
Knowing is not enough; we must apply.
Willing is not enough; we must do.
Johann Wolfgang von Goethe
Hamrock • Fundamentals of Machine Elements
x
y
P
P
l
V M
Example 5.1
Figure 5.1 Cantilevered beam with concentrated force applied at free end. Used in Example 5.1.
Hamrock • Fundamentals of Machine Elements
Deflection by Singularity Functions
1. Draw a free-body diagram showing the forces acting on the system.
2. Use force and moment equilibrium to establish reaction forces acting on the system.
3. Write an expression for the load-intensity function for all the loads acting on the system while makinguse of Table 2.2.
4. Integrate the negative load-intensity function to give the shear force and then integrate the negativeshear force to obtain the moment.
5. Make use of Eq. (5.9) to describe the deflection at any value.
6. Plot the following as a function of x:
(a) Shear
(b) Moment
(c) Slope
(d) Deflection
Hamrock • Fundamentals of Machine Elements
x
a b
y
l
Pb___l
PRA = RB =Pa___l
x
x
a x – a
y
P
V M
RA =Pb___l
(a)
(b)
Figure 5.2 Free-body diagram of force anywhere between simply-supported ends. (a) Complete bar; (b) portion of bar.
Force on Simply Supported Beam
Hamrock • Fundamentals of Machine Elements
y
a
A B
C
b
l
w0
x–yl
RA = w0b
(a)
(b)
(c)
a b
w0
MA = w0b a +b_2( )
a
x
V M
x – a
w0
w0b
w0b a +b_2( )
Figure 5.3 Cantilevered bar with unit step distribution over part of bar. (a) Loads and deflection acting on cantilevered bar; (b) free-body diagram of forces and moments acting on entire bar; (c) free-body diagram of forces and moments acting on portion of bar.
Cantilever with Step Load
Hamrock • Fundamentals of Machine Elements
y
a
A C
b
l
RC
(a)
(b)
MC
B
P
x
RA
a b
P
V
(c)
M
RA
a
x
x – a
P
Figure 5.4 Cantilevered bar with other end simply-supported and with concentrated force acting anywhere along bar. (a) Sketch of assembly; (b) free-body diagram of entire bar; (c) free-body diagram of part of bar.
Cantilever with Concentrated Force
Hamrock • Fundamentals of Machine Elements
Deflection for any x
y = – ( x – a 3 – x3 + 3x2a)P
–—6EI
w0–—EI
bx3–—6
bx2–—2
b–2
1—24
x
bP
y
Type of loading
Concentrated load at any x
Unit step distribution overpart or all of bar
Moment applied to free end
(a)
ly
l
y =
Mx2–——
2EIy = –
– x – a 4
a
x
b
M
y
ly
l
a
x
y
ly
l
w0 +– a
P–—6EI
b–l
y = x 3 – x – a 3 + 3a2x – 2alx –
y
a b
l
P
P
x
w0b–—––24lEI
b–2
b–2
y = –
+ x b3 + 6bc2 + 4b2c + 4c3 – 4l2 c +
4 c + x 3 – x – a 4 –
l–b
(b)
b–l
P a–l
y
a c
l
b
c +
x
b–2
a3x–––
l
x – a – b 4
w0
w0b–––
la +
b–2
w0b–––
l
Beam Deflections
Table 5.1 Deflection for three different situations when one end is fixed and one end is free and two different situations of simply supported ends.
Hamrock • Fundamentals of Machine Elements
y
a
AB
P
l
C
xyl
(a)
Mo
y
a
P
x
x
yl, 1
yl, 2
(b)
(c)
y
Mo
l
Figure 5.5 Bar fixed at one end and free at other with moment applied to free end and concentrated force at any distance from free end. (a) Complete assembly; (b) free-body diagram showing effect of concentrated force; (c) free-body diagram showing effect of moment.
Cantilever with Moment
Hamrock • Fundamentals of Machine Elements
dx
σz
dz
dy
dx
dz
dy
τγγdz
Figure 5.6 Element subjected to normal stress.
Figure 5.7 Element subjected to shear stress.
Stress Elements
Hamrock • Fundamentals of Machine Elements
Strain energy forspecial case whereall three factors are General expression
Loading type Factors involved constant with x for strain energy
Axial P,E,A U =P 2l
2EAU =
l∫0
P 2
2EAdx
Bending M,E, I U =M2l
2EIU =
l∫0
M2
2EIdx
Torson T,G, J U =T 2l
2GJU =
l∫0
T 2
2GJdx
Transverse shear V,G,A U =3V 2l
5GAU =
l∫0
3V 2
5GAdx
(rectangular section)
Strain Energy
Table 5.2 Strain energy for four types of loading.
Hamrock • Fundamentals of Machine Elements
Castigliano’s Theorem
yi =!U
!Qi
∂Qi(5.30)
The load Qi is applied to a particular point of deformation and therefore is not a function of x. Thus, it ispermissible to take the derivative with respect to Qi before integrating for the general expressions for the
The following procedure is to be employed in using Castigliano’s theorem:
1. Obtain an expression for the total strain energy including
(a) Loads (P , M , T , V ) acting on the object (use Table 5.2)(b) A fictitious force Q acting at the point and in the direction of the desired deflection
2. Obtain deflection from y = ∂U/∂Q.
3. If Q is fictitious, set Q = 0 and solve the resulting equation.
Hamrock • Fundamentals of Machine Elements
y
l
l
b
B CA
P
(a)
x
Q
bP
(b)
x
Figure 5.8 Cantilevered bar with concentrated force acting distance b from free end. (a) Coordinate system and important points shown; (b) fictitious force shown along with concentrated force.
Cantilever with Concentrated Force
Hamrock • Fundamentals of Machine Elements
P
A
(a)
θθ
l, A1, E1
l, A2, E2
P
(b)
P2
P1
A Qθθ
Figure 5.9 Linkage system arrangement. (a) Entire assembly; (b) free-body diagram of forces acting at point A.
Linkage System