Fundamentals of Machine Elements_Ch05

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


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.


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


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


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


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


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.


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


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


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.


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.


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


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


PA

BC

l

y

x

Q

h

Example 5.10

Figure 5.10 Cantilevered bar with 90° bend acted upon by horizontal force at free end.

Fundamentals of Machine Elements_Ch05

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