Machinery''''s Handbook 27th Episode 1 Part 4 pps

If moments of inertia, I, are known for a plane area with respect to both x and y axes, then the polar moment for the z axis may be calculated using the equation, A table of formulas f

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RADIUS OF GYRATION 235

Sphere:

Hollow Sphere and Thin Spherical Shell:

Ellipsoid and Paraboloid:

Center and Radius of Oscillation.—If a body oscillates about a horizontal axis which

does not pass through its center of gravity, there will be a point on the line drawn from thecenter of gravity, perpendicular to the axis, the motion of which will be the same as if the

whole mass were concentrated at that point This point is called the center of oscillation The radius of oscillation is the distance between the center of oscillation and the point of

suspension In a straight line, or in a bar of small diameter, suspended at one end and lating about it, the center of oscillation is at two-thirds the length of the rod from the end bywhich it is suspended

oscil-When the vibrations are perpendicular to the plane of the figure, and the figure is pended by the vertex of an angle or its uppermost point, the radius of oscillation of an isos-celes triangle is equal to 3⁄4 of the height of the triangle; of a circle, 5⁄8 of the diameter; of aparabola, 5⁄7 of the height

sus-If the vibrations are in the plane of the figure, then the radius of oscillation of a circleequals 3⁄4 of the diameter; of a rectangle, suspended at the vertex of one angle, 2⁄3 of the diag-onal

Center of Percussion.—For a body that moves without rotation, the resultant of all the

forces acting on the body passes through the center of gravity On the other hand, for a body

that rotates about some fixed axis, the resultant of all the forces acting on it does not pass through the center of gravity of the body but through a point called the center of percus-

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sion The center of percussion is useful in determining the position of the resultant in

mechanics problems involving angular acceleration of bodies about a fixed axis

Finding the Center of Percussion when the Radius of Gyration and the Location of the Center of Gravity are Known: The center of percussion lies on a line drawn through the

center of rotation and the center of gravity The distance from the axis of rotation to the ter of percussion may be calculated from the following formula

cen-in which q = distance from the axis of rotation to the center of percussion; k o = the radius of

gyration of the body with respect to the axis of rotation; and r = the distance from the axis

of rotation to the center of gravity of the body

Moment of Inertia

An important property of areas and solid bodies is the moment of inertia Standard mulas are derived by multiplying elementary particles of area or mass by the squares oftheir distances from reference axes Moments of inertia, therefore, depend on the location

for-of reference axes Values are minimum when these axes pass through the centers for-of ity

grav-Three kinds of moments of inertia occur in engineering formulas:

1) Moments of inertia of plane area, I, in which the axis is in the plane of the area, are

found in formulas for calculating deflections and stresses in beams When dimensions are

given in inches, the units of I are inches4 A table of formulas for calculating the I of

com-mon areas can be found beginning on page238

2) Polar moments of inertia of plane areas, J, in which the axis is at right angles to the

plane of the area, occur in formulas for the torsional strength of shafting When dimensions

are given in inches, the units of J are inches4 If moments of inertia, I, are known for a plane area with respect to both x and y axes, then the polar moment for the z axis may be calcu-

lated using the equation,

A table of formulas for calculating J for common areas can be found on page249 in thissection

When metric SI units are used, the formulas referred to in (1) and (2) above, are valid if the dimensions are given consistently in meters or millimeters If meters are

used, the units of I and J are in meters4 ; if millimeters are used, these units are in millimeters 4

3) Polar moments of inertia of masses, J M*, appear in dynamics equations involving

rota-tional motion J M bears the same relationship to angular acceleration as mass does to linear

acceleration If units are in the foot-pound-second system, the units of J M are ft-lbs-sec2 orslug-ft2 (1 slug = 1 pound second2 per foot.) If units are in the inch-pound-second system,

the units of J M are inch-lbs-sec2

If metric SI values are used, the units of J M are kilogram-meter squared Formulas

for calculating J M for various bodies are given beginning on page250 If the polar moment

of inertia J is known for the area of a body of constant cross section, J M may be calculatedusing the equation,

where ρ is the density of the material, L the length of the part, and g the gravitational

con-stant If dimensions are in the foot-pound-second system, ρ is in lbs per ft3, L is in ft, g is

* In some books the symbol I denotes the polar moment of inertia of masses; J M is used in this handbook

to avoid confusion with moments of inertia of plane areas

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MOMENTS OF INERTIA 23732.16 ft per sec2, and J is in ft4 If dimensions are in the inch-pound-second system, ρ is in

lbs per in3, L is in inches, g is 386 inches per sec2, and J is in inches4

Using metric SI units, the above formula becomes J M = ρLJ, where ρ = the density in

kilograms/meter 3, L = the length in meters, and J = the polar moment of inertia in

meters 4 The units of J M are kg · m 2

Moment of Inertia of Built-up Sections.—The usual method of calculating the moment

of inertia of a built-up section involves the calculations of the moment of inertia for eachelement of the section about its own neutral axis, and the transferring of this moment ofinertia to the previously found neutral axis of the whole built-up section A much simplermethod that can be used in the case of any section which can be divided into rectangularelements bounded by lines parallel and perpendicular to the neutral axis is the so-called

tabular method based upon the formula: I = b(h1 - h3)/3 in which I = the moment of inertia about axis DE, Fig 1, and b, h and h1 are dimensions as given in the same illustration

Example:The method may be illustrated by applying it to the section shown in Fig 2, andfor simplicity of calculation shown “massed” in Fig 3 The calculation may then be tabu-

lated as shown in the accompanying table The distance from the axis DE to the neutral axis

xx (which will be designated as d) is found by dividing the sum of the geometrical moments

by the area The moment of inertia about the neutral axis is then found in the usual way by

subtracting the area multiplied by d2 from the moment of inertia about the axis DE.

Tabulated Calculation of Moment of Inertia

The distance d from DE, the axis at the base of the configuration, to the neutral axis xx is:

The moment of inertia of the entire section with reference to the neutral axis xx is:

3

–

3 -

A

- 0.3150.644 0.49

I N= I DE–Ad2

0.272–0.644×0.492

=0.117

=

Machinery's Handbook 27th Edition

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Square and Rectangular Sections (Continued)

Z I y

12 - = 0.289d

bd33 -

bd23

3 -= 0.577d

bd

6 b( 2 +d2 ) -

d2 cos 2 α +b2 sin 2 α

dcos α +bsin α -

0.289 bd3–hk3

bd–hk

-=

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MOMENT OF INERTIA, SECTION MODULUS

Distance from Neutral

Axis to Extreme Fiber, y

bd2 24

18 - =0.236d

bd3 12

12

6 - =0.408d

d a( b) 2 -

d a( +2b)

3 a( b) - d

3

a2+4ab+b2

36 a( b) - d

2

a2+4ab+b2

12 a( +2b) - d

2

a2+4ab+b2

18 a( b) 2 -

3d2tan 30 °

2 -

0.866d2

=

d

2 -

A

12

d2( 1 + 2 cos 2 30 ° )

4 cos 2 30 ° -

0.06d4

=

A

6 -d 1( +2cos230° )

4 cos 2 30 ° -

0.12d3

=

d2( 1 + 2cos230 ° ) 48cos230 ° -

0.264d

=

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2d2 tan 22 1 ⁄ 2 = 0.828d2

Circular, Elliptical, and Circular Arc Sections

Moments of Inertia, Section Moduli, and Radii of Gyration (Continued)

Section

Area of Section,

A

0.866d2

=

d

2 cos 30 ° - =0.577d

A

12

d2( 1 + 2 cos 2 30 ° )

4 cos 2 30 ° -

0.104d3

=

d2( 1 + 2cos230 ° ) 48cos230 ° -

0.264d

=

d

2 -

A

12

d2( 1 + 2cos2 22 1 ⁄ 2 ° )

4cos2 22 1 ⁄ 2 ° -

0.055d4

=

A

6 -d 1 2cos

2

22 1 ⁄ 2 ° +

4cos222 1 ⁄ 2 ° -

0.109d3

=

d2( 1 + 2cos222 1 ⁄ 2 ° ) 48cos222 1 ⁄ 2 ° -

32 - =0.098d3 d

4 -

0.288d

=

9 π 2 64 –

1152 π -

0.007d4

=

9 π 2 64 –

192 3 ( π 4 – ) -

0.024d3

=

9 π 2 64 –

12 π -

0.7854 D( 2–d2)

=

D

2

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-I y

I A

-πa3b

4 - =0.7854a3b

πa2b

4 - =0.7854a2b a

2 -

π 4

- a( 3b–c3d) 0.7854

= (a3b–c3d)

π a( 3b–c3d)

4a

0.7854a3b–c3d

2sb3 +ht3

6b

- 2sb3+ht3

12 bd[ –h b( –t) ] -

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flange = (h − l)/(b − t)

= 10.0 ⁄ 6 for standard I-beams.

bd − h(b − t)

dt + 2a(s + n)

in which g = slope of flange = (h − l)/(b − t) =1 ⁄ 6 for standard I-beams.

1 ⁄ 12bd3 1

4g - h( 4 –l4 ) –

1

6d - bd3 1

4g - h( 4 –l4 )

4g - h( 4 –l4 ) –

dt+2a s( n) -

d

2 -

bd3 –h3 (b–t) 12 - bd3–h3(b–t)

6d

- bd3–h3(b–t)

12 bd[ –h b( –t) ] -

b

2 -

1 ⁄ 12b3 (d h ) lt3

g

4

- b( 4 –t4 ) + +

1

6b - b3 (d h ) lt3

g

4

- b( 4 –t4 ) + +

I A

-I y

I A

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-MOMENT OF INERTIA, SECTION MODULUS

C–Sections

for standard channels.

1 ⁄ 12bd3 1

8g - h( 4 –l4 ) –

h–l

2 b( –t) -

= = 1 ⁄ 6

1

6d - bd3 1

8g - h( 4 –l4 )

8g - h( 4 –l4 ) –

dt+a s( n) -

b b2s ht

2 2 -

A

÷ –

h–l

2 b( –t) -

h–l

2 b( –t) -

= = 1 ⁄ 6

I y

I A

-d

2 -

bd3 –h3 (b–t) 12 - bd3–h3(b–t)

6d

- bd3–h3(b–t)

12 bd[ –h b( –t) ] -

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2sb3 +ht3 3 - –A b( –y) 2 I

y

I A

-d d2t+s2(b–t)

2 bs( +ht) - –

I y

I A

-bs h T( +t)

2 - +

I y

I A

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-MOMENT OF INERTIA, SECTION MODULUS

a s( n)

+ +

b

2 -

sb3 +mT3 +lt3 12 -

am 2a[ 2 + (2a+3T) 2 ] 36 -

l T( –t ) T t[ ( – ) 2 +2 T( +2t) 2 ] 144 - +

+

I y

I A

-a a2+at–t2

2 2a( –t) - –

I y

I A

-b t 2d( a ) d+ 2

2 d( a) - –

I y

a t 2c( b ) c+ 2

2 c( b) - –

I y

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A

12

- 7 a[ ( 2 +b2) 12y– 2 ]

2ab2 (a b) –

I y

I A

-b

2 -

ab3 –c b( –2t) 3 12 - ab3–c b( –2t) 3

6b

- ab3–c b( –2t) 3

12t b[ +2 a( –t) ] -

2a–t

2 -

b a( +c) 3 –2c3d–6a2cd

12 - b a( +c) 3 –2c3d–6a2cd

6 2a( –t) - b a( +c) 3 –2c3d–6a2cd

12t b[ +2 a( –t) ] -

d

2 -

td3 +s3 (b–t) 12 - td3–s3(b–t)

6d

- td3+s3(b–t)

12 td[ +s b( –t) ] -

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248 MOMENT OF INERTIA, SECTION MODULUS

Polar Area Moment of Inertia and Section Modulus.—The polar moment of inertia, J,

of a cross-section with respect to a polar axis, that is, an axis at right angles to the plane ofthe cross-section, is defined as the moment of inertia of the cross-section with respect to thepoint of intersection of the axis and the plane The polar moment of inertia may be found bytaking the sum of the moments of inertia about two perpendicular axes lying in the plane ofthe cross-section and passing through this point Thus, for example, the polar moment ofinertia of a circular or a square area with respect to a polar axis through the center of gravity

is equal to two times the moment of inertia with respect to an axis lying in the plane of thecross-section and passing through the center of gravity

The polar moment of inertia with respect to a polar axis through the center of gravity isrequired for problems involving the torsional strength of shafts since this axis is usually theaxis about which twisting of the shaft takes place

The polar section modulus (also called section modulus of torsion), Z p , for circular tions may be found by dividing the polar moment of inertia, J, by the distance c from the center of gravity to the most remote fiber This method may be used to find the approxi-

sec-mate value of the polar section modulus of sections that are nearly round For other than

circular cross-sections, however, the polar section modulus does not equal the polar moment of inertia divided by the distance c.

The accompanying table Polar Moment of Inertia and Polar Section Modulus on page

249 gives formulas for the polar section modulus for several different cross-sections Thepolar section modulus multiplied by the allowable torsional shearing stress gives theallowable twisting moment to which a shaft may be subjected, see Formula (7) on page

300

Mass Moments of Inertia *, J M.—Starting on page250, formulas for mass moment ofinertia of various solids are given in a series of tables The example that follows illustrates

the derivaion of J M for one of the bodes given on page250

Example, Polar Mass Moment of Inertia of a Hollow Circular Section:Referring to the

figure Hollow Cylinder on page 250, consider a strip of width dr on a hollow circular tion, whose inner radius is r and outer radius is R.

sec-The mass of the strip = 2πrdrρ, where ρ is the density of material In order to get the mass

of an individual section, integrate the mass of the strip from r to R.

The 2nd moment of the strip about the AA axis = 2πrdrρr2 To find the polar moment of

inertia about the AA axis, integrate the 2nd moment from r to R.

* In some books the symbol I denotes the polar moment of inertia of masses; J M is used in this handbook

to avoid confusion with moments of inertia of plane areas

M r R2πr r( )ρd

∫ 2πρ r2

2

∫ 2πρ r3

r d

( )

r R

∫ 2πρ r4

4

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With reference to axis A − A:

Spherical Segment: With reference to axis A − A:

With reference to axis B − B:

With reference to axis B − B (through the center of

grav-ity):

With reference to axis B − B:

With reference to axis C − C:

20 -+

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252 POLAR MOMENTS OF INERTIA

Cone:

Frustrum of Cone:

Moments of Inertia of Complex Areas and Masses may be evaluated by the addition and

subtraction of elementary areas and masses For example, the accompanying figure shows

a complex mass at (1); its mass polar moment of inertia can be determined by addingtogether the moments of inertia of the bodies shown at (2) and (3), and subtracting that at(4)

Thus, J M1 = J M2 + JM3 − JM4 All of these moments of inertia are with respect to the axis of

rotation z − z Formulas for JM2 and J M3 can be obtained from the tables beginning onpage250 The moment of inertia for the body at (4) can be evaluated by using the following

transfer-axis equation: J M4 = J M4′ + d2M The term J M4′ is the moment of inertia with

respect to axis z ′ − z′; it may be evaluated using the same equation that applies to JM2 where

d is the distance between the z − z and the z′ − z′ axes, and M is the mass of the body (=

weight in lbs ÷ g).

Moments of Inertia of Complex Masses

Similar calculations can be made when calculating I and J for complex areas using the appropriate transfer-axis equations are I = I ′ + d2A and J = J ′ + d2A The primed term, I′ or

J ′, is with respect to the center of gravity of the corresponding area A; d is the distance

between the axis through the center of gravity and the axis to which I or J is referred.

With reference to axis B − B (through the center of

(2)

(4) (1)

(3)

;;

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Moments of Inertia and Section Moduli for Rectangles and Round Shafts

Moments of inertia and section modulus values shown here are for rectangles 1 ter wide To obtain moment of inertia or section modulus for rectangle of given side length,multiply appropriate table value by given width (See the text starting on page238 forbasic formulas.)

millime-Moments of Inertia and Section Moduli for Rectangles (Metric Units)

Moment of Inertia Section Modulus

Length

of Side (mm)

Moment of Inertia Section Modulus

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Section Moduli for Rectangles

Section modulus values are shown for rectangles 1 inch wide To obtain section modulus for angle of given side length, multiply value in table by given width.

rect-Section Moduli and Moments of Inertia for Round Shafts

In this and succeeding tables, the Polar Section Modulus for a shaft of given diameter can be obtained by multiplying its section modulus by 2 Similarly, its Polar Moment of Inertia can be

obtained by multiplying its moment of inertia by 2.

Length

of Side Section Modulus

Length

of Side Section Modulus

of Inertia Dia.

Section Modulus Moment

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Section Moduli and Moments of Inertia for Round Shafts (English or Metric Units)

of Inertia Dia.

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of Inertia Dia.

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of Inertia Dia.

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of Inertia Dia.

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of Inertia Dia.

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260 BEAMS

BEAMSBeam Calculations Reaction at the Supports.—When a beam is loaded by vertical loads or forces, the sum of

the reactions at the supports equals the sum of the loads In a simple beam, when the loadsare symmetrically placed with reference to the supports, or when the load is uniformly dis-tributed, the reaction at each end will equal one-half of the sum of the loads When theloads are not symmetrically placed, the reaction at each support may be ascertained fromthe fact that the algebraic sum of the moments must equal zero In the accompanying illus-

tration, if moments are taken about the support to the left, then: R2× 40 − 8000 × 10 −

10,000 × 16 − 20,000 × 20 = 0; R2 = 16,000 pounds In the same way, moments taken about

the support at the right give R1 = 22,000 pounds

The sum of the reactions equals 38,000 pounds, which is also the sum of the loads If part

of the load is uniformly distributed over the beam, this part is first equally divided betweenthe two supports, or the uniform load may be considered as concentrated at its center ofgravity

If metric SI units are used for the calculations, distances may be expressed in meters

or millimeters, providing the treatment is consistent, and loads in newtons Note: If the load is given in kilograms, the value referred to is the mass A mass of M kilograms has a weight (applies a force) of Mg newtons, where g = approximately 9.81 meters

per second 2

Stresses and Deflections in Beams.—On the following pages Table 1 gives an extensivelist of formulas for stresses and deflections in beams, shafts, etc It is assumed that all thedimensions are in inches, all loads in pounds, and all stresses in pounds per square inch

The formulas are also valid using metric SI units, with all dimensions in millimeters, all loads in newtons, and stresses and moduli in newtons per millimeter 2 (N/mm 2 ).

Note: A load due to the weight of a mass of M kilograms is Mg newtons, where g =

approximately 9.81 meters per second 2 In the tables:

E =modulus of elasticity of the material

I =moment of inertia of the cross-section of the beam

Z =section modulus of the cross-section of the beam = I ÷ distance from neutral

axis to extreme fiber

W =load on beam

s =stress in extreme fiber, or maximum stress in the cross-section considered, due

to load W A positive value of s denotes tension in the upper fibers and pression in the lower ones (as in a cantilever) A negative value of s denotes the

com-reverse (as in a beam supported at the ends) The greatest safe load is that value

of W which causes a maximum stress equal to, but not exceeding, the greatest safe value of s

y =deflection measured from the position occupied if the load causing the

deflec-tion were removed A positive value of y denotes deflecdeflec-tion below this

posi-tion; a negative value, deflection upward

u, v, w, x = variable distances along the beam from a given support to any point

BEAM STRESS AND DEFLECTION TABLES

If l is greater than 2c, the

stress is zero at points

on both sides

of the center.

If cross-section is constant

and if l = 2.828c, the stresses

at supports and center are equal and opposite, and are

Between each support and adjacent end,

deflec-Table 1 (Continued) Stresses and Deflections in Beams

Type of Beam

Stresses Deflections General Formula for Stress

at any Point Stresses at Critical Points General Formula for Deflection at any Point a Deflections at Critical Points a

Z

–

Z

–

-Wa Z

6EI - 3a l[ ( a ) x– 2 ]

=

6EI - 3v l[ ( –v ) a– 2 ]

=

Wa 24EI - 3l( 2 –4a2 )

Wa2

6EI - 3l( –4a)

2Zl - c( u) 2

=

2ZL - c[ 2 –x l( –x) ]

=

2

Wc 2ZL W 2ZL - c( 2 – 1 ⁄ 4l2 )

1 ⁄ 4l2 –c2

WL 46.62Z

±

24EIL - 6c2 (l u)

u2 (4c u)

[ ]

=

y Wx l(–x)

24EIL - x l[ ( –x ) l+ 2 –6c2 ]

=

Wc 24EIL - 3c[ 2 (c+2l ) l– 3 ]

Wl2

384EIL - 5l( 2 –24c2 )

3 ( 1 ⁄ 4l2 –c2 )

W 96EIL

– (6c2 –l2 ) 2

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Case 6 — Both Ends Overhanging Supports Unsymmetrically, Uniform Load

For overhanging end of length c,

Between supports,

For overhanging end of length d,

Stress at support next to

end of length c,

Critical stress between supports is at

and is Stress at support next to

end of length d,

If cross-section is constant, the greatest of these three is the maximum stress.

If x1 > c, the stress is zero

Case 7 — Both Ends Overhanging Supports, Load at any Point Between Between supports:

For segment of length a,

For segment of length b,

Beyond supports s = 0.

Stress at load,

If cross-section is constant,

Between supports, same as Case 3.

For overhanging end of length c,

For overhanging end of length d,

Between supports, same as Case 3.

2ZL - c( u) 2

=

2ZL - c2l–x l

=

Wc2

2ZL

x l2+c2–d22l

- x1

W 2ZL - c( 2 –x1)

6c2u–u2 (4c u) –l3 ] +

[

=

y Wx l(–x)

24EIL - x l( –x)

l2 2 + {

d2 +c2

2

l - d[ 2x+c2 (l–x) ] } –

=

24EIL - 2l c[ ( 2 +2d2 )

=

+6d2w–w2 (4d–w) –l3 ]

Wc 24EIL - 2l d( 2 +2c2 )

3c3 –l3 +

[ ]

Wd 24EIL - 2l c( 2 +2d2 )

3d3 –l3 +

[ ]

Zl

–

-=

Zl

–

-=

Wab Zl

–

6EIl

– (l b)

-=

6EIl

– (l a)

-=

Wabc 6EIl

– (l b)

-Wabd 6EIl

– (l a)

-Machinery's Handbook 27th Edition

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Between load and adjacent support,

Deflection at unloaded end,

Case 9 — Both Ends Overhanging Supports, Symmetrical Overhanging Loads Between each load and adjacent

Between supports the curve is a circle of radius

Deflection at any point x between supports

Deflection at center,

Table 1 (Continued) Stresses and Deflections in Beams

Type of Beam

Z - c( u)

=

Zl - l( –x)

=

Wc Z

Wu 6EI - 3cu( –u2 +2cl)

-=

Wc2

3EI - c( +l)

Wcl2

15.55EI

–

Wcld 6EI

Z - c( u)

6EI - 3c l[ ( u ) u– 2 ]

=

2EI - l( –x) –

Wc2

6EI - 2c( +3l)

Wcl2

8EI

–

-r EI Wc

=

y= r2 – 1 ⁄ 4l2 – r2 – ( 1 ⁄ 2l–x) 2

r2 – 1 ⁄ 4l2 –r

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Case 10 — Fixed at One End, Uniform Load

Stress at support,

Maximum deflection, at end,

Case 11 — Fixed at One End, Load at Other

Stress at support,

Case 12 — Fixed at One End, Intermediate Load Between support and load,

Type of Beam

2Zl - l( –x) 2

=

Wl 2Z

-y Wx224EIl - 2l[ 2 + (2l–x) 2 ]

=

Wl Z

Wx2

6EI - 3l( –x)

Z

6EI - 3l( –x)

=

y Wl26EI - 3v( –l)

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Case 13 — Fixed at One End, Supported at the Other, Load at Center Between point of fixture and load,

Between support and load,

Maximum stress at point

of fixture,

Stress is zero at x = 3 ⁄ 11l

Greatest negative stress at center,

Between point of fixture and load,

Maximum deflection is at v = 0.4472l, and is

Deflection at load,

Case 14 — Fixed at One End, Supported at the Other, Load at any Point Between point of fixture and load,

Greatest positive stress, at point of fixture,

Greatest negative stress, at load,

If a < 0.5858l, the first is the maximum stress If a = 0.5858l, the two are equal

and are If a >

0.5858l, the second is the

maximum stress.

Stress is zero at

Between point of fixture and load,

Deflection at load,

If a < 0.5858l, maximum

deflec-tion is and located between load and support,

16Z - 3l( –11x)

-5 ⁄ 32 – Wl

Z

-y Wx296EI - 9l( –11x)

=

96EI - 3l( 2 –5v2 )

=

Wl3

107.33EI

7 768 -Wl3

=

Wab 2Zl2

- l( b)

Wa2b 2Zl3 - – (3l a)

Wl 5.83Z

- 3n( –mx)

=

y Wa2v 12EIl3

3EIm2l3 -

x 2n m

-=

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Case 15 — Fixed at One End, Supported at the Other, Uniform Load

Maximum stress at point

Deflection at center, Deflection at point of greatest

negative stress, at x = 5 ⁄ 8l is

Case 16 — Fixed at One End, Free but Guided at the Other, Uniform Load

Maximum stress, at port,

sup-Stress is zero at x = 0.4227l

Greatest negative stress, at free end,

Maximum deflection, at free end,

Case 17 — Fixed at One End, Free but Guided at the Other, with Load

Stress at support,

Stress at free end These are the maximum stresses and are equal and opposite.

Stress is zero at center.

Maximum deflection, at free end,

Type of Beam

Z

-y Wx2(l–x)

48EIl - 3l( –2x)

1 ⁄ 2x l

⎝ ⎠ 2 + –

–

-y Wx224EIl - 2l( –x) 2

–

-y Wx212EI - 3l( –2x)

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Stress is zero at x = 1 ⁄ 4l

Maximum deflection, at load,

Case 19 — Fixed at Both Ends, Load at any Point

Stress at end next to

Type of Beam

-Wl 8Z

–

-y Wx248EI - 3l( –4x)

=

Wab2

Zl2 -

Wa2b

Zl2 -

y Wx2b26EIl3

- 2a l[ ( –x ) l a x+ ( – ) ]

=

y Wv2a26EIl3

- 2b l[ ( –v ) l b v+ ( – ) ]

=

Wa3b3

3EIl3 -

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Case 20 — Fixed at Both Ends, Uniform Load

Maximum stress, at ends,

Maximum deflection, at center,

Case 21 — Continuous Beam, with Two Unequal Spans, Unequal, Uniform Loads

Type of Beam

s Wl 2Z

- 1 ⁄ 6 x l

x

l

⎝ ⎠ 2 + –

–

y Wx224EIl - l( –x) 2

=

Wl3

384EI

s l1 –x Z

x l1

W1 - W( 1–R1)

=

R1l12ZW1

–

-u l2

W2 - W( 2–R2)

=

R2l22ZW2

–

Tiêu đề	Radius of Gyration and Moments of Inertia
Trường học	Industrial Press, Inc.
Chuyên ngành	Mechanical Engineering
Thể loại	Reference Handbook
Năm xuất bản	2004
Thành phố	New York

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Số trang	65
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