Beam Formulas

CHAPTER 2BEAM

FORMULAS

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Source: CIVIL ENGINEERING FORMULAS

In analyzing beams of various types, the geometric proper-ties of a variety of cross-sectional areas are used. Figure 2.1gives equations for computing area A, moment of inertia I,section modulus or the ratio S � I/c, where c � distancefrom the neutral axis to the outermost fiber of the beam orother member. Units used are inches and millimeters andtheir powers. The formulas in Fig. 2.1 are valid for bothUSCS and SI units.

Handy formulas for some dozen different types ofbeams are given in Fig. 2.2. In Fig. 2.2, both USCS and SIunits can be used in any of the formulas that are applicableto both steel and wooden beams. Note that W � load, lb(kN); L � length, ft (m); R � reaction, lb (kN); V � shear,lb (kN); M � bending moment, lb � ft (N � m); D � deflec-tion, ft (m); a � spacing, ft (m); b � spacing, ft (m); E �modulus of elasticity, lb/in2 (kPa); I � moment of inertia,in4 (dm4); � � less than; � � greater than.

Figure 2.3 gives the elastic-curve equations for a varietyof prismatic beams. In these equations the load is given asP, lb (kN). Spacing is given as k, ft (m) and c, ft (m).

CONTINUOUS BEAMS

Continuous beams and frames are statically indeterminate.Bending moments in these beams are functions of thegeometry, moments of inertia, loads, spans, and modulus ofelasticity of individual members. Figure 2.4 shows how anyspan of a continuous beam can be treated as a single beam,with the moment diagram decomposed into basic com-ponents. Formulas for analysis are given in the diagram.Reactions of a continuous beam can be found by using theformulas in Fig. 2.5. Fixed-end moment formulas forbeams of constant moment of inertia (prismatic beams) for

16 CHAPTER TWO



BEAM FORMULAS

17

FIG

UR

E 2

.1G

eom

etri

c pr

oper

ties

of s

ectio

ns.



BEAM FORMULAS

18



BEAM FORMULAS

19

FIG

UR

E 2

.1(C

onti

nued

)G

eom

etri

c pr

oper

ties

of s

ectio

ns.



BEAM FORMULAS

20



BEAM FORMULAS

21

FIG

UR

E 2

.1(C

onti

nued

)G

eom

etri

c pr

oper

ties

of s

ectio

ns.



BEAM FORMULAS

22



BEAM FORMULAS

23

FIG

UR

E 2

.1(C

onti

nued

)G

eom

etri

c pr

oper

ties

of s

ectio

ns.



BEAM FORMULAS

24



BEAM FORMULAS

25

FIG

UR

E 2

.1(C

onti

nued

)G

eom

etri

c pr

oper

ties

of s

ectio

ns.



BEAM FORMULAS

26

FIG

UR

E 2

.1(C

onti

nued

)G

eom

etri

c pr

oper

ties

of s

ectio

ns.



BEAM FORMULAS

several common types of loading are given in Fig. 2.6.Curves (Fig. 2.7) can be used to speed computation offixed-end moments in prismatic beams. Before the curvesin Fig. 2.7 can be used, the characteristics of the loadingmust be computed by using the formulas in Fig. 2.8. Theseinclude , the location of the center of gravity of the load-ing with respect to one of the loads; G2 � � Pn/W, wherebnL is the distance from each load Pn to the center of gravity of the loading (taken positive to the right); and S3 �� Pn/W. These values are given in Fig. 2.8 for some com-mon types of loading.

Formulas for moments due to deflection of a fixed-endbeam are given in Fig. 2.9. To use the modified momentdistribution method for a fixed-end beam such as that inFig. 2.9, we must first know the fixed-end moments for abeam with supports at different levels. In Fig. 2.9, the rightend of a beam with span L is at a height d above the leftend. To find the fixed-end moments, we first deflect thebeam with both ends hinged; and then fix the right end,leaving the left end hinged, as in Fig. 2.9b. By noting that aline connecting the two supports makes an angle approxi-mately equal to d/L (its tangent) with the original positionof the beam, we apply a moment at the hinged end to pro-duce an end rotation there equal to d/L. By the definition ofstiffness, this moment equals that shown at the left end ofFig. 2.9b. The carryover to the right end is shown as the topformula on the right-hand side of Fig. 2.9b. By using thelaw of reciprocal deflections, we obtain the end moments ofthe deflected beam in Fig. 2.9 as

(2.1)

(2.2)MRF � K R

F (1 � C LF)

d

L

MFL � K F

L (1 � C FR)

d

L

bn3

bn2

xL

BEAM FORMULAS 27



BEAM FORMULAS

28



BEAM FORMULAS

29

FIG

UR

E 2

.2B

eam

for

mul

as.

(Fro

m J

. C

alle

nder

,Ti

me-

Save

r St

anda

rds

for

Arc

hite

ctur

al D

esig

n D

ata,

6th

ed.,

McG

raw

-Hil

l,N

.Y.)



BEAM FORMULAS

30



BEAM FORMULAS

31

FIG

UR

E 2

.2(C

onti

nued

)B

eam

for

mul

as.



BEAM FORMULAS

32



BEAM FORMULAS

33

FIG

UR

E 2

.2(C

onti

nued

)B

eam

for

mul

as.



BEAM FORMULAS

34



BEAM FORMULAS

35

FIG

UR

E 2

.2(C

onti

nued

)B

eam

for

mul

as.



BEAM FORMULAS

36



BEAM FORMULAS

37

FIG

UR

E 2

.2(C

onti

nued

)B

eam

for

mul

as.



BEAM FORMULAS

38



BEAM FORMULAS

39

FIG

UR

E 2

.2(C

onti

nued

)B

eam

for

mul

as.



BEAM FORMULAS

40

FIG

UR

E 2

.3E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(a)

She

ars,

mom

ents

,de

flect

ions

for

ful

l un

i-fo

rm l

oad

on a

sim

ply

supp

orte

d pr

ism

atic

bea

m.

(b)

Shea

rs a

nd m

omen

ts f

or u

nifo

rm l

oad

over

par

t of

asi

mpl

y su

ppor

ted

pris

mat

ic b

eam

. (c

) Sh

ears

,m

omen

ts,

defle

ctio

ns f

or a

con

cent

rate

d lo

ad a

t an

y po

int

ofa

sim

ply

supp

orte

d pr

ism

atic

bea

m.



BEAM FORMULAS

41

FIG

UR

E 2

.3(C

onti

nued

)E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(d

) Sh

ears

,m

omen

ts,

defle

c-tio

ns f

or a

con

cent

rate

d lo

ad a

t mid

span

of

a si

mpl

y su

ppor

ted

pris

mat

ic b

eam

. (e)

She

ars,

mom

ents

,defl

ec-

tions

for

tw

o eq

ual

conc

entr

ated

loa

ds o

n a

sim

ply

supp

orte

d pr

ism

atic

bea

m.

(f)

Shea

rs,

mom

ents

,de

flect

ions

for

sev

eral

equ

al lo

ads

equa

lly s

pace

d on

a s

impl

y su

ppor

ted

pris

mat

ic b

eam

.



BEAM FORMULAS

42

FIG

UR

E 2

.3(C

onti

nued

)E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(g)

She

ars,

mom

ents

,de

flec-

tions

for

a c

once

ntra

ted

load

on

a be

am o

verh

ang.

(h)

She

ars,

mom

ents

,defl

ectio

ns f

or a

con

cent

rate

d lo

adon

the

end

of a

pri

smat

ic c

antil

ever

. (i)

She

ars,

mom

ents

,defl

ectio

ns f

or a

uni

form

load

ove

r th

e fu

ll le

ngth

of a

bea

m w

ith o

verh

ang.



BEAM FORMULAS

43

FIG

UR

E 2

.3(C

onti

nued

)E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(j)

She

ars,

mom

ents

,defl

ectio

nsfo

r un

ifor

m l

oad

over

the

ful

l le

ngth

of

a ca

ntile

ver.

(k)

Shea

rs,m

omen

ts,d

eflec

tions

for

uni

form

loa

d on

abe

am o

verh

ang.

(l)

She

ars,

mom

ents

,defl

ectio

ns f

or tr

iang

ular

load

ing

on a

pri

smat

ic c

antil

ever

.



BEAM FORMULAS

44

FIG

UR

E 2

.3(C

onti

nued

)E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(m

) Si

mpl

e be

am—

load

incr

eas-

ing

unif

orm

ly to

one

end

.



BEAM FORMULAS

45

FIG

UR

E 2

.3(C

onti

nued

)E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(n)

Sim

ple

beam

—lo

ad in

crea

s-in

g un

ifor

mly

to c

ente

r.



BEAM FORMULAS

46

FIG

UR

E 2

.3(C

onti

nued

)E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(o)

Sim

ple

beam

—un

ifor

m lo

adpa

rtia

lly d

istr

ibut

ed a

t one

end

.



BEAM FORMULAS

47

FIG

UR

E 2

.3(C

onti

nued

)E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(p)

Can

tilev

er b

eam

—co

ncen

-tr

ated

load

at f

ree

end.



BEAM FORMULAS

48

FIG

UR

E 2

.3(C

onti

nued

)E

last

ic-c

urve

equ

atio

ns f

or p

rism

atic

bea

ms.

(q

) B

eam

fixe

d at

bot

h en

ds—

conc

entr

ated

load

at c

ente

r.



BEAM FORMULAS

49

FIG

UR

E 2

.4A

ny s

pan

of a

con

tinuo

us b

eam

(a)

can

be

trea

ted

as a

sim

ple

beam

,as

show

n in

(b)

and

(c)

.In

(c)

,the

mom

ent d

iagr

am is

dec

ompo

sed

into

bas

ic c

ompo

nent

s.



BEAM FORMULAS

50 CHAPTER TWO

FIGURE 2.5 Reactions of continuous beam (a) found by makingthe beam statically determinate. (b) Deflections computed withinterior supports removed. (c), (d ), and (e) Deflections calculatedfor unit load over each removed support, to obtain equations foreach redundant.



BEAM FORMULAS

BEAM FORMULAS 51

FIGURE 2.6 Fixed-end moments for a prismatic beam. (a) For aconcentrated load. (b) For a uniform load. (c) For two equal con-centrated loads. (d) For three equal concentrated loads.

In a similar manner the fixed-end moment for a beam withone end hinged and the supports at different levels can befound from

(2.3)

where K is the actual stiffness for the end of the beam thatis fixed; for beams of variable moment of inertia K equalsthe fixed-end stiffness times .(1 � C L

FC RF)

MF � Kd

L



BEAM FORMULAS

FIGURE 2.7 Chart for fixed-end moments due to any type ofloading.

FIGURE 2.8 Characteristics of loadings.



BEAM FORMULAS

BEAM FORMULAS 53

FIGURE 2.8 (Continued) Characteristics of loadings.

ULTIMATE STRENGTH OFCONTINUOUS BEAMS

Methods for computing the ultimate strength of continuousbeams and frames may be based on two theorems that fixupper and lower limits for load-carrying capacity:

1. Upper-bound theorem. A load computed on the basis ofan assumed link mechanism is always greater than, or atbest equal to, the ultimate load.



BEAM FORMULAS

54 CHAPTER TWO

FIGURE 2.9 Moments due to deflection of a fixed-end beam.

2. Lower-bound theorem. The load corresponding to anequilibrium condition with arbitrarily assumed valuesfor the redundants is smaller than, or at best equal to, theultimate loading—provided that everywhere momentsdo not exceed MP. The equilibrium method, based onthe lower bound theorem, is usually easier for simplecases.



BEAM FORMULAS

BEAM FORMULAS 55

For the continuous beam in Fig. 2.10, the ratio of theplastic moment for the end spans is k times that for the cen-ter span (k � 1).

Figure 2.10b shows the moment diagram for the beammade determinate by ignoring the moments at B and C andthe moment diagram for end moments MB and MC appliedto the determinate beam. Then, by using Fig. 2.10c, equilib-rium is maintained when

(2.4)

The mechanism method can be used to analyze rigidframes of constant section with fixed bases, as in Fig. 2.11.Using this method with the vertical load at midspan equalto 1.5 times the lateral load, the ultimate load for the frameis 4.8MP/L laterally and 7.2MP /L vertically at midspan.

Maximum moment occurs in the interior spans AB andCD when

(2.5)

or if

(2.6)

A plastic hinge forms at this point when the moment equalskMP. For equilibrium,

M � kMP when x �L

2�

kMP

wL

x �L

2�

M

wL

�wL2

4(1 � k)

�wL2

4 � kMP

MP �wL2

4�

1

2MB �

1

2MC



BEAM FORMULAS

56 CHAPTER TWO

FIGURE 2.10 Continuous beam shown in (a) carries twice asmuch uniform load in the center span as in the side span. In (b) areshown the moment diagrams for this loading condition with redun-dants removed and for the redundants. The two moment diagramsare combined in (c), producing peaks at which plastic hinges areassumed to form.



BEAM FORMULAS

57

FIG

UR

E 2

.11

Ulti

mat

e-lo

ad p

ossi

bilit

ies

for

a ri

gid

fram

e of

con

stan

t sec

tion

with

fixe

d ba

ses.



BEAM FORMULAS

leading to

(2.7)

When the value of MP previously computed is substituted,

from which k � 0.523. The ultimate load is

(2.8)

In any continuous beam, the bending moment at anysection is equal to the bending moment at any other section,plus the shear at that section times its arm, plus the productof all the intervening external forces times their respectivearms. Thus, in Fig. 2.12,

Table 2.1 gives the value of the moment at the varioussupports of a uniformly loaded continuous beam over equal

Mx � M3 � V3x � P3a

� P1 (l2 � c � x) � P2 (b � x) � P3a

Mx � R1 (l1 � l2 � x) � R2 (l2 � x) � R3x

Vx � R1 � R2 � R3 � P1 � P2 � P3

wL �4MP (1 � k)

L� 6.1

MP

L

7k2 � 4k � 4 or k (k � 4�7) � 4�7

k2MP2

wL2 � 3kMP �wL2

4� 0

�w

2 � L

2�

kMP

wL � � L

2�

KMP

wL � � � 1

2�

kMP

wL2 � kMP

kMP �w

2x (L � x) �

x

LkMP

58 CHAPTER TWO



BEAM FORMULAS

BEAM FORMULAS 59

FIGURE 2.12 Continuous beam.

FIGURE 2.13 Relation between moment and shear diagrams fora uniformly loaded continuous beam of four equal spans.

spans, and it also gives the values of the shears on each sideof the supports. Note that the shear is of the opposite signon either side of the supports and that the sum of the twoshears is equal to the reaction.

Figure 2.13 shows the relation between the moment andshear diagrams for a uniformly loaded continuous beam offour equal spans. (See Table 2.1.) Table 2.1 also gives themaximum bending moment that occurs between supports,in addition to the position of this moment and the points of



BEAM FORMULAS

60

TA

BLE

2.1

Uni

form

ly L

oade

d C

ontin

uous

Bea

ms

over

Equ

al S

pans

(Uni

form

load

per

uni

t len

gth

�w

; le

ngth

of e

ach

span

�l)

Dis

tanc

e to

poin

t of

Dis

tanc

e to

Shea

r on

eac

h si

dem

ax m

omen

t,po

int o

fN

otat

ion

of s

uppo

rt. L

�le

ft,

mea

sure

d to

infle

ctio

n,N

umbe

rof

R�

righ

t. R

eact

ion

atM

omen

tM

axri

ght f

rom

m

easu

red

toof

supp

ort

any

supp

ort i

s L

�R

over

eac

hm

omen

t in

from

righ

t fro

msu

ppor

tsof

spa

nL

Rsu

ppor

tea

ch s

pan

supp

ort

supp

ort

21

or 2

00

0.12

50.

500

Non

e

31

00

0.07

030.

375

0.75

02

0.07

030.

625

0.25

0

41

00

0.08

00.

400

0.80

02

0.02

50.

500

0.27

6,0.

724

10

00.

0772

0.39

30.

786

52

0.03

640.

536

0.26

6,0.

806

30.

0364

0.46

40.

194,

0.73

41

00

0.07

790.

395

0.78

915� 3

8

2 � 28

13� 2

813

� 28

3 � 28

15� 2

817

� 28

11� 2

8

1 � 10

5 � 10

6 � 10

4 � 10

1 � 85 � 8

5 � 8

3 � 81 � 2



BEAM FORMULAS

61

62

0.03

320.

526

0.26

8,0.

783

30.

0461

0.50

00.

196,

0.80

41

00

0.07

770.

394

0.78

82

0.03

400.

533

0.26

8,0.

790

73

0.04

330.

490

0.19

6,0.

785

40.

0433

0.51

00.

215,

0.80

41

00

0.07

780.

394

0.78

9

82

0.03

380.

528

0.26

8,0.

788

30.

0440

0.49

30.

196,

0.79

04

0.04

050.

500

0.21

5,0.

785

Val

ues

appl

y to

wl

wl

wl2

wl2

ll

The

num

eric

al v

alue

s gi

ven

are

coef

ficie

nts

of th

e ex

pres

sion

s at

the

foot

of

each

col

umn.

12� 1

4271� 1

4272

� 142

11� 1

4270� 1

4267

� 142

15� 1

4275� 1

4286

� 142

56� 1

42

9 � 104

53� 1

0453

� 104

8 � 104

51� 1

0449

� 104

11� 1

0455� 1

0463

� 104

41� 1

04

3 � 38

19� 3

818

� 38

4 � 38

20� 3

823

� 38



BEAM FORMULAS

62 CHAPTER TWO

inflection. Figure 2.14 shows the values of the functions fora uniformly loaded continuous beam resting on three equalspans with four supports.

Maxwell’s Theorem

When a number of loads rest upon a beam, the deflection atany point is equal to the sum of the deflections at this pointdue to each of the loads taken separately. Maxwell’s theo-rem states that if unit loads rest upon a beam at two points,A and B, the deflection at A due to the unit load at B equalsthe deflection at B due to the unit load at A.

Castigliano’s Theorem

This theorem states that the deflection of the point ofapplication of an external force acting on a beam is equal

FIGURE 2.14 Values of the functions for a uniformly loadedcontinuous beam resting on three equal spans with four supports.



BEAM FORMULAS

BEAM FORMULAS 63

to the partial derivative of the work of deformation withrespect to this force. Thus, if P is the force, f is the deflec-tion, and U is the work of deformation, which equals theresilience:

According to the principle of least work, the deforma-tion of any structure takes place in such a manner that thework of deformation is a minimum.

BEAMS OF UNIFORM STRENGTH

Beams of uniform strength so vary in section that the unitstress S remains constant, and I/c varies as M. For rectangu-lar beams of breadth b and depth d, I/c � I/c � bd 2/6 andM � Sbd2/6. For a cantilever beam of rectangular cross sec-tion, under a load P, Px � Sbd2/6. If b is constant, d2 varieswith x, and the profile of the shape of the beam is a parabola,as in Fig. 2.15. If d is constant, b varies as x, and the beamis triangular in plan (Fig. 2.16).

Shear at the end of a beam necessitates modification ofthe forms determined earlier. The area required to resistshear is P/Sv in a cantilever and R/Sv in a simple beam. Dot-ted extensions in Figs. 2.15 and 2.16 show the changes nec-essary to enable these cantilevers to resist shear. The wastein material and extra cost in fabricating, however, makemany of the forms impractical, except for cast iron. Figure2.17 shows some of the simple sections of uniformstrength. In none of these, however, is shear taken intoaccount.

dU

dP� f



BEAM FORMULAS

64 CHAPTER TWO

SAFE LOADS FOR BEAMSOF VARIOUS TYPES

Table 2.2 gives 32 formulas for computing the approximatesafe loads on steel beams of various cross sections for anallowable stress of 16,000 lb/in2 (110.3 MPa). Use theseformulas for quick estimation of the safe load for any steelbeam you are using in a design.

FIGURE 2.15 Parabolic beam of uniform strength.



BEAM FORMULAS

BEAM FORMULAS 65

FIGURE 2.16 Triangular beam of uniform strength.

Table 2.3 gives coefficients for correcting values inTable 2.2 for various methods of support and loading.When combined with Table 2.2, the two sets of formulasprovide useful time-saving means of making quick safe-load computations in both the office and the field.



BEAM FORMULAS

TA

BLE

2.2

App

roxi

mat

e Sa

fe L

oads

in P

ound

s (k

gf)

on S

teel

Bea

ms*

(Bea

ms

supp

orte

d at

bot

h en

ds;

allo

wab

le fi

ber

stre

ss fo

r st

eel,

16,0

00 lb

/in2

(1.1

27 k

gf/c

m2 )

(ba

sis

of ta

ble)

for

iron

,red

uce

valu

es g

iven

in ta

ble

by o

ne-e

ight

h)

Gre

ates

t saf

e lo

ad,l

b‡D

eflec

tion,

in‡

Shap

e of

sec

tion

Loa

d in

mid

dle

Loa

d di

stri

bute

dL

oad

in m

iddl

eL

oad

dist

ribu

ted

Solid

rec

tang

le

Hol

low

rec

tang

le

Solid

cyl

inde

r

Hol

low

cyl

inde

rw

L3

38(A

D2

�ad

2 )

wL

3

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)

L

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L

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1,78

0AD

L

890A

D

L

66



BEAM FORMULAS

67

Eve

n-le

gged

ang

le o

r te

e

Cha

nnel

or

Z b

ar

Dec

k be

am

I be

am

* L�

dist

ance

bet

wee

n su

ppor

ts,f

t (m

);A

�se

ctio

nal a

rea

of b

eam

,in2

(cm

2 );D

�de

pth

of b

eam

,in

(cm

);a

�in

teri

or a

rea,

in2

(cm

2 );d

�in

teri

or d

epth

,in

(cm

); w

�to

tal w

orki

ng lo

ad,n

et to

ns (

kgf)

.

wL

3

93A

D2

wL

3

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3,39

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3

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1.77

0AD

L

885A

D

L



BEAM FORMULAS

68

TA

BLE

2.3

Coe

ffici

ents

for

Cor

rect

ing

Val

ues

in T

able

2.2

for

Var

ious

Met

hods

of

Sup

port

and

of

Loa

ding

†

Max

rel

ativ

ede

flect

ion

unde

rM

ax r

elat

ive

max

rel

ativ

e sa

feC

ondi

tions

of

load

ing

safe

load

load

Bea

m s

uppo

rted

at e

nds

Loa

d un

ifor

mly

dis

trib

uted

ove

r sp

an1.

01.

0L

oad

conc

entr

ated

at c

ente

r of

spa

n0.

80Tw

o eq

ual l

oads

sym

met

rica

lly c

once

ntra

ted

l/4c

Loa

d in

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sing

uni

form

ly to

one

end

0.97

40.

976

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d in

crea

sing

uni

form

ly to

cen

ter

0.96

Loa

d de

crea

sing

uni

form

ly to

cen

ter

1.08

3 � 23 � 41 � 2



BEAM FORMULAS

69

Bea

mfix

ed a

t one

end

,can

tilev

erL

oad

unif

orm

ly d

istr

ibut

ed o

ver

span

2.40

Loa

d co

ncen

trat

ed a

t end

3.20

Loa

d in

crea

sing

uni

form

ly to

fixe

d en

d1.

92

Bea

m c

ontin

uous

ove

r tw

o su

ppor

ts e

quid

ista

nt f

rom

end

sL

oad

unif

orm

ly d

istr

ibut

ed o

ver

span

1.If

dis

tanc

e a

�0.

2071

ll2 /

4a2

2.If

dis

tanc

e a

�0.

2071

ll/(

l–4a

)3.

If d

ista

nce

a�

0.20

71l

5.83

Two

equa

l loa

ds c

once

ntra

ted

at e

nds

l/4a

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ngth

of

beam

; c�

dist

ance

fro

m s

uppo

rt to

nea

rest

con

cent

rate

d lo

ad; a

�di

stan

ce f

rom

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port

to e

ndof

beam

.

3 � 81 � 81 � 4



BEAM FORMULAS

70



BEAM FORMULAS

71

FIG

UR

E 2

.17

Bea

ms

of u

nifo

rm s

tren

gth

(in

bend

ing)

.



BEAM FORMULAS

72



BEAM FORMULAS

73

FIG

UR

E 2

.17

(Con

tinu

ed)

Bea

ms

of u

nifo

rm s

tren

gth

(in

bend

ing)

.



BEAM FORMULAS

74



BEAM FORMULAS

75

FIG

UR

E 2

.17

(Con

tinu

ed)

Bea

ms

of u

nifo

rm s

tren

gth

(in

bend

ing)

.



BEAM FORMULAS

76



BEAM FORMULAS

77

FIG

UR

E 2

.17

(Con

tinu

ed)

Bea

ms

of u

nifo

rm s

tren

gth

(in

bend

ing)

.



BEAM FORMULAS

78

FIG

UR

E 2

.17

(Con

tinu

ed)

Bea

ms

of u

nifo

rm s

tren

gth

(in

bend

ing)

.



BEAM FORMULAS

BEAM FORMULAS 79

ROLLING AND MOVING LOADS

Rolling and moving loads are loads that may change theirposition on a beam or beams. Figure 2.18 shows a beamwith two equal concentrated moving loads, such as twowheels on a crane girder, or the wheels of a truck on abridge. Because the maximum moment occurs where theshear is zero, the shear diagram shows that the maximummoment occurs under a wheel. Thus, with x � a/2:

M2 max when x � a

M1 max when x � a

Figure 2.19 shows the condition when two equal loadsare equally distant on opposite sides of the center of thebeam. The moment is then equal under the two loads.

Mmax �Pl

2 �1 �a

2l �2

�P

2l �l �a

2 �2

3�4

1�4

M1 �Pl

2 �1 �a

l�

2a2

l 2 �2x

l

3a

l�

4x2

l 2 �

R2 � P �1 �2x

l�

a

l �

M2 �Pl

2 �1 �a

l�

2x

l

a

l�

4x2

l 2 �

R1 � P �1 �2x

l�

a

l �



BEAM FORMULAS

80

FIGURE 2.18 Two equal concentrated moving loads.

FIGURE 2.19 Two equal moving loads equally distant on oppo-site sides of the center.



BEAM FORMULAS

BEAM FORMULAS 81

If two moving loads are of unequal weight, the condi-tion for maximum moment is the maximum moment occur-ring under the heavier wheel when the center of the beambisects the distance between the resultant of the loads andthe heavier wheel. Figure 2.20 shows this position and theshear and moment diagrams.

When several wheel loads constituting a system are on abeam or beams, the several wheels must be examined inturn to determine which causes the greatest moment. Theposition for the greatest moment that can occur under a given

FIGURE 2.20 Two moving loads of unequal weight.



BEAM FORMULAS

wheel is, as stated earlier, when the center of the spanbisects the distance between the wheel in question and theresultant of all loads then on the span. The position for max-imum shear at the support is when one wheel is passing offthe span.

CURVED BEAMS

The application of the flexure formula for a straight beamto the case of a curved beam results in error. When all“fibers” of a member have the same center of curvature,the concentric or common type of curved beam exists(Fig. 2.21). Such a beam is defined by the Winkler–Bachtheory. The stress at a point y units from the centroidalaxis is

M is the bending moment, positive when it increases curva-ture; y is positive when measured toward the convex side; Ais the cross-sectional area; R is the radius of the centroidalaxis; Z is a cross-section property defined by

Analytical expressions for Z of certain sections are givenin Table 2.4. Z can also be found by graphical integrationmethods (see any advanced strength book). The neutralsurface shifts toward the center of curvature, or inside fiber,an amount equal to e � ZR/(Z � 1). The Winkler–Bach

Z � �1

A� y

R � ydA

S �M

AR �1 �y

Z (R � y) �

82 CHAPTER TWO



BEAM FORMULAS

83

TA

BLE

2.4

Ana

lytic

al E

xpre

ssio

ns f

or Z

Sect

ion

Exp

ress

ion

A�

2 [(

t�

b)C

1�

bC2]

Z�

�1

�R A�b l

n �R

�C

2

R�

C2��

(t�

b) ln

�R�

C1

R�

C1��

and

A�

tC1

�(b

�t)

C3

�bC

2

Z�

�1

�R A�t ln

(R

�C

1)�

(b�

t) ln

(R

�C

0)�

b ln

(R

�C

2)�

Z�

�1

�2�R r

��R r

�√�

R r�2

�1�

Z�

�1

�R h�ln

R�

C

R�

C�



BEAM FORMULAS

theory, though practically satisfactory, disregards radialstresses as well as lateral deformations and assumes purebending. The maximum stress occurring on the inside fiberis S � Mhi/AeRi, whereas that on the outside fiber is S �Mh0/AeR0.

The deflection in curved beams can be computed bymeans of the moment-area theory.

The resultant deflection is then equal to in the direction defined by Deflections canalso be found conveniently by use of Castigliano’s theorem.It states that in an elastic system the displacement in thedirection of a force (or couple) and due to that force (or cou-ple) is the partial derivative of the strain energy with respectto the force (or couple).

A quadrant of radius R is fixed at one end as shown inFig. 2.22. The force F is applied in the radial direction atfree-end B. Then, the deflection of B is

tan � � y / x.0 � √2

x � 2y

FIGURE 2.21 Curved beam.

84 CHAPTER TWO



BEAM FORMULAS

By moment area,

y � R sin � x � R(1 � cos �)

ds � Rd� M � FR sin �

and

at

By Castigliano,

Bx �FR3

4EI By � �

FR3

2EI

� 32.5�

� tan�1 2

�x � tan�1 ��FR3

2EI�

4EI

FR3 �

B �FR3

2EI √1 �2

4

Bx �FR3

4EI By � �

FR3

2EI

FIGURE 2.22 Quadrant with fixed end.

BEAM FORMULAS 85



BEAM FORMULAS

Eccentrically Curved Beams

These beams (Fig. 2.23) are bounded by arcs having differ-ent centers of curvature. In addition, it is possible for eitherradius to be the larger one. The one in which the sectiondepth shortens as the central section is approached may becalled the arch beam. When the central section is thelargest, the beam is of the crescent type.

Crescent I denotes the beam of larger outside radius andcrescent II of larger inside radius. The stress at the centralsection of such beams may be found from S � KMC/I.In the case of rectangular cross section, the equationbecomes S � 6KM/bh2, where M is the bending moment, bis the width of the beam section, and h its height. The stressfactors, K for the inner boundary, established from photo-elastic data, are given in Table 2.5. The outside radius

86 CHAPTER TWO

FIGURE 2.23 Eccentrically curved beams.



BEAM FORMULAS

BEAM FORMULAS 87

TABLE 2.5 Stress Factors for Inner Boundary at Central Section

(see Fig. 2.23)

1. For the arch-type beams

(a)

(b)

(c) In the case of larger section ratios use the equivalent beamsolution

2. For the crescent I-type beams

(a)

(b)

(c)

3. For the crescent II-type beams

(a)

(b)

(c) K � 1.081 � h

Ro � Ri�

0.0270 if Ro � Ri

h� 20

K � 1.119 � h

Ro � Ri�

0.0378 if 8 �Ro � Ri

h� 20

K � 0.897 � 1.098h

Ro � Ri

if Ro � Ri

h� 8

K � 1.092 � h

Ro � Ri�

0.0298 if Ro � Ri

h� 20

K � 0.959 � 0.769h

Ro � Ri

if 2 �Ro � Ri

h� 20

K � 0.570 � 1.536h

Ro � Ri

if Ro � Ri

h� 2

K � 0.899 � 1.181h

Ro � Ri

if 5 �Ro � Ri

h� 10

K � 0.834 � 1.504h

Ro � Ri

if Ro � Ri

h� 5

is denoted by Ro and the inside by Ri. The geometry ofcrescent beams is such that the stress can be larger in off-center sections. The stress at the central section determinedabove must then be multiplied by the position factor k,



BEAM FORMULAS

88 CHAPTER TWO

given in Table 2.6. As in the concentric beam, the neutralsurface shifts slightly toward the inner boundary. (SeeVidosic, “Curved Beams with Eccentric Boundaries,”Transactions of the ASME, 79, pp. 1317–1321.)

ELASTIC LATERAL BUCKLING OF BEAMS

When lateral buckling of a beam occurs, the beam under-goes a combination of twist and out-of-plane bending(Fig. 2.24). For a simply supported beam of rectangularcross section subjected to uniform bending, buckling occursat the critical bending moment, given by

where L � unbraced length of the member

E � modulus of elasticity

Iy � moment of inertial about minor axis

G � shear modulus of elasticity

J � torsional constant

The critical moment is proportional to both the lateralbending stiffness EIy /L and the torsional stiffness of themember GJ/L.

For the case of an open section, such as a wide-flange orI-beam section, warping rigidity can provide additional tor-sional stiffness. Buckling of a simply supported beam of opencross section subjected to uniform bending occurs at thecritical bending moment, given by

Mcr �

L√EIyGJ



BEAM FORMULAS

TABLE 2.6 Crescent-Beam Position Stress Factors

(see Fig. 2.23)†

Angle �,k

degree Inner Outer

10 1 � 0.055 H/h 1 � 0.03 H/h20 1 � 0.164 H/h 1 � 0.10 H/h30 1 � 0.365 H / h 1 � 0.25 H/h40 1 � 0.567 H / h 1 � 0.467 H/h

50 1 � 0.733 H/h

60 1 � 1.123 H/h

70 1 � 1.70 H/h

80 1 � 2.383 H/h

90 1 � 3.933 H/h

†Note: All formulas are valid for 0 � H/h 0.325. Formulas for the innerboundary, except for 40 degrees, may be used to H/h 0.36. H � distancebetween centers.

2.531 �(0.2939 � 0.7084 H/h)1/2

0.3542

2.070 �(0.4817 � 1.298 H/h)1/2

0.6492

1.756 �(0.2416 � 0.6506 H/h)1/2

0.6506

1.521 �(0.5171 � 1.382 H/h)1/2

1.382

where Cw is the warping constant, a function of cross-sectional shape and dimensions (Fig. 2.25).

In the preceding equations, the distribution of bendingmoment is assumed to be uniform. For the case of a nonuni-form bending-moment gradient, buckling often occurs at

Mcr �

L √EIy �GJ � ECw

2

L2

BEAM FORMULAS 89



BEAM FORMULAS

90 CHAPTER TWO

FIGURE 2.24 (a) Simple beam subjected to equal end moments.(b) Elastic lateral buckling of the beam.

a larger critical moment. Approximation of this criticalbending moment, Mcr� may be obtained by multiplying Mcr

given by the previous equations by an amplification factor

where Cb �12.5Mmax

2.5Mmax � 3MA � 4MB � 3MC

M �cr � Cb Mcr



BEAM FORMULAS

BEAM FORMULAS 91

and Mmax � absolute value of maximum moment in theunbraced beam segment

MA � absolute value of moment at quarter point ofthe unbraced beam segment

MB � absolute value of moment at centerline of theunbraced beam segment

MC � absolute value of moment at three-quarterpoint of the unbraced beam segment

FIGURE 2.25 Torsion-bending constants for torsional buckling.A � cross-sectional area; Ix � moment of inertia about x–x axis;Iy � moment of inertia about y–y axis. (After McGraw-Hill, NewYork). Bleich, F., Buckling Strength of Metal Structures.



BEAM FORMULAS

Cb equals 1.0 for unbraced cantilevers and for memberswhere the moment within a significant portion of theunbraced segment is greater than, or equal to, the larger ofthe segment end moments.

COMBINED AXIAL AND BENDING LOADS

For short beams, subjected to both transverse and axialloads, the stresses are given by the principle of superpositionif the deflection due to bending may be neglected withoutserious error. That is, the total stress is given with sufficientaccuracy at any section by the sum of the axial stress andthe bending stresses. The maximum stress, lb/in2 (MPa),equals

where P � axial load, lb (N )

A � cross-sectional area, in2 (mm2)

M � maximum bending moment, in lb (Nm)

c � distance from neutral axis to outermost fiber atthe section where maximum moment occurs,in (mm)

I � moment of inertia about neutral axis at thatsection, in4 (mm4)

When the deflection due to bending is large and the axial load produces bending stresses that cannot be neglect-ed, the maximum stress is given by

f �P

A�

Mc

I

92 CHAPTER TWO



BEAM FORMULAS

BEAM FORMULAS 93

where d is the deflection of the beam. For axial compres-sion, the moment Pd should be given the same sign as M;and for tension, the opposite sign, but the minimum valueof M � Pd is zero. The deflection d for axial compressionand bending can be closely approximated by

where do � deflection for the transverse loading alone, in(mm); and Pc � critical buckling load 2EI / L2, lb (N).

UNSYMMETRICAL BENDING

When a beam is subjected to loads that do not lie in a planecontaining a principal axis of each cross section, unsym-metrical bending occurs. Assuming that the bending axis ofthe beam lies in the plane of the loads, to preclude torsion,and that the loads are perpendicular to the bending axis, topreclude axial components, the stress, lb/in2 (MPa), at anypoint in a cross section is

where Mx � bending moment about principal axis XX,in lb (Nm)

My � bending moment about principal axis YY,in lb (Nm)

f �Mx y

Ix

�My x

Iy

d �do

1 � (P/Pc)

f �P

A� (M � Pd)

c

I



BEAM FORMULAS

94 CHAPTER TWO

x � distance from point where stress is to becomputed to YY axis, in (mm)

y � distance from point to XX axis, in (mm)

Ix � moment of inertia of cross section about XX,in (mm4)

Iy � moment of inertia about YY, in (mm4)

If the plane of the loads makes an angle � with a principalplane, the neutral surface forms an angle � with the otherprincipal plane such that

ECCENTRIC LOADING

If an eccentric longitudinal load is applied to a bar in theplane of symmetry, it produces a bending moment Pe,where e is the distance, in (mm), of the load P from thecentroidal axis. The total unit stress is the sum of thismoment and the stress due to P applied as an axial load:

where A � cross-sectional area, in2 (mm2)

c � distance from neutral axis to outermost fiber, in(mm)

f �P

A�

Pec

I�

P

A �1 �ec

r 2 �

tan � �Ix

Iy

tan �



BEAM FORMULAS

BEAM FORMULAS 95

I � moment of inertia of cross section about neutralaxis, in4 (mm4)

r � radius of gyration � , in (mm)

Figure 2.1 gives values of the radius of gyration for severalcross sections.

If there is to be no tension on the cross section undera compressive load, e should not exceed r2/c. For a rectangularsection with width b, and depth d, the eccentricity, there-fore, should be less than b/6 and d/6 (i.e., the load shouldnot be applied outside the middle third). For a circular crosssection with diameter D, the eccentricity should not exceedD/8.

When the eccentric longitudinal load produces a deflec-tion too large to be neglected in computing the bendingstress, account must be taken of the additional bendingmoment Pd, where d is the deflection, in (mm). This deflec-tion may be closely approximated by

Pc is the critical buckling load 2EI/L2, lb (N).If the load P, does not lie in a plane containing an axis

of symmetry, it produces bending about the two principalaxes through the centroid of the section. The stresses, lb/in2

(MPa), are given by

where A � cross-sectional area, in2 (mm2)

ex � eccentricity with respect to principal axis YY,in (mm)

f �P

A�

Pexcx

Iy

�Peycy

Ix

d �4eP/Pc

(1 � P/Pc)

√I/A



BEAM FORMULAS

96 CHAPTER TWO

ey � eccentricity with respect to principal axis XX,in (mm)

cx � distance from YY to outermost fiber, in (mm)

cy � distance from XX to outermost fiber, in (mm)

Ix � moment of inertia about XX, in4 (mm4)

Iy � moment of inertia about YY, in4 (mm4)

The principal axes are the two perpendicular axes throughthe centroid for which the moments of inertia are a maxi-mum or a minimum and for which the products of inertiaare zero.

NATURAL CIRCULAR FREQUENCIESAND NATURAL PERIODS OF VIBRATION OF PRISMATIC BEAMS

Figure 2.26 shows the characteristic shape and gives con-stants for determination of natural circular frequency � andnatural period T, for the first four modes of cantilever,simply supported, fixed-end, and fixed-hinged beams. Toobtain �, select the appropriate constant from Fig. 2.26 and multiply it by . To get T, divide the appropriate constant by .

In these equations,

� � natural frequency, rad/s

W � beam weight, lb per linear ft (kg per linear m)

L � beam length, ft (m)

√EI/wL4√EI/wL4



BEAM FORMULAS

97

FIG

UR

E 2

.26

Coe

ffici

ents

for

com

putin

g na

tura

l ci

rcul

ar f

requ

enci

es a

nd n

atur

al p

erio

ds o

f vi

brat

ion

ofpr

ism

atic

bea

ms.



BEAM FORMULAS

98 CHAPTER TWO

E � modulus of elasticity, lb/in2 (MPa)

I � moment of inertia of beam cross section, in4 (mm4)

T � natural period, s

To determine the characteristic shapes and natural peri-ods for beams with variable cross section and mass, use theRayleigh method. Convert the beam into a lumped-masssystem by dividing the span into elements and assuming themass of each element to be concentrated at its center. Also,compute all quantities, such as deflection and bending moment,at the center of each element. Start with an assumed charac-teristic shape.]



BEAM FORMULAS

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