steel designers manual (6th edition) - bending moment, shear and deflection tables

CA NTIL EV&S

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c b—4 aL I' L -

Wx2

WaMmax 7

W

cwrved—raight I—Frjp Izçjg° W(&+/8a2b+/2ab3b3)

dmaxfJ('24E1

+Ja) _______________________2W

___ I/A J/Q

a b— c— ____ ____L. L

M

N ______RAW

4.— curved .4strai'ht f_-i ajdnaxjx c/C /5(15bumax. ,(/#

Bending moment, shear and deflection 1077

Steel Designers Manual (6th Edition) - Bending moment, shear and deflection tables

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http://sefie.steelbiz.org/DiscussSteelbizContent.aspx?ResourceID=1000052

CA NT/LEVERS

9 w I 2W

________ ________ ________ bL -I L 'I

M _T()_2] Mmax = w( 4)

i-.--— curved —H straight k— IIillWp—j1°'C 60E1 W(2#SOo2b,4Oab2-H/b)

/SbI dmox 60E1dmax Jf(i+ i7)

IA a b— aL L s.j

Mx IX Mx = MCM,,,x— P a

A [I

No shears

1N. B. For ant/—clockwise moments

the deflect/on is upwards.

—curved k— fCU/hLkSEX C 2E1

d=E2'(, Sb'

1078 Bending moment, shear and deflection


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SIMPLY SUPPORTED BEAMS

w/z

L

//— T

N

_ 0122°I'ZL/

when x1—

'RB

When xa

djmt2n(2n)d#n2(242m]When xod4

where mx/L and naa/L

IA BT L

RB

R8W

RA=RB T

= S WL'"max 384 El

94 R5

Wa fSL2— c2jdà,cx.= 96(1

2 /ien .;—

RAfl\_______ RB

dmax j.j (8?_4Lb2#b3)



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MA4nqx. —O•/28WL

when x,=0•5774L

L 84p8

Mx = Wx( 2)—WL/6

II .x;. I= WIS

R8

RB 2W/S

4,dmcx

0.0/304 wi!dinax

when x =05/PJL51AA BAT.—a 'F b -F a

L"A B

I'"mar. 4 ('- 3J

RA=RB =

Jdmax

WL560(1

Ij x 2X2Mx L"YLJ

Mmax'Wi/i2

RR8 wft

j dc,s

A5

d, ur=ëi(85,7aLL4c2L4I)

i,dmax.

—3iffk1



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SIMPLY SUPPORT(O BEAMS

,l. ti.

£k"\M -'"cx. 6

N'A=B= W/2

NRa

lOab# 5b2)

6

_____ RB

RA[— R8—w/2

4/max.

dmxJi(/5a2#2Oabi'.5b2)

N yzW/a-

Lb

Mmcx _(_m# 'j)iW,en x—

WmR8 —

z_when x= a/i—

_______IA

B .7



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P p

Lpa

VA4nc P

I I

a>c

,,Ppib*Zc)"C LPc(b#2a)

MD— L

I I

L HP(b *)

LPot central deflection

odd the values for each Pderived from the formulain the adjacent diagram.

1

M,,,— T

I I

I IB'4—R8— .

J/nax.

PL3°;nax.

RA-R8-P

J/nax.

-

— Pb/L Po/L

always occurs within00774 L of tfie centre of the beo,nWhen ba,

d PL3rsa Ia i7centre 48E1L L (LII

This value is a/ways with/nS % of the maximum value.



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PLMmax rPL

MC Mf —SPL

MD

RAI I

I JB

dp

d 23PL3mczx. 648(1

/9 P1!dmax = 304(1

— — 2P

4/PL3768(1

= P

dinax.

d SJPLin ax. —

MC=ME

A*C



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SiMPLY SUPPOATED BEAMS

pPppPP P

LA (n—I) forces BA

'9

pp p pb F AB

MCflMDME!fL

A4 —A8 "2P

;,2'max.

When n is odd.(nL /) p1

Mmax —When n is even.

Mmcx. — n. PL/8

R8'4B("')'%

When n is oddPL3 r i7r it /

When n /5 even

dmaxiuu,p4,. . nft_ :# )J

d 6JPLmax. 1000E%

TOTAL LOAD

When n >10, consider the load uniformly distributedThe reaction at the supports = W/2, but the maximum SE

at the ends of the beam — W(n;/)11,AW

The value of the maximum bending moment — C. WL

The value of the deflection at the centre of the span — k.

Value otn A C k2 0 250O 0.1250 O• 0/05S 0.3333 0/I/I 001184 03750 0.1250 00/245 Q.4 01200 0.0/266 04/67 O•IZSO OO/277 O4286 0/224 0•0/288 0•4375 0•1250 00/289 04444 0/236 00l29/0 O•4500 0.1250 0•0/29



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L '.1

A CI BA MA(,a + b—I ___________

1 L MA ®M4>M5 'M8

MA Al8

_ Al4 1jt'—M8Al. GIL MC8 — Al . bIL (Al8 antic/ockw,M8

_________________ Shear diagram when MA "M8

PAl Va R4( Vs

A4 A5 M/L Al4 — MB

As shown

M.cb 'a b' When M4M8.dc-31 (zz) ML2For anti-clockwise moments dmcxaii —ythe deections are reversed ____________________________2nd degree_parabola. W Complement of parabola.KiI '°1

L L

Mx -i (m4-2m#m) Mx (mJm2#4m1_2m4)— Mmq .LfL

/6RAfl ______J RA4—R5—W/2 ,IA_R8_W/z

dmqx

a' 61WL3 28W1maz — dmax —



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w unit /oao'

C A 0 B

-

AAI

_

dc=o_4(jn3,Ln2_ i)(s-)

Where

/W.unitboo

CiA BAD-HNI L '1QH-

___ W(L#N)(LN)

m.x/L ,,1.N/L

/W.unFt /oaoçCADBflL NHA5

MA=MB=-

—4..',,rTA

RDL,

RA=RBWNd4f(fC

wL2 N/6(1

w— unit

_4J4

—j O•5774L0 BEL IQH

AMax. upward deflection is at 0.

NIA



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8(J/LT—/N BEAMS

1p1

WL

WLMC

4B WI?

—.10g/L .—O s8L—.12/L p—

WI!=384(1

V4—a-4..—b- "+ c—4

MA

MA,b [e(4L-Je)- c 31'4L —Jc)]

B,bLY4t--7°" (4L —sajRA

When r is the sinp/e support reactionM-M M-MA4 L _r3# L

/W wI2 W12

-- b -H L -4

ENMA=MB

= = w/2

____ I /dflfl

-J Wa / I

vinax=4f% (L—OJ

'-jvl'A

k—- L— -in

MA=-.in (Jnr"—em+o)M8_ jrn2 (4-3m) ,'Mmax'L,,2( 3si,)/2

k—x- When x=2(n,3—2m42)2

RA= W(m-2) 8 = Wm(2-m)2 2ri,

dna.

uWhen a=4/2 and x1=O-445L

WI!3JJ(%

WI!c/C = 384(1

When a = c.

.Ey(L3#2Lba #4Lc?—8a)



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B/JILT-/N BEAMS

LA L8 A C ___

M

MA ___________M8 MA V NMB- WL 7/Qx 9x ) M - SWL

#Mmax WL/233 N5Qn x=OS5L MWL//6MA = —WI//S M8 =— WL/iO

-

R8 RB

R=O.SW R8=O7W RA=RB=W/2

—4J•22LF—OS6L —Ø22L

WI! — 1.4W!!max. - 384E1

when x, = 0• 525L

W/2 2W/L HI'a

La- b—+_aJ ____CL—4 F I.

MA M8 (SL44aL_402) MA = M8 -WL//6

M=WL/48

JRBR8 W/2 A =R3 W/Z

. L/2 —H ¼

O-WL3364(1



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B(//LT-/N BEAMS

fwTw/a____L

MAV NMaMA

AB W/2

____ I /d________4

Wa /dmax =

2W/a

bL

MA =-J'2(Jo/obL)M3

mAC. Mx =P.X+M3-2W(x.-bP

InCB.Mx RB.X#M5

A = (/OL—sLa42a)2, prO,, —

W/2 W/a W/2-Y HJ—a I b -f

L

1NMBMA Maz_R(4L_3a)

= = W/2

dmax = (/SL—Ma)

W

bI. L .1

MAMB

MA — _!10L_15#502)

M3 _lOL2(5L4t

= (/oL-/sLa'#8a)

R = '/5L —ec)



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BUILT—/N BEAMSparabolic total /oao W

JRa'!4 W/2

F4axJ1.3 WL3

d,,, 384E!

Any symmetrical load WrA ______

symmetrical diagram

NMBMA = M8 —A,JL

where A5 is the area of the 'free'bending moment diagram

ARB

A =R8 = W/2

A,- - I The fic.re shown

is ha/f the bending* + moment diagram

C 1* and +7fr- X, -H-X--- LareC.GI

A5x —AIX1lmaxatC ZETWhere A' is the area of the fixing

moment diagram

RA R8 W/2

rA

O•4WL3dma 384 (2'

a rb2br

2 -2br----—-

aab r&_

CM. f(Ja—L)M8c -M(3b-L)When

'4I=P8 = slope of moment diagramM#M M#M

EabWhen '2/L = in,

M. L2m2(/_m)2(1_2m)2E%

For ant/clockwise momentsreverse the deflect/ons

con,c*mentparabo/p

JM8A -M5=-WL//O

MAV NMBAq rM8i.._WL/2O



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BUILT—/N BEAMSp p

L/2 L/2 I-. b 8JJF. L

MA L—' Al5 MA LZ EEE3 M8- MA = - M5 = M = PL/8 MA=- M5 = —

_______ Mc2Pa2b

__________ 1!A1_____ _____ A__________

=P (iT) (I#2 *)= = /,,21Q8 = z

4mcx __I.— Xp9 d — Po3b3dinax = /92 El C —

d _ZPa2b3 WñQnX="3EI(3L2a)2 JL-Za

p p p p

Pa MA/ \MB- __Pa(L-q) MA=MB=_JPLh6MA—MB— L

MC MD = P02,'L MC =MD =

RAfl AI ILJ I IPA=P'P

PLdmax = /30(a)7



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BUILT—/N BEAMSP p p p______ IrgfA C D

'.—L/y —1--L/j —.—L/j.

MA MB MA LF- 5N M8MA =M = —/9PL/72

MD =1/

PArJ__I

LJARA = JP/2

p p p P P p pII + + 1I

MAVNMB MAL/L NEiNMBMA =M3=—I/PL/j2MD = Mf = SPL/32

max. = 96(1

l,L/3 P.-L/3L2

MA =M8 = — 2PL/9

MC =MD = PL/p

AL I

S PL3dmax 548?1

4/PL3

A C D £ BLL/4--L/4 L/4 L/4

34C D E PBL/L/4 +L/4 +L/4 4L14

MA=MB =r—SPL/,6= JPL/15

2P

— PL3dmax — 96(1



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BUILT—/N BEAMS

HJ-

Y1F,ennisoø6 ,I / /Cimox 114-4c (,ji

Whennis even,

•:.:i•• O...o.&. ,. 4;.' •O4•4•.EfIiiiitiIIIn CONTINUOUS BEAM

}d L/n#-+-L/n4L/n+-L/-#L1-+-44L

L. - When n >10, consider the load .nitorrrdy oYstiibuted -The load on the outside stringers is carried c'/rect/y by the supportsThe continuous beam Is assumed to be horizontal at each supportThe reaction at the supports for each s,oan = W/2. but the maximum

shear force in any span of tM continuous beam = V/J.IAWThe value of the fixing moment at each support = — B. WL

The value of the maximum positive moment for each span = C. W4The value of the maximum deflection for each span —0'0O26

Value ofn A B C2 02500 00625 0•0 6253 0•3333 0074/ 003704 03750 0078/ 00469S 04000 0-0800 0•04006 04/67 Qc// 0•04397 04286 0O8/5 00408B Q•4375 00820 00 4309 04444 00823 004/310 04500 00825 00425

p pp pA C V t PL/5 +Lg +L,/s+L4

ppppppp4 (n-i) forces B

I I sôcev,naLJ,

MA LNMB MAMA M52PL/5MD wM — PL/5

AArLMA=MB=— P1,'/)

'A B=2'°

d = 13PL3max. 1000 El

COLUMN LOAD PER SA4NW



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WL= — ---

— WL3dmax. - ____

PROPPED CAN T/L(VERS

/WA C B

WebL

BAb .f- c—.JL°'

MA7tSMA =— d2-c2)(29c2_c12)

-____L R8=r5—

Where and r3 are the simplesupport reactions for the beam

(MA being considered positive)

w/

MA

W

A C B

3L/8 -— 9W!.C —

___ Ca -f-b

MA=— f (2n)2wñere a/Ln

#MmaxRA -n (4-n)]

RAi4J[8n2(4n)]1

P3 Hn2( fl)

dC(/_I2n#7n2_n)

—--_J P8

P3 =fx/Lm

LA'max.

d=j(m —Jm3#2m)

MA

'P

MA=—

A I

J2 (o - n 2)

= f (n'-n-pi.e)

a, d=J42/n3(3n?-6b5When x a.

dj[2p4—p3n (n-6n,'.8)+_________ pn2(jn2_8n#6)]



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PROPPED CANTILEVERS

W/2 W/21'

C U BA—f— a -H

L

MA

if = % then between B and U.Al,. = ftx4xa(4 -Sm+Zm9]+M,n0x. when x=I'4Jm#2rn2,

—20)

RB

MAZ_1027SLf

X/L = mx —HMx(2Qm'—27m#7)A 7WL —'''max.

'9:4=0 67L]

= (2i."#soi. 42)p8 = (2L2_30L#40')

Vx i(9_rn2)A PB =

- OO0/WL3El

When x=0•598L

W/? W/2

-L

MA

RB

3WLMA

#MmaxO0454 WL[When x 0 283L]

4W= T RB = * p_LA - /3WB - 32

- 0•0047WL'dmax (I

When x=O•447L—ii

When x=0404L



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PROPPED CANT/LEVERS

a_____ C B A A _____a

L iR8 L A5___ ___ 0•577bI S

0/28 Wa

MAMA

_________________ JLBQt we en

CandA, Mx=R5.x-j!('x-b) M5=R5.x -(JaL/5aL#2OL2) =_ j9(si.2_j,2)

#Mmox when xb# fI/-

___ RAt NJR5A8 =('5L —a) = 9 (si!—,/)

W P5 A8 = (t+SaL2)

2W2W W

__________ ____________ -____ C ______ ____a " b—' 4—a .. b I

L R3 L

H0577aH 0/28 Waf—x —Wab H042Jb—

MA

When ,n=a/L 2 3m#2 I

MC=AB.b MA=_f/0L2-3b2)Between AandC RA I

=RAW52/a2 I_______________ Between C and B \

}.—x--H \ IRS Vx=RA_Wrc24Z N

A8 = '('/SL —4a) A3 = [L(//L -/S,i#(5L- a)]RA=W-RB RA=W-RB



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fAA 51 CD1. L

IWa2

M5=—2M— F

c-'3'L 8TN

IJA 8+ CD

'1. L

M'184 'C 0

1.

—-Pa=

q

- 8 =-

= j4p2,pq#Jp #Jq)d,n0x 27(1

PROPPED CANTILEVERS

W.iw4,

.jwap

d0 WL4f2(ep#4?q#dp3(p#i)J-Omax. —

54(1

TflM3 -2M4 =—M

4I

— dmax

IdoI0'O

a#b)#a2(2# VJ'-dmax"



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PROPPED CANT/LEVERS

W/2

AB

Wa

MA _—4(2L_a)

NBA -(4L?#ZaL— az)

W

L kO.4/SLtq

MA

#MmaxOO948 WLA4flR,

A32 B j/

+

na/L *7/q

MA=— f-(4L—3a)W'?Qn X< a,tl =(9,r?x —/2nx# /2x— 4xg2)+Mmax occurs whQn

A r—\

(4L2 #4aL—JoT

R8 W—

wIa —+—b —+-—a —4

I— —L——-——--R8

MA7ZMArn (5L2+40L—4G2)

RAI\ I '?&

A477(2/1!#4aL 4a2)A8 —



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SWL

M .('/Om-ZOm#7m)#My,OO888WL,when X

0 •3965L

7PL7(%

dmax'OOOPJ2 ft3

coIr4oleraQnt of parabola!F051

— SWL

Mx =(-4om+8om-om4I7m)#Mmax*O0399WL,whQn x —

02343L

A6

MAW-' lz' ,nax.MA-O./93PLwMn baO577L

M=('Z-# p)maxA1-0./74PL

A3=f3b#2L) RAPR5

PROPPED CANT/LEVERS

7WB —

MA

A51•• —

F9



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PROPPED CANT/LEVERS

AC D EB*L/L/3 Lb

Pp P

A C D E¼+V4+L/4 +-/4

ISPLMA = —

/7PL JJPLD "E7PAL

I

PAJ3 J2

dmax00209 zr

P PP PLj.4C DEFBA

0nczx. — O.OZdS

MA

'4

MA

4P 2P

dx 00152

P P P

MA MA

/9PLMA=— 48M -. M ....SJPL0 96 £ 288

dxOQ/69



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PROPPED CANILEVEPS

p p p pD E F19

L/4tL/4tL/44L/MA

P PPPP Ppforces B

MAE7ZTJPL(,,?l)

MA= 8nJSPL

MA=— -—Ad /S7PLrT(— 5/2

P8

GO•OZZl

RAEL (5n2-4n-i)

RB = (34z_4fl#/)

whennis!or9e, dmax. /85(1

Any symmetrical load W

AreaR L AreoS'B

MAVAr000 X

If AArea of free B.MDiagramM-A8

P =!—82 L

JA ____I.

a=L

a>O423L® _r—cO423L

a< O423L

M1 9(2_6n#sn')MCA (2— 6n# Pn?3ni

-Sn

It'B

LIn Case!, R — SM/ZL

CaseS, A= M/L

dmax occurs at point correspondingto Xon M diagram, the area Abeing equal to the area 0

Area SXxVmax =



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steel designers manual (6th edition) - bending moment, shear and deflection tables

Documents

deflection tablesdiscuss

b3bending moment

dinaxwhen x

o28wlwhen x

f b f ala bimar

x p aa

ia jqa b c

straiht f