tables - bending moments shear forces and deflection tables

CA NTIL EV&S

'iwc b4 a

L I' L -

Wx2

WaMmax 7

W

cwrvedraight IFrjp Izjg W(&+/8a2b+/2ab3b3)

dmaxfJ(' 24E1+Ja)_______________________

2W___

I/A J/Qa 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 tablesDiscuss me ...

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CA NT/LEVERS

9 w I 2W

________ ________________

bL -I L 'I

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

i-.-- curved H straight k IIillWpj1'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 shears1 N. B. For ant/clockwise moments

the deflect/on is upwards.curved k fCU/hLk

SEX C 2E1d=E2'(, Sb'

1078 Bending moment, shear and deflection


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SIMPLY SUPPORTED BEAMS

w/zL

// T

N

_0122I'ZL/

when x1

'RB

When xadjmt2n(2n)d#n2(242m]When xod4

where mx/L and naa/L

IA BT LRB

R8WRA=RB T

= S WL'"max 384 El

94 R5

Wa fSL2 c2jd,cx.= 96(1

2 /ien

.;

RAfl\_______

RB

dmax j.j (8?_4Lb2#b3)



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MA4nqx. O/28WL

when x,=05774L

L 84p8

Mx = Wx( 2)WL/6

II.x;. I

= WISR8

RB 2W/S4,dmcx

0.0/304 wi!dinaxwhen x =05/PJL51AA BA

T.a 'F b -F aL"A B

I'"mar. 4 ('- 3J

RA=RB =

JdmaxWL5

60(1

Ij x 2X2Mx L"YLJ

Mmax'Wi/i2

RR8 wftj dc,s

A5

d, ur=i(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[ R8w/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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SIMPLY SUPPORTED BEAMSP p

L pa

VA4nc P

I I

a>c

,,Ppib*Zc)"C L

Pc(b#2a)MD L

I I

L HP(b *)L

Pot central deflectionodd the values for each Pderived from the formulain the adjacent diagram.

1

M,,, T

I I

I IB'4R8 .

J/nax.PL3

;nax.

RA-R8-PJ/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 r PLMC Mf SPLMD

RAI II JB

dpd 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 BEAMSpPppPP P

LA (nI) forces BA'9

pp p pb F AB

MCflMDME!fL

A4 A8 "2P;,2'max.

When n is odd. (nL /) p1Mmax When n is even.

Mmcx. n. PL/8

R8'4B("')'%

When n is oddPL3 r i7r it /When n /5 even

dmaxiuu,p4,. . nft_ :# )Jd 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,AWThe 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 OIZSO OO/277 O4286 0/224 00/288 04375 01250 00/289 04444 0/236 00l29/0 O4500 0.1250 00/29



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L '.1

A CI BA MA(,a + bI ___________

1 L MA M4>M5 'M8MA Al8

_

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

_________________

Shear diagram when MA "M8

PAl Va R4( VsA4 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 '1L L

Mx -i (m4-2m#m) Mx (mJm2#4m1_2m4) Mmq .LfL/6RAfl ______J R

A4R5W/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.unit booCiA BAD

-HNI L '1QH-

___ W(L#N)(LN)

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

/W.unFt /oaoCADBflL NHA5

MA=MB=-

4..',,rTA

RDL,

RA=RBWNd4f(fC

wL2 N/6(1w unit

_4J4j O5774L0 BE

L IQHAMax. 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

V4a-4..b- "+ c4

MA

MA,b [e(4L-Je)- c 31'4L Jc)]B,bLY4t--7" (4L saj

RA

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 / Ivinax=4f% (LOJ

'-jvl'Ak- L -

in

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

kx- When x=2(n,32m42)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 L 8 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/24J22LFOS6L 22L

WI!

1.4W!!max. - 384E1

when x, = 0 525L

W/2 2W/L HI'a

La- b+_aJ____CL4 F I.

MA M8 (SL44aL_402) MA = M8 -WL//6M=WL/48

JRBR8 W/2 A =R3 W/Z

. L/2 H

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.-bPInCB.Mx RB.X#M5

A = (/OLsLa42a)2, prO,,

W/2 W/a W/2-Y HJ

a I b-f

L

1NMBMA Maz_R(4L_3a)

= = W/2

dmax = (/SLMa)

W

bI. L .1

MA MB

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 WL3d,,, 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/2A,

- - I The fic.re shownis ha/f the bending

* + moment diagramC 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

O4WL3dma 384 (2'

a rb2br

2 -2br-----

aab r&_

CM. f(JaL)M8c -M(3b-L)When

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

_______

Mc 2Pa2b__________ 1!

A1_____ _____ A__________=P (iT) (I#2 *)= = /,,2

1Q8 = z

4mcx__

I. Xp9 d Po3b3dinax = /92 El C d _ZPa2b3 WQnX="3EI(3L2a)2 JL-Za

p p p p

Pa MA/ \MB- __Pa(L-q) MA=MB=_JPLh6MAMB L

MC MD = P02,'L MC =MD =RAfl AI ILJ I I

PA=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/9MC =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 =rSPL/,6= JPL/15

2P

PL3dmax 96(1



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BUILT/N BEAMS

HJ-

Y1F,enniso6 ,I / /Cimox 114-4c (,jiWhennis even,

:.:i O...o.&. ,. 4;.' O44.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 0'0O26Value ofn A B C

2 02500 00625 00 6253 03333 0074/ 003704 03750 0078/ 00469S 04000 0-0800 004006 04/67 Qc// 004397 04286 0O8/5 00408B Q4375 00820 00 4309 04444 00823 004/3

10 04500 00825 00425

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

ppppppp4 (n-i) forces B

I I scev,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

Web LBA

b.f- c.J

L'

MA7tSMA = d2-c2)(29c2_c12)

-____

L R8=r5Where 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)2were a/Ln#MmaxRA -n (4-n)]

RAi4J[8n2(4n)]1

P3 Hn2( fl)

dC(/_I2n#7n2_n)

--_J P8P3 =f

x/LmLA'max.

d=j(m Jm3#2m)

MA

'P

MA=

A IJ2 (o - n 2)= f (n'-n-pi.e)

a, d=J42/n3(3n?-6b5When x a.dj[2p4p3n (n-6n,'.8)+_________

pn2(jn2_8n#6)]



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PROPPED CANTILEVERS

W/2 W/21' C U BA

f a-HL

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 H

Mx(2Qm'27m#7)A 7WL

'''max.

'9:4=0 67L]

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

Vx i(9_rn2)A PB =

- OO0/WL3ElWhen x=0598L

W/? W/2

-LMA

RB

3WLMA

#MmaxO0454 WL[When x 0 283L]

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

- 00047WL'dmax (IWhen x=O447L

iiWhen x=0404L



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PROPPED CANT/LEVERS

a

_____

C B A A _____aL iR8 L A5___ ___

0577bI S

0/28 Wa

MAMA

_________________

JLBQt we enCandA, 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)2W

2W W__________ ____________ -

____

C______ ____a " b' 4a .. b IL R3 L

H0577aH 0/28 Wafx 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 NA8 = '('/SL 4a) A3 = [L(//L -/S,i#(5L- a)]

RA=W-RB RA=W-RB



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fAA 51 CD1. L

I Wa2M5=2M Fc-'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

dmaxIdo

I0'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-(4L3a)W'?Qn X< a,tl =(9,r?x /2nx# /2x 4xg2)+Mmax occurs whQn

A r\

(4L2 #4aLJoTR8 W

wIa +b +-a 4

IL---R8MA7Z

MArn (5L2+40L4G2)

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 PA C D E+V4+L/4 +-/4

ISPLMA = /7PL JJPLD "E7

PALI

PAJ3 J2dmax00209 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 CANILEVEPSp p p pD E F19

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

P PPPP Ppforces B

MAE7ZTJPL(,,?l)MA= 8nJSPLMA= -

Ad /S7PLrT( 5/2

P8

GOOZZl

RAEL (5n2-4n-i)RB = (34z_4fl#/)

whennis!or9e, dmax. /85(1

Any symmetrical load W

AreaR L AreoS'BMA

VAr000 XIf AArea of free B.MDiagramM-

A8

P =!82 L

JA ____I.

a=L

a>O423L _rcO423La< O423L

M1 9(2_6n#sn')MCA (2 6n# Pn?3ni

-SnIt'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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tables - bending moments shear forces 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