Steel Designers Manual (6th Edition) - Bending Moment, Shear and Deflection Tables

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

Bending moment, shear and deflection

1077

CA NTIL EV&S

'iw L

c b—4

a

I'

-

L

Wx2 Wa

Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Mmax 7 W

Frjp Izçjg cwrved—raight I—

°

W(&+/8a2b+/2ab3b3)

dmaxfJ(' +Ja) ___ a

b— c— L.

24E1

_______________________ 2W

I/A J/Q

____ ____ L

M

N RAW

i aj ______

4.— curved

dnaxjx

.4strai'ht f_-

c/C /5(1

umax. ,(/#

5b


1078

Bending moment, shear and deflection CA NT/LEVERS

w

9

2W

I

________L________ -I

________L


M _T()_2]

1

/SbI

dmax Jf(i+ i7)

IA

a

b—

L

'I

Mmax = w( 4)

i-.--— curved —H straight k—

llWp—j °'C 60E1

b

Mx IX

IIi

dmox

W(2#SOo2b,4Oab2-H/b) 60E1

a L

s.j

Mx = MC

M,,,x— P a

No shears

A[

I

1

N. B. For ant/—clockwise moments

the deflect/on is upwards.

k—

—curved

SEX

d=E2'(,

Sb'

fCU/hLk C 2E1



SIMPLY SUPPORTED BEAMS

w/z

IA

T

B

L

L


RB

//

—

N

T

R8 W

RA=RB T

= S WL' "max 384 El

94 R5

dà,cx.=

Wa fSL2— c2j

_

ien

2/

96(1

012

2°I'ZL/

when x1—

.;—

RAfl\

'RB

_______ RB

When xa

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

dmax j.j (8?_4Lb2#b3)

d4where mx/L and naa/L

1079


1080


L

M Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

x,=0•5774L

II

.x;.

— WL/6

I

R8

= WIS RB 2W/S

RA=RB =

Jdmax

4,dmcx dinax

2)

Mx = Wx(

A4nqx. —O•/28WL when

84 p8

0.0/304 wi!

WL5

60(1

51

when x =05/PJL

AA'F T.—a

"A

b L

BA

-F a

B

I' 4 ('- 3J "mar.

Mx

Ij x 2X2

L"YLJ

Mmax' Wi/i2 A5

RR8 wft j dc,s d, ur=ëi(85,7aLL4c2L4I)

i,dmax. —3iffk1



SIMPLY SUPPORT(O BEAMS

-

N yzW/a ,l.

ti.

£k"\ MCreated on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

'"cx. 6

N

Lb

Mmcx _(_m# 'j) iW,en x—

NRa

'A=B= W/2

R8 —

Wm

lOab# 5b2)

6

_____

RB

z_ RA[

when x= a/i—

_______IA

— R8—w/2

4/max. B

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

.7

1081


1082



p

P

1 L

pa

V

A4nc P


M,,,— T I

I

I

I

IB

I

RA-R8-P

'4—R8— .

J/nax.

J/nax. °;nax.

-

PL3

a>c

"C ,,Ppib*Zc) L MD—

L

I

I

— Pb/L

Pc(b#2a)

Po/L

P(b *)

LH

L

always occurs within

00774 L of tfie centre of the beo,n

When ba, d

PL3rsa

Ia i7

centre 48E1L L (LII This value is a/ways with/n

S % of the maximum value.

Pot central deflection odd the values for each P derived from the formula in the adjacent diagram.



PL

Mmax r Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

RAI

I I

=P

MC Mf —

PL

MD

JB

dp

dinax.

dmczx. 23PL3 648(1

din ax. — SJPL

A*C

MC=ME

— — 2P

/9 P1!

dmax = 304(1

4/PL3

768(1

SPL

1083


1084


SiMPLY SUPPOATED BEAMS

pp p p A

F

b

pPppPP

AB

LA

P

(n—I) forces B

'9

MDME!fL Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

MCfl

When n is odd. (nL /) p1 Mmax — When n is even. Mmcx. — n. PL/8

'4 B("')'%

A4 —A8 "2P

;,2'max.

it /

When n is odd

6JPL dmax. 1000E%

PL3 r

R8

i7r

When n /5 even

dmaxiuu,p4,. . nft_ :# )J TOTAL LOAD

When n >10, consider the load uniformly distributed The 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

2 S 4 5

0 250O 0.3333

7 8 9

O4286

6

/0

03750

C

k

0.1250

O• 0/05

0/I/I

00118 00/24 0.0/26

Q.4 04/67

0.1250

0•4375

0•1250

04444

0/236

OO/27 0•0/28 00/28 00l29

O•4500

0.1250

0•0/29

01200

O•IZSO

0/224



SIMPLY SUPPORTED BEAMS L

CI BA

A

a 1

L

+ b—I

'.1

MA(, MA

___________'M8 ®M4>M5

MA

Al8

_ Al. GIL MC8 — Al . bIL Al4 1jt'—M8

(Al8 antic/ockw,M8

_________________

Va R4(

PAl Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Shear diagram when MA "M8

Vs Al4 — MB

A4 A5 M/L As shown

M.cb 'a b' dc-31 (zz)

When M4M8. ML2

dmcxaii —y

For anti-clockwise moments the deections are reversed 2nd degree_parabola. W

____________________________ Complement of parabola.

L

L

KiI

Mx -i (m4-2m#m) —

RAflJ

'°1

Mx Mmq

R

A4—R5—W/2

(mJm2#4m1_2m4) .LfL

/6

______ ,IA_R8_W/z

dmqx

a'maz — 61WL3

dmax —

28W1

1085


1086



w unit /oao' C

A

0

B

/W.unFt /oaoç

CADBfl L NH A5


-

MA=MB=RD L,

AAI

—4..',,

rTA

_

dc=o_4(jn3,Ln2_ i) Where

(s-)

/W.unit

CiA

-HNI

d4f( RA=RBWN

L

boo

BAD

'1QH-

fC

wL2 N

/6(1 w— unit

0 BE IQH

—j O•5774L

L _4J4 A Max. upward deflection is at 0.

NIA ___

W(L#N)(LN)

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


Bending moment, shear and deflection 8(J/LT—/N BEAMS

wI2

1p1/W

L

-4

EN

WL

MA=MB

WL

MC Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

b-

--

H

4B —.10g/L

W12

= = w/2

WI?

.—O s8L—.12/L p—

I /dflfl

____ = WI!

Wa

-J

384(1

/

I

vinax=4f% (L—OJ

V4—a-4..—b- "+ c—4

'-jvl

'Ak—-

-

L— in

MA

MA,b [e(4L-Je)- c 31'4L —Jc)]

MA=-.in (Jnr"—em+o)

B,bLY4t--7°" (4L —saj M8_ jrn2 (4-3m) ,'Mmax

'L,,2( 3si,) 2 /2

RA

When r is the

A4

sinp/e

M-M L

k—x- When x=2(n,3—2m42)

support reaction

M-M

_r3# L

RA =

W(m-2) 8 = Wm(2-m) 2

2ri,

dna.

u When a = c.

When

a=4/2 and x1=O-445L WI! 3JJ(%

.Ey(L3#2Lba #4Lc?—8a) c/C =

WI!

384(1

1087


1088

Bending moment, shear and deflection B/JILT-/N BEAMS

LA

A

8

L

C

___

M NMB

MA ___________M8 MA V - WL 7/Qx 9x

#Mmax

WL/233 N5Qn

SWL MMWL//6

)

x=OS5L


MA = —WI//S M8 =— WL/iO -

RB

R8

RA=RB=W/2

R=O.SW R8=O7W

—4J•22LF—O S6L —Ø22L

WI! max. when x, = 0• 525L

1.4W!!

—

- 384E1

W/2 2W/L

La- b—+_aJ L—4 MA

M8

(SL44aL_402)

HI'a

____C F

I.

MA = M8 -WL//6

M=WL/48

JRB

R8

W/2

A = R3 W/Z

. L/2 —H ¼ O-WL3 364(1



B(//LT-/N BEAMS 2W/a

fwTw/a L

____

b L

NMa

MAV

MA =-J'2(Jo/obL)

MA

M3 mAC. Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

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

InCB.Mx RB.X#M5

AB W/2 ____

A = (/OL—sLa42a)

I /d____ ____ 4

dmax =

2 , prO,,

—

Wa /

-Y HJ W/2

W/a

—a I

b

W

W/2

b

-f

L

I.

1NMB MA Maz_R(4L_3a)

L

MA

.1

MB

MA — _!10L_15#502)

M3 _lOL2(5L4t

= = W/2 =

(/oL-/sLa'#8a)

R = '/5L —ec) dmax =

(/SL—Ma)

1089


1090


BUILT—/N BEAMS parabolic total /oao W

JM8


NMB

MAV

A -M5=-WL//O

Aq rM8i.._WL/2O

JRa

RA R8 W/2

W/2

'!4

con,c*ment

parabo/p

F4axJ

O•4WL3

1.3 WL3

d,,, 384E!

dma 384 (2'

r

Any symmetrical load W

a rb 2br rA

______

A

symmetrical diagram

NMB MA = M8 —A,JL where A5 is the area of the 'free' bending moment diagram

2 -2br----—-

aab r&_

CM. f(Ja—L)M8c -M(3b-L) When

A RB

A =R8 = W/2 - - I The fic.re shown

A,

is ha/f the bending

* + moment diagram fr- X, -

H-X--lmaxatC

C 1* and +7 LareC.GI

A5x —AIX1 ZET

Where A' is the area of the fixing moment diagram

'4I = P8 = slope of moment diagram

M#M M#M

Eab

When '2/L = in,

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

For ant/clockwise moments reverse the deflect/ons



BUILT—/N BEAMS

p

p

I-.

L/2

L/2

b

MA L—'

Al5

- MA = - M5 = M = PL/8

A1_____ _____ Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

MA LZ EEE3 M8 M5 = —

MA=Mc

_______ __________

= = /,,2

4mcx p9 dinax = /92 El

8JJ

L

F.

2Pa2b 1!

A__________ =P (iT) (I#2 *)

z

1Q8 =

I.— X

dC— —

__ Po3b3

d _ZPa2b3 WñQnX= JL-Za "3EI(3L2a)2

p

p

p

p

Pa MA/ \MB

- __Pa(L-q) L

MA=MB=_JPLh6

MA—MB—

MC MD = P02,'L

RAfl

LJ

dmax PL = /30(a)7

AI

MC =MD = I

II

PA=P'P

1091


1092


BUILT—/N BEAMS

p

______ fA C P

D '.—L/y —1--L/j —.—L/j.

p

Irg

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

MB MA LF- 5N M8

MA

MA =M8 = — 2PL/9

MC =MD = PL/p

AL Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

p

MA =M = —/9PL/72 MD =1/

PArJ__LJA

I

I

RA =

S PL3

dmax 548?1

pp + + C D £ BL L/4--L/4 L/4 L/4 II A

p

JP/2

4/PL3

PPpp

1I 34C D E PB L/L/4 +L/4 +L/4 4L14

MAVNMB MAL/L NEiNMB MA=MB =r—SPL/,6 = JPL/15

MA =M3=—I/PL/j2 MD = Mf = SPL/32

2P

— PL3 dmax —

96(1

max. = 96(1



BUILT—/N BEAMS

p pp p A CVt L/5 +Lg +L,/s+L4

P

ppppppp

4 (n-i) forces I

I

B

sôcev,naLJ,

MAMALNMB MA M52PL/5

MA=MB=—

P1,'/)

MD wM — PL/5


AArL

HJ

'A B=2'°

Y1F,ennisoø6 13PL3 dmax.=1000 El

Cimox

Whennis even,

/ / 114-4c (,ji

,I

EfIiiiitiIII •:.:i••

COLUMN

LOAD PER SA4NW

4;.' •O4•4•.

O...o.&. ,.

n CONTINUOUS BEAM

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

L. - When n >10, consider the load .nitorrrdy oYstiibuted

-

The load on the outside stringers is carried c'/rect/y by the supports The continuous beam Is assumed to be horizontal at each support The reaction at the supports for each s,oan = W/2. but the maximum

shear force in any span of tM continuous beam = V/J.IAW The value of the fixing moment at each support = — B. WL The value of the maximum positive moment for each span = C. W4 The value of the maximum deflection for each span —0'0O26

Value ofn

A

2 3

02500

4 S

6 7 B 9 10

0•3333

03750 04000 04/67 04286 Q•4375 04444 04500

B

C 0•0 625

00625 0074/ 0078/

00370 00469

Qc//

0•0439

0-0800

0O8/5 00820 00823 00825

0•0400

00408 00 430 004/3 00425

1093


1094


PROPPED CAN T/L(VERS

/w A

W

___ a

B

C

C

-f-b

MA

3L/8 =

—WL ---

— 9W!.

C—

MA =—

f (2n)2wñere a/Ln

#Mmax -n (4-n)]


RA

—--_J P8

P3 =f

1

RAi4J[8n2(4n)] fl)

P3 Hn2(

x/Lm LA'max.

d=j(m —Jm3#2m)

dC(/_I2n#7n2_n)

WL3 dmax.— - ____

/W A

C Web

B

L

MA 'P

MA7tS MA =—

MA=—

AI J2 (o - n 2)

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

a, d=J42/n3(3n?-6b5

When x a.

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

BA

.f- c—.J L°'

b

d2-c2)(29c2_c12)

-____ R8=r5— L Where and r3 are the simple support reactions for the beam (MA being considered positive)


Bending moment, shear and deflection PROPPED CANTILEVERS

W/2 1'

W/2

CU BA —f— a -H L

MA

if = % then between B and U.

Al,. = ftx4xa(4 -Sm+Zm9] +M,n0x. when x=I'4Jm#2rn2, Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

—20)

MAZ_ —H X/L = m Mx(2Qm'—27m#7) 1027SLf x

A 7WL '''max.—

=0 67L] '9:4

RB

Vx i(9_rn2)

= (2i."#soi. 42)

PB =

A

p8 = (2L2_30L#40') - OO0/WL3

El

When x=0•598L W/?

W/2

-L RB MA MA

3WL

#MmaxO0454 WL [When x 0 283L]

=4W T

dmax

RB = *

p_L A-

- 0•0047WL'

(I

When x=O•447L

/3W

B - 32

—ii When

x=0404L

1095


1096

Bending moment, shear and deflection PROPPED CANT/LEVERS

a C

_____ L

BA

iR8 ___S___

A

_____

a L

A5

0•577b

I

0/28 Wa MA

MA

JL

_________________

BQt we en


CandA, Mx=R5.x-j!('x-b)

M5=R5.x =_ j9(si.2_j,2)

(JaL/5aL#2OL2) #Mmox when xb# fI/-

___

RAt NJR5 = 9 (si!—,/)

A8 =('5L —a) W P5

A8 = 2W

__________ C

____ a

" b—'

R3

L

(t+SaL2) W

2W

____________ 4—a______ .. L b____I

H0577aH 0/28 Wa

f—x — Wab

H042Jb—

MA

When ,n=a/L

2

3m#2 I

MC=AB.b Between AandC

_______________ =RAW52/a2 IRS }.—x--H \

A8 = '('/SL —4a) RA=W-RB

MA=_f/0L2-3b2) RA I

Between C and B \ Vx=RA_Wrc24Z

I

N

A3 = [L(//L -/S,i#(5L- a)] RA=W-RB



PROPPED CANTILEVERS

W.iw4,

51 CD

fAA

1.

I

L

Wa2

M5=—2M— F


c-'3'L

8TN

.jwap

d0

WL4f2(ep#4?q#dp3(p#i)J

- Omax.

I

—

54(1 M '1 84 'C 0

8+ CD

JA

L

'1.

1.

—-Pa

M3 -2M4 =—M

=

Tfl

q

4I

- 8 =-

— dmax

I0'O

Ido

= j4p2,pq#Jp #Jq) d,n0x

27(1

a#b)#a2(2# VJ

'-dmax"

1097


1098


PROPPED CANT/LEVERS

W/2

+ AB

na/L *7/q

Wa


MA _—4(2L_a)

MA=— f-(4L—3a)

W'?Qn X< a,

tl =(9,r?x —/2nx# /2x— 4xg2) +Mmax occurs whQn

NB

A r—\

A -(4L?#ZaL— az)

(4L2 #4aL—JoT

R8 W—

I

W

L kO.4/SLtq MA

32

/

Bj

A

a —+—b —+-—a —4

—L——-——--R8

MA7Z RAI\ I—

MArn (5L2+40L—4G2)

#MmaxOO948 WL

A4fl

w

R,

I '?&

A477(2/1!#4aL 4a2) A8

—



1099

PROPPED CANT/LEVERS

!F051

coIr4oleraQnt of parabola

— SWL

SWL

M .('/Om-ZOm#7m) #My,OO888WL,when X Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

0 •3965L

B—

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

02343L

A6

7W

MA

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

M=('Z-# p)maxA1-0./74PL 1•• —

F9 7PL

7(%

dmax'OOOPJ2

ft3

A5

A3=f3b#2L) RAPR5


1100


PROPPED CANT/LEVERS

Pp

'4

P

ACDE ¼+V4+L/4 +-/4

MA

MA MA = —

ISPL

D "E7 Created on 24 July 2007 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

/7PL

JJPL

I

PAL 4P

AJ3 J2

2P

P

Lj

dx 00152 P

P

dmax00209 zr

P PP P .4C DEFBA

P

AC D

L/L/3 Lb

EB*

MA

MA

/9PL

MA=— 48

M0 -.96 M ....SJPL £ 288

dxOQ/69

0nczx. — O.OZdS



PROPPED CANILEVEPS

pppp

P PPPP Pp

D E F19 L/4tL/4tL/44L/ MA

MA=— Ad

forces

B

MAE7ZTJ PL(,,?l)

JSPL

-—

MA=

8n

/S7PL


rT(— 5/2

P8

RAEL (5n2-4n-i)

RB = (34z_4fl#/) whennis

GO•OZZl

!or9e, dmax.

/85(1

Any symmetrical load W

JA

AreaR L AreoS'B MA

____ I.

a=L

® _r—

a>O423L

VAr000 X

M-

If AArea of free B.MDiagram

cO423L

a< O423L A8

P82 =!— L

M1 9(2_6n#sn') MCA (2— 6n# Pn?3ni -Sn

It'B dmax occurs at point corresponding

to Xon M diagram, the area A being equal to the area 0 Vmax = Area SXx

L In Case!, R — SM/ZL

CaseS, A= M/L

1101

Steel Designers Manual (6th Edition) - Bending Moment, Shear and Deflection Tables

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