M259 WELDMENT STRENGTH EXCEL CALCULATIONS John An Andre drew P. P.E.
Course Content
Revised 22 July 201 2016 6
SUMMARY: This 2 DH ,ourse can ,e used to calculate the/ shear# ,endin# and torsion strenth of many types of welds. The calculations within this live spreadsheet will facilitate the initial desin and enineerin of many types of weldments. !hen various factors are typed into the "nput cells# E$cel will calculate results automatically. automatically. "n addition# add ition# the powerful tool# 9:oal %ee39 can ,e used to optimi;e each C(,u(t&on C(,u(t&on.. CONTENTS: 1. !elds in Tension Tension and %hear 2. !elds in &endin and tor'ue
Spre! S"eet Met"o!: #$ T%pe &n '(ues )or t"e &nput !t$ 2$ Enter$ (. Anser: X &(( 1e ,(,u(te!$ <. Automatic calculations are 1o(! t%pe. t%pe .
(. )on !elds 0UTT WELDS IN TENSION A full full penetration 60 deree *+,utt *+,utt weld is illustrated here. The applied load -P is constant tension.
W"en us&n* E+,e(-s Go( See./ unprote,t t"e spre! s"eet 1% se(e,t&n*: Drop !on 3enu: Too(s 4 rote,t&on 4 Unprote,t S"eet 4 O6$ W"en E+,e(-s Go( See. &s not nee!e!/ restore prote,t&on &t": Drop !on 3enu: Too(s 4 rote,t&on 4 rote,t S"eet$
The strenth of the of this weld is determined ,y the properties of the electrode and the 'uality of the weldin process. The American !eldin %ociety - A!% has esta,lished standards includin weld/ types# sym,ols# dimensions# strenth# endurance# test procedures# and welder classifications. Applied load - P is reacted ,y a tensile tensile stress - %t in the ,utt weld.
Note: E+,e( uses t"e s %31o( 7 8 )or 3u(t&p(&,t&on$
We(! Stren*t" !eld tension allowa,le# %t !eld thic3ness# w !eld )enth# ) Efficiency# e M+ (o!/ rt Stren*t" Part tension allowa,le# 8t Part thic3ness# T Part )enth# ) M+ (o!/
Input 1(600 0.(45 5.000 07 C(,u(t&on St 8 8 L 8 e 2;<;; Input (6000 0.(45 5 C(,u(t&on =t 8 T 8 L
l,sin2 in in
(1s
l,sin2 in in
>?5;;
(1s #
0UTT WELDS IN SHEAR Applied shear load - * is reacted ,y a shear stress - %s in the ,u tt weld.
We(! Stren*t" !eld allowa,le shear# %s !eld thic3ness# w !eld )enth# ) Efficiency# e M+ s"er (o!/ @ @ rt Stren*t" Part allowa,le shear# 8s Part thic3ness# T Part )enth# ) M+ s"er (o!/ @ @ =ILLET WELDS IN SHEAR ;B o) e(!s re )&((et e(!s$ Applied shear load - * is reacted ,y a shear stress stress - %s in the the ,utt weld. !eld allowa,le shear stress %s - l,sin2 l,sin2 8illet weld si;e - w is specified in the weld sym,ol. 8illet weld effective thic3ness - t is measured on the plane at <5 derees as shown ,elow.
Input 1(600 0.500 5 07 C(,u(t&on Ss 8 T 8 L 8 e 2?2;; Input 22000 0.45 5 C(,u(t&on =s 8 T 8 L 25;;
psi in in
(1s
psi in in
(1s
>?5;;
(1s #
0UTT WELDS IN SHEAR Applied shear load - * is reacted ,y a shear stress - %s in the ,u tt weld.
We(! Stren*t" !eld allowa,le shear# %s !eld thic3ness# w !eld )enth# ) Efficiency# e M+ s"er (o!/ @ @ rt Stren*t" Part allowa,le shear# 8s Part thic3ness# T Part )enth# ) M+ s"er (o!/ @ @ =ILLET WELDS IN SHEAR ;B o) e(!s re )&((et e(!s$ Applied shear load - * is reacted ,y a shear stress stress - %s in the the ,utt weld. !eld allowa,le shear stress %s - l,sin2 l,sin2 8illet weld si;e - w is specified in the weld sym,ol. 8illet weld effective thic3ness - t is measured on the plane at <5 derees as shown ,elow.
Input 1(600 0.500 5 07 C(,u(t&on Ss 8 T 8 L 8 e 2?2;; Input 22000 0.45 5 C(,u(t&on =s 8 T 8 L 25;;
psi in in
(1s
psi in in
(1s
2 We(! Stren*t" !eld allowa,le shear# %s !eld si;e# w )enth# ) Efficiency# e S"er An*(e/ A We(! t"&,.ness/ t t M+ s"e s"err (o (o!/ !/ @ @ =&((et We(!s &n S"er =a$imum shear stress is on the plane of minimum weld thic3ness. !eld si;e w %hear Anle# A <5 -de !eld thic3ness# t w > %"?-A t ;$?;? 8
rt Stren*t" Part allowa,le shear# 8s Part thic3ness# T1 Part )enth# ) M+ s"er (o!/ @ @
Input 1(#600 0.25 5 07 C(,u(t&on <5 8 SIN7A ;$#?? 2 8 Ss 8 t 8 L 8 e #92 Input 1000 0.45 < C(,u(t&on =s 8 T# 8 L 5<;;;
WELDS IN COM0INED SHEAR TENSION
Tens&on ens&on (o! (o! =n We(! tens&on tens&on )or,e )or,e =s We(! s"er )or,e )or,e Eu&(&1r&u3 o) =or,es @ori;ontal forces actin riht 2>8n>B%2>8n>B%-C0 C0 + A 8n *ertical *ertical forces actin actin up P
@ori;ontal forces actin left 2>8s>B% 2>8s>B% -A 8s>B% 8s>B% -A -A %"? -A *ertical *ertical forces actin actin down 2>8s>%"?2>8s>%"?-A A D 2>8n>%"?-C0 2>8n>%"?-C0 + A
psi in
!e* &n (1s
psi in in
(1s
P 2>8s>%"? -A D 2>8n>B%-A P 2>8s>%"?-A 2>-8s>B%-A %"?-A>B%-A 8s P>%"? -A 2 Rev 2C ?ov 04
=r,ture (ne An*(e 7 A F !eld shear plane thic3ness !eld si;e# w !eld thic3ness# t !eld shear stress# %s %s
t -in t>B% -A D T>%"? -A w -B% -A D %"? -A 8s -t > ) - P>%"? -A >- -B% -AD%"? -A - w > )
M+ S"er Stress Applied tension# P !eld si;e# w !eld lenth# ) !eld shear anle# A
Input 16C00 l,s 0.( in <.00 in <5.00 de C(,u(t&on %s -P>%"? -A>--B% -AD%"? -A - w > ) Ss ##2>? (1s&n2
E+,e( So('er "f %olver is not installed# select the 9Tools9 drop down menu# pic3 9Add+"ns...9. ou my have to insert the FP disc. The Add+"ns menu shown on the riht will open. hec3 %olver Add+in. %tudy the =a$ %hear %tress "nputs and alculations a,ove. %et !eld shear anle# A <5 as shown ,elow. The !eld shear stress - %s will chane to/ %s 11264
Pic3/ So('er 4 ell containin ##2>? as taret cell 4 &y hanin ell <5$;; %elect M+ 4 So('e < E+,e( So('er C(,u(tes S"er An*(e )or M+ We(! Stress
Con,(us&ons: 1. =a$imum weld shear stress occurs on the plane that is 64.5 derees to the hori;ontal. 2. %olver can ,e used to find the ma$imum weld shear stress and the shear plane anle. 5 AWS Co!e A((o1(e stress &n tens&on n! s"er/ St #/>;; 7 (1&n2 un(ess ot"er&se spe,&)&e!$ =&((et We(! Stren*t" &n Tens&on Input Allowa,le tension stress# %t 1(#600 l,sin2 !eld si;e# w 0.266 in !eld lenth# ) < in C(,u(t&ons %hear Anle# A 64.5 de !eld thic3ness# t w -B% -A D %"? -A t 0.20< in %t -Pa>%"? -A>-B% -AD%"? -A - w > ) %t 1.204 > P -w > ) Allowa,le weld tension load# Pa %t > w > ) 1.204 #2;;; (1s We(! s&e 7 &(( 1e ;$< &n,"es to support *o( (o! o) 22;;; (1s$ %elect/ Too(s 4 Go( See.J4 Pic3 allowa,le load cell &26 a,ove# #2;;; 4 Type To value# 22;;; G &y chanin cell/ Pic3 weld si;e cell &25C# ;$2>> See e+3p(e Go( see. 1e(o:
We(! Tens&on Stren*t" Allowa,le weld tension# %s !eld si;e# w !eld lenth# )
Input 1(#600 0.(45 <
l,sin2 in in
Efficiency# e !eld shear anle# A !eld thic3ness# t t Allowa,le weld shear load# *a
@ rt Stren*t" Part allowa,le shear# 8s Part thic3ness# T1 Part )enth# ) M+ s"er (o!/ @ @
07 C(,u(t&ons <5 de w - %"? -A D B% -A 0.265 in %s > t > ) > e ##5<; (1s Input (6000 1.000 5 C(,u(t&on =s 8 T# 8 L #;;;;
>
psi in in
(1s
WELDS IN COM0INED SHEAR TENSION 0Y CALCULUS C"e,. so('er so(ut&on )or e(! s"er n*(e 7A$ Eu&(&1r&u3 o) =or,es 7 Re)err&n* to t"e !K,ent )&*ure @ori;ontal forces actin riht @ori;ontal forces actin left 8n>%"? -A 8s>B% -A 8n 8s>B% -A %"? -A *ertical forces actin up P P 8s
*ertical forces actin down 2>8n>B% -A D 2>8s>%"? -A 2>-8s>B% -A %"? -A>B% -A D 2>8s>%"? -A P>%"? -A 2
!eld thic3ness# t w -B% -A D %"? -A %hear stress# %s 8s -t > ) %s 8s > %"? -A > - -B% -A D %"? -A - w > ) =a$imum stress when# d%sHA 0 %u,stitutin# B% -2A and# %"? -A>B% -A :ives# %"? -2A TA? -2A 2A A (u* We(! Stren*t" !eld allowa,le shear stress# %s %pot diameter# H =aterial thic3ness# T ?um,er of %pot welds# ?
0 8s I -%"? -A>-- +%"? -A D B% -AD-B% -AD%"? -A > B% -A w>) B% -2A -12>%"?-2A +B%-A2 +1 1(5 de >?$5 !e* %ame as o,tained ,y usin :oal %ee3. Input 1(600 0.( 0.25 5
l,sin2 in in
Efficiency# e
07 C(,u(t&on %pot weld area# A > H2 < A 0.11( in2 (u* e(! s"er stren*t"/ ;$?5 8 Ss 8 A 8 N 8 e <>2? (1s Note/ The pea3 shear stress in a round area is < ( times the averae stress# therefore the strenth reduction factor is 0.45.
(u* We(!
?
Spre! S"eet Met"o!: ne E+,e( 'ers&on 1. Type in values for the Input Dt. 2. E$cel will ma3e the C(,u(t&ons .
E+,e(-s GOAL SEE6 E$celKs# 9:oal %ee39 adLusts one "nput value to cause a alculated formula cell to e'ual a iven value. !hen usin E$celKs :oal %ee3# unprotect the spread sheet ,y selectin/ Hrop down menu/ @ome G 8ormat G Mnprotect %heet G BN !hen E$celKs :oal %ee3 is not needed# restore protection with/ Hrop down menu/ @ome G 8ormat G Protect %heet G BN
GOAL SEE6 EXAMLE We(! Stren*t" !eld tension allowa,le# %t !eld thic3ness# w !eld )enth# ) Efficiency# e M+ (o!/
Input 1(600 0.(45 4.(5( 07 C(,u(t&on St 8 8 L 8 e ;;;;
GOAL SEE6 METHOD Step# &,. ,e(( &t" )r3u( 7H<2 Step2 &,.: Dt 4 W"tI) An(%s&s Step &,.: Go( See. Step< To '(ue: t%pe ;;;; Step5 0% ,"n*&n* ,e((: &,. ,e(( H
l,sin2 in in
(1s
Spre! S"eet Met"o!: E+,e(9? 2;; o(! 'ers&on 1. Type in values for the Input Dt. 2. E$cel will ma3e the C(,u(t&ons.
E+,e(-s GOAL SEE6 E$celKs# 9:oal %ee39 adLusts one "nput value to cause a alculated formula cell to e'ual a iven value. !hen usin E$celKs :oal %ee3# unprotect the spread sheet ,y selectin/ Hrop down menu/ Tools G Protection G Mnprotect %heet G BN !hen E$celKs :oal %ee3 is not needed# restore protection with/ Hrop down menu/ Tools G Protection G Protect %heet G BN
WELDMENT STRENGTH EXCEL CALCULATIONS Course Content o"n An!re $E$ =ILLET WELDS IN 0ENDING AND TORSION @ert&,( L&ne We(! &n 0en!&n* The section modulus of any thin area can ,e appro$imated to that of a line multiplied ,y the width. Input 100 0.(45 (.0 C(,u(t&ons w > H( 12 0.<( H2 1.500 =>c "$ #? %, > w >?
&endin moment# = %inle weld si;e# w !eld lenth# H Area moment of inertia# "$ "$ =a$ distance from line : to end# c c =a$ stress due to ,endin# %, S1 =a$ weld stress per inch due to ,endin# 8w
=
in+l,s in in
(1s&n2 7(1&n2 &n e(! Rev 2C?ov04
L&ne We(! &n Tors&on %ee the fiure# riht. Input Applied Tor'ue# T 6< 8illet weld si;e# w 0.(45 !eld lenth# H (.0 Referrin to the vertical line , elow/ C(,u(t&ons Polar moment of vertical line weld# Jv H>w>- w2DH2 12 Jv 0.56C =a$ distance from line : to end# c H2 c 1.500 =a$ stress due to tor'ue# % T>c Jv S ##2
in+l,s in in
in< in (1s&n2
Re,tn*u(r L&ne We(! &n 0en!&n* 1out XX %ee the fiure# riht. &endin moment# = 8illet weld si;e# w !eld lenth# H !eld width# @
Input <00 in+l,s 0.(45 in <.0 in (.0 in C(,u(t&ons - @ > H( 12 16.0000 in< - -@+2>w > -H+2>w( 12 6.<(65 in<
Buter area moment of inertia# "o "o "nner area moment of inertia# "o "o The area moment of inertia is e'ual to that of the inner su,tracted from the outer# "$ " outside + " inside "$ C.56(5
in<
=a$ distance from line : to end# c c =a$ stress due to ,endin# %, S1
Re,tn*u(r L&ne We(! &n Tors&on 1out t"e CG: %ee the fiure# riht. Applied tor'ue# T 8illet weld si;e# w !eld lenth# H !eld width# @
H2 2.000 = > c "$ <
Input <000 0.25 <.0 (.0 C(,u(t&ons
#
(1s&n2
in+l,s in in in
Polar moment of outer weld area is Jo/ Jo @ > H > - @2 D H2 12 Jo 25.0000 in< Ji - -@+2>w > -H+2>w > - -@+2>w2 D -H+2>w2 12 Ji 1(.<C6 in< Total polar moment of inertia of the < line welds of the rectanle is Jt/ Jt Jo D Ji Jt (.<C6 in< =a$ distance from line : to corner# R - H 2 2 D - @ 2 2 0.5 R 2.500 =a$ stress due to tor'ue# %t T > R Jt St 2>; (1s&n2
EXAMLE# The rectanular cantilever pictured at the riht is fillet welded to a vertical plate. 8ind the fillet weld si;e# re'uired to support a load# P 5#000 l,s. The dimensions of the ,eam are/ ) < in# H ( in# @ 2 in. %olution/ =a$imum moment# = P > ) 5#000 > < 20#000 in+l, !eld area moment of inertia is "w/ "w "1 D "2 "w 2 > w > @ > -H 22 D w > H( 12 "w 2.2500 in< =a$imum ,endin stress is %m/ %m = > - H2 "w
%m 1(#((0 - l,in2 %ee calculation ,elow/ =&((et We(! Stren*t" &n 0en!&n* antilever ,eam end load# &eam lenth# &eam width# &eam depth# 8illet weld si;e#
2
P ) @ H w
=a$ moment# = = )eft O riht weld area moment of inertia/ "1 "1 Top O ,ottom weld area moment of inertia/ "2 "2 !eld ma$ ,endin stress# %m %m !eld area# Aw Aw Averae shear stress# %s %s !eld resultant stress# %r Sr =a$ weld load per inch# 8w =
Cnt&(e'er 0e3 Stren*t" &n 0en!&n* antilever ,eam end load from a,ove# P =a$ moment from a,ove# = &eam area moment of inertia# ", ", &eam ma$ ,endin stress# %, tension at top# compression at ,ot# S1 &eam ma$ shear stress# %, at the neutral a$is# S1
Input (000 6 2 ( 0.450 C(,u(t&ons P>) 1#000 2 >w > H( 12 (.(450 2 > w > @ > -H 22 6.4500 = > -H2 "2 <000 w> -2 >H D- 2> @ .5 P Aw (5( - %m2 D %s2 0.5 <;#> %r > w ;#2
C(,u(t&ons (000 1#000 @ > H( 12 <.5000 = > -H2 ", >;;; 1.5 > P - @ > H ?5;
RO0LEM# :iven that the ma$ weld stress is to ,e less than 1(#600 l,sin2# use :oal %ee3 to calculate the weld si;e of weld -w re'uired to support an end load - P of 10#000 l,s on the cantilever a,ove. %olution/ Pic3/ &126 - %r G Tools G :oal %ee3 G %et cell to value# - 1(600 G G pic3 cell &10C - w G BN.
l,s in in in in
in+l,s
in<
in< l,in2 in2 l,in2 (1&n2 (1&n
l,s in+l,s in< (1&n2 (1&n2
Anser: We(! s&e/ ;$<9 s% ;$5 &n EXAMLE2 The rectanular cantilever pictured at the riht is fillet welded to a vertical plate. 8ind the fillet weld si;e# re'uired to support a load# P 5#000 l,s. The dimensions of the ,eam are/ ) < in# H ( in. %olution/ =a$imum moment# = P > ) 5000 > < 20#000 in+l, !eld area moment of inertia is "w/ "w (.1<16 > - -H D w < + - H < 6< "w 2.2500 in< =a$imum ,endin stress is %m/ %m = > - H2 "w %m 1(#((0 - l,in2 %ee calculation ,elow/ =&((et We(! Stren*t" &n 0en!&n* antilever ,eam end load# &eam lenth# &eam depth# 8illet weld si;e#
P ) H w
=a$ moment# = = !eld area moment of inertia# "w "w !eld ma$ ,endin stress# %m %m !eld area# Aw Aw Averae shear stress# %s %s !eld resultant stress# %r Sr =a$ weld load per inch# 8w = Cnt&(e'er 0e3 Stren*t" &n 0en!&n* antilever ,eam end load from a,ove# P =a$ moment from a,ove# = &eam area moment of inertia# ", ",
Input 500 l,s 6 in 1 in 0.250 in C(,u(t&ons P>) (#000 in+l,s (.1<16 > - -H D w < + - H < 6< 0.040 in< = > -H2 "w 21200 l,in2 (.1<16 > - -H D w 2 + - H 2 1.4642 in2 P Aw 2( l,in2 - %m2 D %s2 0.5 2#2;2 (1&n2 %r > w 5;; (1&n C(,u(t&ons 500 (#000 (.1<16 > H< 6< 0.0
l,s in+l,s in<
&eam ma$ ,endin stress# %, = > -H2 ", tension at top# compression at ,ot# S1 ;55 (1&n2 &eam ma$ shear stress# %, 1.((( > P - (.1<16 > H2 < at the neutral a$is# S1 <9 (1&n2 EXAMLE The rectanular cantilever ,rac3et pictured at the riht is fillet welded to a vertical plate. 8ind the fillet weld si;e# re'uired to support a load# P 5#000 l,s. The dimensions of the ,eam are/ ) ( in# H ( in# @ < %olution/ =a$imum moment# = P > ) 5#000 > ( 15#000 in+l, !eld area moment of inertia is "w/ "w - -H D w < + - H < 6< "w 2.2500 in< =a$imum ,endin stress is %m/ %m = > - H2 "w %m 1(#((0 - l,in2 %ee calculation ,elow/ 0r,.et We(! Stress The vertical load - P applied at the free end of the ,rac3et is replaced with an e'uivalent tor'ue - T and shear - * at the center of ravity of the weld area - A . !eld area# A - H>w D- 2>- @ D w >w &y symmetry# n - HDw 2 Fn - - 2 > @ > w > - @2 D - H D 2>w > 0 A
<
5 The applied shear load - * is ,alanced ,y weld shear force - 8p l,sin. * 8p>- H D- 2>- @ D w The tor'ue - T is ,alanced ,y shear forces - 8$ l,sin# and - 8y l,sin. T - 2>8$>-@Dw> n D- 8y > H > Fn 0r,.et =&((et We(! Stren*t" antilever ,rac3et end load# &eam lenth# !eld width# !eld depth# 8illet weld si;e#
P ) @ H w
!eld area# A A !eld : ,y symmetry# n n !eld :# Fn Fn * * T T M+ e(! s"er stress t po&nt 7 M : %v S' =a$ weld shear per inch# 8m =3
Input 000 20 5 10 0.(<4 C(,u(t&ons - H>w D- 2>- @ D w >w 4.11 - HDw 2 5.14< - - 2>@>w > - @2 D - H D 1.20 P 000 P>- )D@+Fn 15612 (>* - 2>A #>?# %m > w 5;
l,s in in in in
in2 in 2>w > 0 A in l,s in+l,s
(1&n2 7(1&n2 &n e(! Rev 2C?ov04
M+ e(! tors&on s"er stress t po&nt 7 N : Polar moment of vertical line weld# Jv - w>H>- w2DH2 12 D - H>w>Fn2 Jv (<.016 Polar moment of 2 hori;ontal welds# Jh - 2>@>w>- @2Dw2 12 D - 2>@>w>n2 Jh 100.1(C Total polar moment of welds# Jt Jv D Jh Jt 1(<.15<4 Radius# R1 - - @ + Fn 2 D - H 2 2 0.5 R1 6.245 %n T>R1 Jt Sn ?<#9 (1&n2 !eld shear per inch# 8n %n > w =n 25?< (1&n
>
Note: T"ree !&))erent )or3u(s )or 3o3ent o) &nert& I+ 1o'e$ 1. 8irst moment of inertia/ Rectanular Area "$ w > H( 12 2. %econd moment of inertia/ !eld Around a Rectanular " outer area - @ > H( 12 " inner area - -@+2>w > -H+2>w( 12 "$ " outer area + " inner area (. Third moment of inertia/ !eld Around a Rectanular Allows two different weld -w si;es. 2 *ertical !elds# "1 2 > w > @ > -H 22 2 @ori;ontal welds# "2 w > H( 12 "$ "1 D "2 Rev 2C ?ov 04
?
WELDMENT STRENGTH EXCEL CALCULATIONS Course Content o"n An!re $E$ LONG WELDS
We(!e! (te 0e3 Stren*t" &eam end load# * !eld allowa,le shear stress# %a "ntermittent weld lenth# N antilever ,eam lenth# ) &eam depth# H 8lane width# @ 8lane thic3ness# T2 !e, thic3ness# T1
Input 200000 l,s 1(600 l,sin2 ( in 50.00 in (6.00 in 12.00 in 2.500 in 0.625 in C(,u(t&on &eam top flane area# A @ > T2 A (0 in2 w - H2 + T2 w 15.5 - H2 + T22 16.45 in %ection moment of area# " - @>H( 12 + - - @+T1>-2>w( 12 " 1<14 in< We1 s"er stress t Y Y/ Ss @ 8 A 8 7 T# 8 I Ss ?# (1s&n2 T"ere re 2 )&((et e(!s t Y Y: "nitial weld si;e# wi 1.00 in !eld shear force -l,s per in# 8w %s > 0.404 > wi 2 = ;> (1&n (te "nitial weld stress# %i 8w wi T"&,.ness %i (06 (1s&n2 up to 0.59 8inal weld stress# from input# %a 1(600 (1s&n2 0.59 to 0.459 Theoretical weld si;e# wt wi>%i %a 0.459 to 1.259 wt 0.224 in 1.259 to 29 Enter 3&n&3u3 e(! s&e/ ;$?5 &n 29 to 69 See t"e !K,ent t1(e )or 3&n&3u3 7 )(n*e e(! s&e 7 a,ove 69
M&n&3u3 We(! S&e 0.19 0.259 0.(1(9 0.(459 0.59 0.6259
est1(&s"e! 1% t"e A3er&,n We(!&n* So,&et%. Inter3&ttent We(! S&e C(,u(t&on !eld si;e ratio# R - =in continuous weld si;e - "ntermittent weld si;e R wt w R 617 "ntermittent weld total lenth# )w R>) )w (0 in Total of aps ,etween welds# ) ) + )w ) 20 in G&'en &nter3&ttent e(! (en*t"/ 6 &n ?um,er of intermittent welds# ?w )w N ?w 10 + Inter3&ttent e(! p&t,"/ 7 L N L* 7 N # 5$2 &n Su33r%: The a,ove ,eam shear stress distri,ution is illustrated here. The ,eamKs we, shear stress - %s is calculated at the top of the we, at w. This we, shear stress is carried ,y 2 f illet welds havin an initial si;e - w of 1.00 inch each and a minimum thic3ness of 0.404 inches. -%ee# Tension+%hear ta, pae ( The minimum weld si;e - w is e'ual to the ratio of applied shear stress to allowa,le shear stress.
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2
