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W~1 ~~ 9F MSl~ ~ -n~ J~ ~sA4i~1 A~ A -l0l- ~ T~~~~1

1 I

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~ ArtAMA~1 ~(S~~ ~) -n~G (Ob~

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J~i~A~ ~~Y t~ fbJcb fN~~ ~SSlre AJ1E

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it-PZshy

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b~

~~

Pz A-T - I (U ~ ltgt -IgtJJ)tD-Zcgtlt -3str5J~f~~ -------- s G3 kN

4-P2 A ) [5 ~ n~5 2 krJ (EX~~~j JELF VIT ~ 3~~ _ ~ ++lcrJ1r p~ 1Jrr s~~

~ImiddottJtt CAJA~ CV (s)

t-A~- CPl) 8 - 2~152ybI

-n~~tAlt- EIltT1l cu~ ~

0 ~ [~-r1l(j+ ~ (C~)] -s

C III - Igt~Jsfo 0 bull t CJ~ SJ~+VI~- t QC 51 vttJ-1 ~ J - -tshy

0-5 f- =11Z~ ~1 ~tlrr[(1[nQtgtlt I at)raquo)(Zoo ltlt10 IltI-Sll~f~J] -=2ij C1lt1Jy~ _

---- -~-~

i -lTJtir JX3~_ )82shy-0

T -1c _ ~ T-CJ

1- Zcr ~Cgt -25 (F~ ~~I$ 9~Atzt ~w DIt1II)~-r)

~tt~Mo~ AT UlS lv5 1icl JaF WT (23 7t ~-Ii) x1-2

~ ImiddotZ~W

20middot -r ( 141 gt (h middot8q~ 1-2)WAr

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~~ CS-S)

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5_~ 18gt T-t xfo) x 0003shy

s Si--xllo)cI3 ~ 21CJltJ t

=--(1 --94- ~a -jbY1oW

3 o~ 4- S I laquof

fl 8~ h(7Wf-r5fj W~~ e= ~ ~ b)( middot15 _ W~wJC blC J+

VlllV~I~~ ~1Ei1

~WtNb ~ ~~~~

_ ~T J~S~ $ -= ZX_(O o~~ = - CdJqJ 2shy

w-middott lCl 2 2middott~bl

~ (ftc 1)1__~+ (r~~45 ) == IdJJ~_ ~ c ----

(NIC~_~I~ _ ~~ - ~1~~ ~~ -Bf pound4J

~--- ~

__ ( =~_[c t~-Z-- A == 1tZIt)W (1QJ= ~~~t 5- 3-__ ~

A - ~~Z~t-

Jl~ _ t~YQ16 amp

Z ~ C)~ CfiQM ampJ~~middot~~ __ 9-l~l_IN____~~-r)

bullbull tzIl omiddot o~x6034- Le-J

-= I sl5CtrJ gt_lo1-b krJ h _)S_Zr9)cZ93_X_~b_ ts SrfmiddotSrJyen~Wf

aQciF -ii~Gt h ~Jcw

J1TIQ s-~ ~ WA-~

_~~ ~__is T)e C~ SfA-l 8-rN~ ~~ 2 (r b-~) (~

~ LeNI4I

([ fnlM = --S

- 31Q~ gtC1O NoM 12~2 fI1to f~_1 ca

)Jegt t51-)(52 X ~3- -~M -= til ~ 1 1--M ~~_~__kIJM)~ ~ - ~

b)_

e

JZ-= middotc-tT1

d l =68

-JCt euroltNA WOJfAA lQJ6 = Nonv oiS~l~I1gt It Nlrv~t

~ It ( w f Kz 1- we r~) = 2 t1~p [~ t6l) + 7J 1 ]

J ~ 2 ~JI- l ~~ r

L i 4~ - I JiQtUlt ~ Bi

rtAmiddotSN~ 0- HAt=~ ~- ~H ~Gio-J

~E ~

M~n= WCft~) (zi + ~)

~

-- --

J - 11~_I~t ~~ I~ ~~__~~ _

e~ tl~l~ ~_b53~

j-=- ~l (~)+ It)=~~l~~~

C= lb - (1l1J2) =- 10-UMIl

~~ __~_~ t

CD a~-K9-Cfgtx3x-S) + ~8]

35middot +LcrJ~ bull i

____tz ~

umiddott c ~ 10middot IN If hi

== 13imiddot3~

QboU ~~~ ~lhkeP

~ ~ ~~f- ~I[Qumiddotq-l 3= 1~Jmiddot3 )Cl9 =-2+-~ L

tlS~ Z ~2_ tI~1

~ J c 7lt_ =fl~~

~ 1 - J1p - C1d -t d~ 3~ bfmi1l4

~

~~2 )( qlI sfq) Xl-Q~

_~_l-~ kNlIIA gt SIS-1shy

bull1

J)

-me 7-c1N) fVlIJ n -rnAt-Is Fat TllG

B9Vb1~ NroNl6f1Jl M 4J= ~ JiIC11 ~o ~ ~ TIE flA Ftaj HAIJw4

~ TE crovMIJ

lgteS1~ CLtC lt)~ M (ftAF1gtampl 4VfVCt jOl S07S ASoJT 0) bull cJrC-l WE-lO I N ~( t otJ glt-r~ nA FIC3L ANto poundNb- plATe-

2 ~ ~SIoJ iN D~~ I PAZtmiddotCJA U1 I t1 A~ 2~ I1JoIN

Fl1rO tV 1ltJ r= bull

3 CLeo-( trb1 wOe I ~~rvfJr OF ell)-p-ce ~ ooVMIJ P-ArVlitS

f J I~ ~yen= flt)P ~1Vamptamp OF kA~

f ~ ~lt~ Q6fAU~ cJF= GoViAIi tVc=f8 IN 1tJmiddotfi~ OF

-0 ~c~ OF ttA~11 (N~Ce ~JImiddotQl JII~4-z F rvea ~)

i cA~ B-Jc~(fV~ ltgtP Vo~ ~ N VI~lVi1 OF

~M 91= fovNCLl (NiVrQ~CE VOrtP~HIQV JlI~

J= ~Ji~)

~tGN ~~ ~

cN~ S~ ltAtA ti1 OF W~

2 ltJ~ r~~ ~HA1 -Qr ampL7S (Mrr1E 8o7J -u ~ )JS Jd~ ~ (it)(1$IMb S ~~ ~ sbJ)

3 otoLA 8~~ hrGc- CAvAci1 Ol= tlViI-~ AVO ltiltJ v IrfrJ

~ ~G A 8o7 p

-n~~tAlt- EIltT1l cu~ ~

0 ~ [~-r1l(j+ ~ (C~)] -s

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0-5 f- =11Z~ ~1 ~tlrr[(1[nQtgtlt I at)raquo)(Zoo ltlt10 IltI-Sll~f~J] -=2ij C1lt1Jy~ _

---- -~-~

i -lTJtir JX3~_ )82shy-0

T -1c _ ~ T-CJ

1- Zcr ~Cgt -25 (F~ ~~I$ 9~Atzt ~w DIt1II)~-r)

~tt~Mo~ AT UlS lv5 1icl JaF WT (23 7t ~-Ii) x1-2

~ ImiddotZ~W

20middot -r ( 141 gt (h middot8q~ 1-2)WAr

qg Aa~ ~ lO2-7C2 3 is SA-rS~~Y01

~~ CS-S)

~MAc =b~~

5_~ 18gt T-t xfo) x 0003shy

s Si--xllo)cI3 ~ 21CJltJ t

=--(1 --94- ~a -jbY1oW

3 o~ 4- S I laquof

fl 8~ h(7Wf-r5fj W~~ e= ~ ~ b)( middot15 _ W~wJC blC J+

VlllV~I~~ ~1Ei1

~WtNb ~ ~~~~

_ ~T J~S~ $ -= ZX_(O o~~ = - CdJqJ 2shy

w-middott lCl 2 2middott~bl

~ (ftc 1)1__~+ (r~~45 ) == IdJJ~_ ~ c ----

(NIC~_~I~ _ ~~ - ~1~~ ~~ -Bf pound4J

~--- ~

__ ( =~_[c t~-Z-- A == 1tZIt)W (1QJ= ~~~t 5- 3-__ ~

A - ~~Z~t-

Jl~ _ t~YQ16 amp

Z ~ C)~ CfiQM ampJ~~middot~~ __ 9-l~l_IN____~~-r)

bullbull tzIl omiddot o~x6034- Le-J

-= I sl5CtrJ gt_lo1-b krJ h _)S_Zr9)cZ93_X_~b_ ts SrfmiddotSrJyen~Wf

aQciF -ii~Gt h ~Jcw

J1TIQ s-~ ~ WA-~

_~~ ~__is T)e C~ SfA-l 8-rN~ ~~ 2 (r b-~) (~

~ LeNI4I

([ fnlM = --S

- 31Q~ gtC1O NoM 12~2 fI1to f~_1 ca

)Jegt t51-)(52 X ~3- -~M -= til ~ 1 1--M ~~_~__kIJM)~ ~ - ~

b)_

e

JZ-= middotc-tT1

d l =68

-JCt euroltNA WOJfAA lQJ6 = Nonv oiS~l~I1gt It Nlrv~t

~ It ( w f Kz 1- we r~) = 2 t1~p [~ t6l) + 7J 1 ]

J ~ 2 ~JI- l ~~ r

L i 4~ - I JiQtUlt ~ Bi

rtAmiddotSN~ 0- HAt=~ ~- ~H ~Gio-J

~E ~

M~n= WCft~) (zi + ~)

~

-- --

J - 11~_I~t ~~ I~ ~~__~~ _

e~ tl~l~ ~_b53~

j-=- ~l (~)+ It)=~~l~~~

C= lb - (1l1J2) =- 10-UMIl

~~ __~_~ t

CD a~-K9-Cfgtx3x-S) + ~8]

35middot +LcrJ~ bull i

____tz ~

umiddott c ~ 10middot IN If hi

== 13imiddot3~

QboU ~~~ ~lhkeP

~ ~ ~~f- ~I[Qumiddotq-l 3= 1~Jmiddot3 )Cl9 =-2+-~ L

tlS~ Z ~2_ tI~1

~ J c 7lt_ =fl~~

~ 1 - J1p - C1d -t d~ 3~ bfmi1l4

~

~~2 )( qlI sfq) Xl-Q~

_~_l-~ kNlIIA gt SIS-1shy

bull1

J)

-me 7-c1N) fVlIJ n -rnAt-Is Fat TllG

B9Vb1~ NroNl6f1Jl M 4J= ~ JiIC11 ~o ~ ~ TIE flA Ftaj HAIJw4

~ TE crovMIJ

lgteS1~ CLtC lt)~ M (ftAF1gtampl 4VfVCt jOl S07S ASoJT 0) bull cJrC-l WE-lO I N ~( t otJ glt-r~ nA FIC3L ANto poundNb- plATe-

2 ~ ~SIoJ iN D~~ I PAZtmiddotCJA U1 I t1 A~ 2~ I1JoIN

Fl1rO tV 1ltJ r= bull

3 CLeo-( trb1 wOe I ~~rvfJr OF ell)-p-ce ~ ooVMIJ P-ArVlitS

f J I~ ~yen= flt)P ~1Vamptamp OF kA~

f ~ ~lt~ Q6fAU~ cJF= GoViAIi tVc=f8 IN 1tJmiddotfi~ OF

-0 ~c~ OF ttA~11 (N~Ce ~JImiddotQl JII~4-z F rvea ~)

i cA~ B-Jc~(fV~ ltgtP Vo~ ~ N VI~lVi1 OF

~M 91= fovNCLl (NiVrQ~CE VOrtP~HIQV JlI~

J= ~Ji~)

~tGN ~~ ~

cN~ S~ ltAtA ti1 OF W~

2 ltJ~ r~~ ~HA1 -Qr ampL7S (Mrr1E 8o7J -u ~ )JS Jd~ ~ (it)(1$IMb S ~~ ~ sbJ)

3 otoLA 8~~ hrGc- CAvAci1 Ol= tlViI-~ AVO ltiltJ v IrfrJ

~ ~G A 8o7 p

3 o~ 4- S I laquof

fl 8~ h(7Wf-r5fj W~~ e= ~ ~ b)( middot15 _ W~wJC blC J+

VlllV~I~~ ~1Ei1

~WtNb ~ ~~~~

_ ~T J~S~ $ -= ZX_(O o~~ = - CdJqJ 2shy

w-middott lCl 2 2middott~bl

~ (ftc 1)1__~+ (r~~45 ) == IdJJ~_ ~ c ----

(NIC~_~I~ _ ~~ - ~1~~ ~~ -Bf pound4J

~--- ~

__ ( =~_[c t~-Z-- A == 1tZIt)W (1QJ= ~~~t 5- 3-__ ~

A - ~~Z~t-

Jl~ _ t~YQ16 amp

Z ~ C)~ CfiQM ampJ~~middot~~ __ 9-l~l_IN____~~-r)

bullbull tzIl omiddot o~x6034- Le-J

-= I sl5CtrJ gt_lo1-b krJ h _)S_Zr9)cZ93_X_~b_ ts SrfmiddotSrJyen~Wf

aQciF -ii~Gt h ~Jcw

J1TIQ s-~ ~ WA-~

_~~ ~__is T)e C~ SfA-l 8-rN~ ~~ 2 (r b-~) (~

~ LeNI4I

([ fnlM = --S

- 31Q~ gtC1O NoM 12~2 fI1to f~_1 ca

)Jegt t51-)(52 X ~3- -~M -= til ~ 1 1--M ~~_~__kIJM)~ ~ - ~

b)_

e

JZ-= middotc-tT1

d l =68

-JCt euroltNA WOJfAA lQJ6 = Nonv oiS~l~I1gt It Nlrv~t

~ It ( w f Kz 1- we r~) = 2 t1~p [~ t6l) + 7J 1 ]

J ~ 2 ~JI- l ~~ r

L i 4~ - I JiQtUlt ~ Bi

rtAmiddotSN~ 0- HAt=~ ~- ~H ~Gio-J

~E ~

M~n= WCft~) (zi + ~)

~

-- --

J - 11~_I~t ~~ I~ ~~__~~ _

e~ tl~l~ ~_b53~

j-=- ~l (~)+ It)=~~l~~~

C= lb - (1l1J2) =- 10-UMIl

~~ __~_~ t

CD a~-K9-Cfgtx3x-S) + ~8]

35middot +LcrJ~ bull i

____tz ~

umiddott c ~ 10middot IN If hi

== 13imiddot3~

QboU ~~~ ~lhkeP

~ ~ ~~f- ~I[Qumiddotq-l 3= 1~Jmiddot3 )Cl9 =-2+-~ L

tlS~ Z ~2_ tI~1

~ J c 7lt_ =fl~~

~ 1 - J1p - C1d -t d~ 3~ bfmi1l4

~

~~2 )( qlI sfq) Xl-Q~

_~_l-~ kNlIIA gt SIS-1shy

bull1

J)

-me 7-c1N) fVlIJ n -rnAt-Is Fat TllG

B9Vb1~ NroNl6f1Jl M 4J= ~ JiIC11 ~o ~ ~ TIE flA Ftaj HAIJw4

~ TE crovMIJ

lgteS1~ CLtC lt)~ M (ftAF1gtampl 4VfVCt jOl S07S ASoJT 0) bull cJrC-l WE-lO I N ~( t otJ glt-r~ nA FIC3L ANto poundNb- plATe-

2 ~ ~SIoJ iN D~~ I PAZtmiddotCJA U1 I t1 A~ 2~ I1JoIN

Fl1rO tV 1ltJ r= bull

3 CLeo-( trb1 wOe I ~~rvfJr OF ell)-p-ce ~ ooVMIJ P-ArVlitS

f J I~ ~yen= flt)P ~1Vamptamp OF kA~

f ~ ~lt~ Q6fAU~ cJF= GoViAIi tVc=f8 IN 1tJmiddotfi~ OF

-0 ~c~ OF ttA~11 (N~Ce ~JImiddotQl JII~4-z F rvea ~)

i cA~ B-Jc~(fV~ ltgtP Vo~ ~ N VI~lVi1 OF

~M 91= fovNCLl (NiVrQ~CE VOrtP~HIQV JlI~

J= ~Ji~)

~tGN ~~ ~

cN~ S~ ltAtA ti1 OF W~

2 ltJ~ r~~ ~HA1 -Qr ampL7S (Mrr1E 8o7J -u ~ )JS Jd~ ~ (it)(1$IMb S ~~ ~ sbJ)

3 otoLA 8~~ hrGc- CAvAci1 Ol= tlViI-~ AVO ltiltJ v IrfrJ

~ ~G A 8o7 p

_~~ ~__is T)e C~ SfA-l 8-rN~ ~~ 2 (r b-~) (~

~ LeNI4I

([ fnlM = --S

- 31Q~ gtC1O NoM 12~2 fI1to f~_1 ca

)Jegt t51-)(52 X ~3- -~M -= til ~ 1 1--M ~~_~__kIJM)~ ~ - ~

b)_

e

JZ-= middotc-tT1

d l =68

-JCt euroltNA WOJfAA lQJ6 = Nonv oiS~l~I1gt It Nlrv~t

~ It ( w f Kz 1- we r~) = 2 t1~p [~ t6l) + 7J 1 ]

J ~ 2 ~JI- l ~~ r

L i 4~ - I JiQtUlt ~ Bi

rtAmiddotSN~ 0- HAt=~ ~- ~H ~Gio-J

~E ~

M~n= WCft~) (zi + ~)

~

-- --

J - 11~_I~t ~~ I~ ~~__~~ _

e~ tl~l~ ~_b53~

j-=- ~l (~)+ It)=~~l~~~

C= lb - (1l1J2) =- 10-UMIl

~~ __~_~ t

CD a~-K9-Cfgtx3x-S) + ~8]

35middot +LcrJ~ bull i

____tz ~

umiddott c ~ 10middot IN If hi

== 13imiddot3~

QboU ~~~ ~lhkeP

~ ~ ~~f- ~I[Qumiddotq-l 3= 1~Jmiddot3 )Cl9 =-2+-~ L

tlS~ Z ~2_ tI~1

~ J c 7lt_ =fl~~

~ 1 - J1p - C1d -t d~ 3~ bfmi1l4

~

~~2 )( qlI sfq) Xl-Q~

_~_l-~ kNlIIA gt SIS-1shy

bull1

J)

-me 7-c1N) fVlIJ n -rnAt-Is Fat TllG

B9Vb1~ NroNl6f1Jl M 4J= ~ JiIC11 ~o ~ ~ TIE flA Ftaj HAIJw4

~ TE crovMIJ

lgteS1~ CLtC lt)~ M (ftAF1gtampl 4VfVCt jOl S07S ASoJT 0) bull cJrC-l WE-lO I N ~( t otJ glt-r~ nA FIC3L ANto poundNb- plATe-

2 ~ ~SIoJ iN D~~ I PAZtmiddotCJA U1 I t1 A~ 2~ I1JoIN

Fl1rO tV 1ltJ r= bull

3 CLeo-( trb1 wOe I ~~rvfJr OF ell)-p-ce ~ ooVMIJ P-ArVlitS

f J I~ ~yen= flt)P ~1Vamptamp OF kA~

f ~ ~lt~ Q6fAU~ cJF= GoViAIi tVc=f8 IN 1tJmiddotfi~ OF

-0 ~c~ OF ttA~11 (N~Ce ~JImiddotQl JII~4-z F rvea ~)

i cA~ B-Jc~(fV~ ltgtP Vo~ ~ N VI~lVi1 OF

~M 91= fovNCLl (NiVrQ~CE VOrtP~HIQV JlI~

J= ~Ji~)

~tGN ~~ ~

cN~ S~ ltAtA ti1 OF W~

2 ltJ~ r~~ ~HA1 -Qr ampL7S (Mrr1E 8o7J -u ~ )JS Jd~ ~ (it)(1$IMb S ~~ ~ sbJ)

3 otoLA 8~~ hrGc- CAvAci1 Ol= tlViI-~ AVO ltiltJ v IrfrJ

~ ~G A 8o7 p

-JCt euroltNA WOJfAA lQJ6 = Nonv oiS~l~I1gt It Nlrv~t

~ It ( w f Kz 1- we r~) = 2 t1~p [~ t6l) + 7J 1 ]

J ~ 2 ~JI- l ~~ r

L i 4~ - I JiQtUlt ~ Bi

rtAmiddotSN~ 0- HAt=~ ~- ~H ~Gio-J

~E ~

M~n= WCft~) (zi + ~)

~

-- --

J - 11~_I~t ~~ I~ ~~__~~ _

e~ tl~l~ ~_b53~

j-=- ~l (~)+ It)=~~l~~~

C= lb - (1l1J2) =- 10-UMIl

~~ __~_~ t

CD a~-K9-Cfgtx3x-S) + ~8]

35middot +LcrJ~ bull i

____tz ~

umiddott c ~ 10middot IN If hi

== 13imiddot3~

QboU ~~~ ~lhkeP

~ ~ ~~f- ~I[Qumiddotq-l 3= 1~Jmiddot3 )Cl9 =-2+-~ L

tlS~ Z ~2_ tI~1

~ J c 7lt_ =fl~~

~ 1 - J1p - C1d -t d~ 3~ bfmi1l4

~

~~2 )( qlI sfq) Xl-Q~

_~_l-~ kNlIIA gt SIS-1shy

bull1

J)

-me 7-c1N) fVlIJ n -rnAt-Is Fat TllG

B9Vb1~ NroNl6f1Jl M 4J= ~ JiIC11 ~o ~ ~ TIE flA Ftaj HAIJw4

~ TE crovMIJ

lgteS1~ CLtC lt)~ M (ftAF1gtampl 4VfVCt jOl S07S ASoJT 0) bull cJrC-l WE-lO I N ~( t otJ glt-r~ nA FIC3L ANto poundNb- plATe-

2 ~ ~SIoJ iN D~~ I PAZtmiddotCJA U1 I t1 A~ 2~ I1JoIN

Fl1rO tV 1ltJ r= bull

3 CLeo-( trb1 wOe I ~~rvfJr OF ell)-p-ce ~ ooVMIJ P-ArVlitS

f J I~ ~yen= flt)P ~1Vamptamp OF kA~

f ~ ~lt~ Q6fAU~ cJF= GoViAIi tVc=f8 IN 1tJmiddotfi~ OF

-0 ~c~ OF ttA~11 (N~Ce ~JImiddotQl JII~4-z F rvea ~)

i cA~ B-Jc~(fV~ ltgtP Vo~ ~ N VI~lVi1 OF

~M 91= fovNCLl (NiVrQ~CE VOrtP~HIQV JlI~

J= ~Ji~)

~tGN ~~ ~

cN~ S~ ltAtA ti1 OF W~

2 ltJ~ r~~ ~HA1 -Qr ampL7S (Mrr1E 8o7J -u ~ )JS Jd~ ~ (it)(1$IMb S ~~ ~ sbJ)

3 otoLA 8~~ hrGc- CAvAci1 Ol= tlViI-~ AVO ltiltJ v IrfrJ

~ ~G A 8o7 p

~~ __~_~ t

CD a~-K9-Cfgtx3x-S) + ~8]

35middot +LcrJ~ bull i

____tz ~

umiddott c ~ 10middot IN If hi

== 13imiddot3~

QboU ~~~ ~lhkeP

~ ~ ~~f- ~I[Qumiddotq-l 3= 1~Jmiddot3 )Cl9 =-2+-~ L

tlS~ Z ~2_ tI~1

~ J c 7lt_ =fl~~

~ 1 - J1p - C1d -t d~ 3~ bfmi1l4

~

~~2 )( qlI sfq) Xl-Q~

_~_l-~ kNlIIA gt SIS-1shy

bull1

J)

-me 7-c1N) fVlIJ n -rnAt-Is Fat TllG

B9Vb1~ NroNl6f1Jl M 4J= ~ JiIC11 ~o ~ ~ TIE flA Ftaj HAIJw4

~ TE crovMIJ

lgteS1~ CLtC lt)~ M (ftAF1gtampl 4VfVCt jOl S07S ASoJT 0) bull cJrC-l WE-lO I N ~( t otJ glt-r~ nA FIC3L ANto poundNb- plATe-

2 ~ ~SIoJ iN D~~ I PAZtmiddotCJA U1 I t1 A~ 2~ I1JoIN

Fl1rO tV 1ltJ r= bull

3 CLeo-( trb1 wOe I ~~rvfJr OF ell)-p-ce ~ ooVMIJ P-ArVlitS

f J I~ ~yen= flt)P ~1Vamptamp OF kA~

f ~ ~lt~ Q6fAU~ cJF= GoViAIi tVc=f8 IN 1tJmiddotfi~ OF

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(~~)

Examiners comments to be attached to the crib

A disappointing paper with raw marks in all questions lt50 The candidates were clearly under time pressure but this does not explain the lack of logical thought in the bits they did do

Qu 1 Steel design for a grandstand

The main problem here was that most of the candidates only checked for one of the three conditions (strength stiffness and buckling) although the first two were specifically asked for and they were given a strong hint about the need to think about the buckling Quite a large number could not determine the load in the column at the base either leaving out the wind load or leaving out the snow load or simply being unable to apply equilibrium properly The last part where they were asked (implicitly) to find an alternative load path for the wind load was very poorly done - very few seemed able to think in three dimensions

Qu 2 Steel Portal Frame

The most popular question The methods adopted by the candidates were largely correct but there were several errors and omissions in the calculations The first part asked the candidates to estimate the size of the rafter by assuming a fixed-ended condition Most candidates successfully derived (or recalled) that for a fixed ended beam subjected to UDL Mp= wP16 but several candidates used the full span between the columns rather than the clear span between the haunches Part b of the question asked the candidates to derive the plastic moment at the plastic hinges of a portal frame Most students set this up correctly by equating the external work done to the work dissipated in the hinges but several candidates failed to notice that the total rotation in the hinge closest to the haunch was (fA + Ih) rather than OJ The free body diagrams required to solve Part c of the question were generally correct but the most common error was an inconsistent sign convention leading to an incorrect direction of the bending moment at the apex haunch Despite these errors most candidates were able to sketch a sensible bending moment diagram The final part of the question asked the candidates to consider the design of the column-rafter haunch There were easy marks to be gained in this final part but few candidates attempted it Those that did secured most of the marks allocated to this part

Qu 3 Composite GlassSteel beam

Very disappointing question It could easily have been set to the lA students as a revision exercise since it is essentially the application of lA principles to a new material but it is almost as though they had put that knowledge away never to be used again They were unable to distinguish between loads per unit length and loads per unit area and a significant number tried to use the density of the steel and the glass to transform the section when determining the second moment of area rather than Youngs modulus Some simply ignored the distinction between two materials Even if they got it right they seemed incapable of sketching the expected stress distribution over the section The crib expected them to take account of transverse bending ofthe flange when determining the stresses over the support

but no one attempted that and it wasnt penalised given the overall difficulty Two candidates included it in their answers to part (c) - what else would you check Very few of them made any serious attempt at calculating the deflections

Qu 4 Moment curvature relationship of a reinforced concrete beam

Quite a few candidates did not appear to know what a moment-curvature relationship was and several did not distinguish between the uncracked and cracked behaviour The biggest problem was that almost nobody used axial equilibrium in the beam for determining where the neutral axis was located which was essential for determining the cracked-elastic and ultimate load stress states and wrote various degrees ofnonsense depending on where they thought the neutral axis might be Several used the design equations to determine the ultimate moment capacity for which they were given some credit The overall impression from this question was that they were trying to remember the formula for rather than applying fundamental principles

C J Burgoyne

May 2013

Examiners comments to be attached to the crib

A disappointing paper with raw marks in all questions lt50 The candidates were clearly under time pressure but this does not explain the lack of logical thought in the bits they did do

Qu 1 Steel design for a grandstand

The main problem here was that most of the candidates only checked for one of the three conditions (strength stiffness and buckling) although the first two were specifically asked for and they were given a strong hint about the need to think about the buckling Quite a large number could not determine the load in the column at the base either leaving out the wind load or leaving out the snow load or simply being unable to apply equilibrium properly The last part where they were asked (implicitly) to find an alternative load path for the wind load was very poorly done - very few seemed able to think in three dimensions

Qu 2 Steel Portal Frame

The most popular question The methods adopted by the candidates were largely correct but there were several errors and omissions in the calculations The first part asked the candidates to estimate the size of the rafter by assuming a fixed-ended condition Most candidates successfully derived (or recalled) that for a fixed ended beam subjected to UDL Mp= wP16 but several candidates used the full span between the columns rather than the clear span between the haunches Part b of the question asked the candidates to derive the plastic moment at the plastic hinges of a portal frame Most students set this up correctly by equating the external work done to the work dissipated in the hinges but several candidates failed to notice that the total rotation in the hinge closest to the haunch was (fA + Ih) rather than OJ The free body diagrams required to solve Part c of the question were generally correct but the most common error was an inconsistent sign convention leading to an incorrect direction of the bending moment at the apex haunch Despite these errors most candidates were able to sketch a sensible bending moment diagram The final part of the question asked the candidates to consider the design of the column-rafter haunch There were easy marks to be gained in this final part but few candidates attempted it Those that did secured most of the marks allocated to this part

Qu 3 Composite GlassSteel beam

Very disappointing question It could easily have been set to the lA students as a revision exercise since it is essentially the application of lA principles to a new material but it is almost as though they had put that knowledge away never to be used again They were unable to distinguish between loads per unit length and loads per unit area and a significant number tried to use the density of the steel and the glass to transform the section when determining the second moment of area rather than Youngs modulus Some simply ignored the distinction between two materials Even if they got it right they seemed incapable of sketching the expected stress distribution over the section The crib expected them to take account of transverse bending ofthe flange when determining the stresses over the support

but no one attempted that and it wasnt penalised given the overall difficulty Two candidates included it in their answers to part (c) - what else would you check Very few of them made any serious attempt at calculating the deflections

Qu 4 Moment curvature relationship of a reinforced concrete beam

Quite a few candidates did not appear to know what a moment-curvature relationship was and several did not distinguish between the uncracked and cracked behaviour The biggest problem was that almost nobody used axial equilibrium in the beam for determining where the neutral axis was located which was essential for determining the cracked-elastic and ultimate load stress states and wrote various degrees ofnonsense depending on where they thought the neutral axis might be Several used the design equations to determine the ultimate moment capacity for which they were given some credit The overall impression from this question was that they were trying to remember the formula for rather than applying fundamental principles

C J Burgoyne

May 2013

but no one attempted that and it wasnt penalised given the overall difficulty Two candidates included it in their answers to part (c) - what else would you check Very few of them made any serious attempt at calculating the deflections

Qu 4 Moment curvature relationship of a reinforced concrete beam

Quite a few candidates did not appear to know what a moment-curvature relationship was and several did not distinguish between the uncracked and cracked behaviour The biggest problem was that almost nobody used axial equilibrium in the beam for determining where the neutral axis was located which was essential for determining the cracked-elastic and ultimate load stress states and wrote various degrees ofnonsense depending on where they thought the neutral axis might be Several used the design equations to determine the ultimate moment capacity for which they were given some credit The overall impression from this question was that they were trying to remember the formula for rather than applying fundamental principles

C J Burgoyne

May 2013