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CE 433 REINFORCED ONCRETE ESIGN Fall2007
Probiem Se t No . I
Due Aueust 29th
The exterior balcony shown n the sketch below s to be supported n two simply-supported dge
beams.. Determine he oads hat would be used or the design of the beams by answering he
following questions:
l. Calculate he dead oad of the slab (in psf). Assume hat the unit weight of the concrete
slab (including reinforcement) s 150 pcf'
2. From Table 1. 1 n your text, select n appropriate ervice iv e load or th e design of th e
slab.
3. Recognizing hat load acting on the slab s carried by the beams, determine he dead oad
and he ive load (in lb/ft) are acting on the beams'
4. Determine he additional component f dead oad representing he self weight of the
beam in lb/ft).
5. Determine he factored otal load (in lb/fi) that he beam must be designed o carry.
Apply load actors, iven n Table 1.2 n th e ext, as appropriate.
6. Draw the shear and bending mornent cliagrams,ased on total factored oad, for the beam'
wl
Or'
-\| .. ;i
a .Y-a
s\o-b
b eo -m r s up po rl in g stab
|-SYbI io
I i ; Ff| = ;o-I N q- ' s ]
| 1' r !
Fr rbi -T Nt.(--.-
colurnnss up por*rngbec..n:
- Nor ro 3cs\z -
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CE 433 REINFORCED ONCRETE ESIGN Fall2007
Probiem Se t No . I
Due Aueust 29th
The exterior balcony shown n the sketch below s to be supported n two simply-supported dge
beams.. Determine he oads hat would be used or the design of the beams by answering he
following questions:
l. Calculate he dead oad of the slab (in psf). Assume hat the unit weight of the concrete
slab (including reinforcement) s 150 pcf'
2. From Table 1. 1 n your text, select n appropriate ervice iv e load or th e design of th e
slab.
3. Recognizing hat load acting on the slab s carried by the beams, determine he dead oad
and he ive load (in lb/ft) are acting on the beams'
4. Determine he additional component f dead oad representing he self weight of the
beam in lb/ft).
5. Determine he factored otal load (in lb/fi) that he beam must be designed o carry.
Apply load actors, iven n Table 1.2 n th e ext, as appropriate.
6. Draw the shear and bending mornentcliagrams, ased on total factored oad, for the beam'
wl
Or'
-\| .. ;i
a .Y-a
s\o-b
b eo -m r s up po rl in g stab
|-SYbI io
I i ; Ff| = ;o-I N q- ' s ]
| 1' r !
Fr rbi -T Nt.(--.-
colurnnss up por*rngbec..n:
- Nor ro 3cs\z -
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c,E r+33
\ . S t ab d ead load
?robte..^ 5 e F
/ . t .)o \uttovt
:
= lO in . X
No. t
\R Y \5olus
tL\n . f+?
4vo.* *ex+ TaLla
Fa-lt Zool
= IZS \t ,
f+z
Fh.= -i .5 St
tt"/F+
\u/Tf
Ex-fen'o. b o-\cang + t oo tbsstz
--^--
$+=
+[ ="
=tsL
q3e
!+
L.
Bear,', c.'rc2*1" ec.hon at
k>ecr,nr DL :
?o tn, x l2 fn, : 24O
^ z*ain** !+tt=
fq$t ;"1
in ,z
?5o (Fgf+
Fac.i ,o*qi l ,**e. \ ? t ,2 (qTB + 26o) + \,b (rso) ? ZbZ6,P,
Loo*d 5
5heqr
- tg.-l t( :- - Zb.3l(
- zz . gK
13Atf . \ (
W ,'d t. al gld"b supporled b.{ eaej^ bec.*
DL bea.^
b gc^-*
\ 25 pss
roo plF
Area. =
t 5c) \bsf[r
x *1.5
( 1"9
7LZ6 rts /+L
Mo"qenf
' \31. j sf ,k
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CE 433 RbINFORCED CONCRETE ESIGN Fall2007
Problem Se t No. 2
Flexural Beam Analysis
Due September
The rectangular einforced concrete eam shown below s to support a uniformly distributed load,w, as shown n th e sketch. Th e beam will made of normal weight concrete w" : 145 pcf) with f '":4000 psi and a modulus of rupture 47 5 psi, and he reinforcing steel o be used n the beam willhave a specified ield strength f 60,000 psi. Negiect he self-weight of the beam n all calculations.
1. Determine he value of the load w which will cause he beam o begin cracking. Calculateth e stresses n both he concrete nd steel orresponding o this oad.
2. If w:2.5l
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CE 433 RbINFORCED CONCRETE ESIGN Fall2007
Problem Se t No. 2
Flexural Beam Analysis
Due September
The rectangular einforced concrete eam shown below s to support a uniformly distributed load,w, as shown n th e sketch. Th e beam will made of normal weight concrete w" : 145 pcf) with f '":4000 psi and a modulus of rupture 47 5 psi, and he reinforcing steel o be used n the beam willhave a specified ield strength f 60,000 psi. Negiect he self-weight of the beam n all calculations.
1. Determine he value of the load w which will cause he beam o begin cracking. Calculateth e stresses n both he concrete nd steel orresponding o this oad.
2. If w:2.5l
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cs +33 t-{W }-l.,.Z Soltrfiorl
Fir ' 'dILrto pco cf Je' c rfl.c L: ,5 :
?Q, ooo, croo psr
E3 ( +g) 's {-{-..
t lq/ -
Fatt ?-CI*T
3,4oo,Qoa psr
on b oth:""
k]T
psi
So* Afl -\., bo*o*o
\,tf
E5
Ec
Ca\c*t"ohe T +r 4ao
'n(3"
' l
/v' 2\.6 l \ E V6l ; i - 13r y_e
0r,qcr,*cked gh*n:
(n- r )As k-t)
( g" \ ,2? )
. ' l / ' -2L6,61 to-
(n- r ; A* , ?b, fu1 v,L
Cz (t j*zB)(r+)+ zL.r*l zs) s r+.-1 (n,( r3
xza\ - \ -zL.Aj
Eso a (F)(za)r * (r t ) lze) ( \+, '1s \ ,+) t + zb,6l (zE-\+, tS)L\ \\ t-
= z6,_1qo inLt
o ec"ur \A)ke* tanr ion StregS
rear[res
g:+f
t-fr =-1.5\+ooo
=
\"J/ I : drgia.nre-
l,^urackirg vJi t \
nro t D .aru
Ft=
Lat"^d M:
n?*wx
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zls\ nus:
2L,rqo in*
Solvinq 1^J- - I 3 3 \bS = \ .60 krpcrG,*
5t 're6se $ ..
Fop Fcr = (brl?B)(zorrz)a (r+-1 5)2L,-] q o
Fcr a 51-1 FsL*
battorn Fc z = $n a 4f ut p g,
Fc al- (evet aF shee.t Fc*-
(Zb-t+ ,15) x +r *
tza -*+. t t
$c+ = 361 Frl
jn $s-- nScs= A(aar)
Fs ' ?1?,b psi%
\ r* psl = ( b "r)( .0$r* ,
kt (r * to, r+,-rb,n
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zIs?, For rnr= ?,S k/Fb, \dhoF are sFresses
6tngg VI: 2 .5V/\+Fr
Sec-hon rg ero.c-ked
, - ' (' a ' t- l-{--T__
7r,,l l-JtY_-. ' - n As = 8 (3xr,zr) : 3o,4b in ,
+ 3o.r t rB (ZS-c) :O
B.-13 t. ' .
*( 'g)(a . -18)= * (r :XB,-18) I e , r3\a'L tz /
+ jo "r{B(zq - g.- tg)e
\o,Q 5D in+
/ a\ / \ /.
\( t J ) [c ) L c)z
t'bOlUrnQ L;
J
-T- , r ,r- far
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+153. 6hres* f i 'n , ' t s ;
$. ( o,L!glL: \AOopr;
J+t ./rs . 7) . ooo P (
l . t l
AttowaJate- SFre ses a.pp[t rJnder. ecLrr 'ce oad ecmd' 'h 'onS
!, ose e ackeJ gaehan {n r an*\s\!
I'F conccebe s fe
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s lstf. W h&F \^r
be.o...t
r^JnJd pnoduce {a^ f r^e
is undec. fei.,F, = f s el ftruUrt db=\"
) / l"t
eaeiQ ^neft /2"
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2, Repeo* problewr wi+,^ ( Na,l bars As=6Xo,6
e) Pp.ou,= 3.6 = c.org ( Poo,l ro) (ro)
ono -,"i: jej^I594
b) Aspnou, 3. L t'ni ) As *i.
of Po"r. lure Q*: (g.a)(ao) : a ,35i . , ,offil
c : 6,35 = ?,+J in,o, a5
a-1]Zo
Bl"'- :9f nct_."qg,."T
v-
-I-l
3-6 (ao) (zo
36 34 i . , . :
o-374
'o' tt ) o, oos
6,35\-t 2-)
3oz,q Pl .Ltpr
( 1us*)
c) Mn
Mn
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\o"
,l73,75
No, 3 9{u\rup
l ' lL" c l , zove , .
o, gJS "["tu'
cl ,CoVe,/-
),du>/
5 No.?Ica-rS
/ v \
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3.
ed
fiClC:J
ci:'ir,1
fiflrl
v
-.{t'\m-x\),'3
sfisirnp\1^ 5uPt-. aea* u niForrn load : $e\F W\ | O'S ta. bu
6pau, = lL Fl ,
b = \* i^ 3 lJr.-l \ocnll
+LL
lrr l
ta : tooo pst
c{ = \1 ,5 o,(
At = 3 (o.a) L{ :. 4 o,boa q.r l"
h = 22 in, : \ ,BO i ,*L
serF*!,= (b ) (h) ( r++ leug+h)\ 5ok
b gc.r,n nomtns,l nncr*.e,r'h (Fver,.,gtr^ : Mn
G.= t l .?) (+o) . Z,o2f ,^o.ss (s)( t*)
V\n" (\ 'B)(Uo)(f l .E- tgl)= \33\ i" ' \bs = llo.g+f 'kipt
+ l.lrn / Mu
o.q(r ro .q) : ! .1*t ru) ' \i
ot t tTur. : 3 . l? . k tp . l / r+-
3' z
Or f'-SV. \iUe \oa.d z t . t l Ylf+f
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*. , a) *tovgr-rrn$ere-edbeao, ' , pp.* )"puo) c"oncrehe- rughes be.fore S Fe et $, qtct:
e. britHe, *oi [+re,'urrder-
rei^Sorced beo*t, _ ?gcuv+ 5 t-ee.l gietgg before consce,Fe rusher
: ci *ch le $o.i [or,Tg (e,zz) (ec:) 41. t = ?q .J kier\ ll ,
+lln= \3,) + Zq,-? _ +1,i
Ireid) 4v. a o,-?s (z) {E; (s+B)(eo) = \q,o \ . .p ,
rsclo
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=l+Z.
1'.*fg
-? .+\'
- 58.1 (
V* o-F d c''\^ro,.t Svo* Sup\' c-
6 v" = 6 15 z) {I& (ra) (go)
- -13.Lt
(rr- ?"lz)-1r
=' 5l .Z b
' ' /l ( '
-A l 4-c.y vc -- 4 >,e
dtsb, F.ro*
d,tsF. Pront
StJpporL heq Vu --
gugporf t"Jhen u.
4'V. =lrz- ( H-x rz) = 7.g3f+
+* , ,. -(r#r* 't)=-r,8(P
t - lz4 s+.
fl= t-fooopll
\ = 6o,ooo sr
VJ*,r= \ ,?(2, '1 ) + l .b( t ,g \ ? 6, tZ k/*
5rr.[ rn1 sheoc d*'a6eerr* Soc bga* t
tz58"1 t Vu {Lfd
bL: ?,1 r/+r 'l:\;tr,LL s l ,& vl $*
pof nkShvcu Ps
b= 18"
As = 6 t-Ja,9
, .. l3 '* : 59, \u '
r,rjkeY{-
aa
-
no \ nn5er h+edeJ
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"i!J-;:,:: ;:'r l:.
=bTt1 * 3 Shrr^u;*r
af Vu s.Fd' : 58.1 k
e.tr/ , 'J fecu = e.-19 Z*o, t r ) ( 6o) ( 3a ) _ u3 in.-- ,---**--.: ***.-i*-*;' I
59, \ - 5\ ,L
noF +o ercceed 5*of 4. d = Zo?z( Z4 u.
r ls rel id = A.Q 4 = (o, tz)G")Qd 9,ZL5+3
r \q L*d ? +G6- ( ,a)(ao) \3- tktC3()ofo cLbclfe S*oor cr k-
oc Av: *io)
s { A'rfi_ s (:,t#g*H = \s.E no.rs{Sfb^r o,1sffi (rA)
, lh( &h a ( o,ea)_( ?-S) :- ({,'? nSohr to (r a)
.'o R'""r S*rnr = \\I t,.,. la*;o};*, ^,*+ a-n/ Aq**)
LSc'Le ar \.l. n.
Pot'*b ulLe*,e vr,o sh'rrupr re{J = l. g t +f ,
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=:j.rii;r55a:_l
+frl1.
"lt1Yf dp p { o.ce rvr r'l '.
t +L lf *-$
1 lJ; r.+
Fr-*sl d eg rgn '.
a
3
f ter.
t
l* t ' ,n
i" i ' )
II
- l -
l l iIr l
. J ,J- .1I
-l --t- -
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CE 433 REINFORCED CONCRETE DESIGN
Problem Set No. 6
Integrated Beam Design
Due October 15
Fal l2007
The simply-supported beam shown below spans between wo masonry walls with a wall centerline-to-centerline pacing of 25 ft. The beam s to carry a superimposed istributed dead oad of 1.0kips/ft, a distributed ive load of 1.0 kips/ft, and a concentrated ive load of 42 kips applied at themidspan ofthe beam. All loads are service oads and do not include the selfiareight fthe beam. Forarchitectural easons, he beam will have dimensions f h: 32in.and b :20 in. A complete designof the beam s to be made using , = 4000 ps i and i:60,000 psi. A clear cover of 1.5 n. isrequired.
I ' Determine the required flexural reinforcement or the beam. Provide the steel using No. g2+ bars n two layers. Remember o consider he beam self:weight.
Vet a) Cutoff 2 5Yoof the reinforcement where permitted. Include calculations ofthe exact cutoff' locatron.
4l- b) Check to be sure that adequate embedded ength is provided for the continued anddiscontinued ars.
6pr '
|-t
d) If required, pecifu pecial einforcing etails n thevicinityof the ebar utoffs.Assume' t thatNo . 4 stimrpswill be used. tzlo.6
4t c) Check special equirements tthe supportwhere he moment s zero.
Design the shear reinforcement for the beam using No. 4reinforcement sing open stimrps with 135o ooks.
Make a sketch of the beam summarizing alr design details. b, 2o!'H:
T]1J.
stimrps. Detail the shear
tc'
h=32
Wa
twa.tl
q,
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Ce 433 l.\W NJo,A l-alt'/g2ooJ
2D"r*f
t ,o V/ptr deo.d Ise(P.:l-
ft . o k / +f l ive )
*Lv'of rnidspan lrt .or.."nlvc,kd {
o. 6-1 u /+t
Tc.
\
c[ ,
(za- IE) + ?3tLt tz
hr
tlruce.4, :
--
!
: t$ooo
= 4o,ooo
Covec =
PJ(
Pt .
\ .9 r 'n ,
l oc^dE :
xlio:too()
ve
oad
SelFv. l t = (zo)a3)r**
\ .2 (o .61 * t ,o) + l .L( l ,o)
l . L ($Z) = L1 z I \l
Oi - hr.|tC SFan =.
ccl culohbns, b ecr* S pan
c Ieoc k pa-n + beo,r,n deflL -_
z,3+Ff- fb e oF suppavf5 = ZS
.'o hI*
t>tul t ,
r.o(
m,.uniPoccnloc.d
3,6o vl *
+F,
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zlqE sl-d,bt, h monncnF o.nd sheo.r dt.agcome
Vut tV= 16,6
33.Lu
!
'iI
5clpF.c - 33.6(
supl-EI
e g.6k
Yu**' t (e .so\(zs)?+
t (6r,2 (zS) = Jot $h.\
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