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Unformatted text preview: Print name 60L U T1 0 M
l have abided by the Virginia Tech Honor Code.
Sign name CEE 3424 — Reinforced Concrete Structures I
Fall Semester 2005 — November 27, 2007
Exam #3 1. Name the three factors with the greatest influence on crack widths [3].
Swain in ﬁne rebar Coggp over We. (doctr 59cmth 0 ‘9 Felon“? 2. Give two reasons why the 4) factor for shear (0.75) is smaller than the 4) factor for tension
controlled ﬂexure (0.90) [2]. ' Less ("taunt a‘e Calculation EFKPHQ {OCAUVQ 3. Name 4 of the 6 requirements that a oneway slab system must meet in order to use the
ACI approximate method to determine slab moments [2] +wo or more, spans unitormw dust“ buteﬁ LDMS
New“: is pm‘ smmﬁg spams more or“ less ecbwti
LL. 6351... braced (:r‘mwc 4. For the indicated Tbeam, what is the effective flange width? [3] IOft—Oin ~~~~~~~~ \ determine . H '2 effective _
'2'” m flange F4— I81n
spon length — 20ft width for
~ this beam ~~~~ 5. Is the following Tbeam adequate to support a factored moment, Mu= 500 kft? [20] #— I8"——1 f’c = 3500 psi 26" fy = 60,000 psi “.990. No.8 bars 1
As’q 5.7911} 3 Zlhn
I e0. No. 8 — As = 0.79 in2 4.5..  in. clear between layers
{5 a ‘m £\om<3e:) ° 45: 0L6 + 333600332002)
7.1lvbo _ 797“.
£1,893,645 ’ no 65; 0,76
Dawda amd Cahabver'. u?) x 79—. 4E4VH
CF: bx(t8103',%6'3I5 . 142.3: CPMn ‘ 9
(men:  . 7 r . LSOO‘V"
'A‘SF= bolts:  2'38"“ ASN : 2~38 : 4‘7ginz £92
a: 4:134“) : 9.34m
#833,540
qés M“ = 142.5(2153)+ 4.7sxeo(2t.15‘ 2
Mn : 73‘90 ‘m'L ’ 9‘3 k'g c: ———= 11.22pm
4,6 = 2m + a" +921 235;“ .007: :
gt  (Tamas—H.227 .0037: 'h'ams PH 0 n 6. For the following slab layout [20}: . . . _ slob depth = 6.5"
distributed live load _ 75 psf Fe : 4000 psi additional distributed dead Iood = l0 psf fy = 60,000 psi Fm?
" Elm Wand—J L—l3‘O"—— L—l4'—0"—~l L—t2'—O"——l Uzi/0* a) What is the maximum moment (absolute value) in the slab, and where does it occur? . D
W§¢\(’ = (726 M50 : ‘4; u‘/
LU“: I.L,(7s)+t.z (towns) = 224.5 5? m
7." + S
( zzctsY'B’Q‘) — 3.30%”, larges L) @E MK: 4 /
2H Z ’
C? C M“, f/O(ZZQ.6)(L€3) = 3,69 l: «Cl b) At the location of maximum moment, what As is required for strength? cl = (as ,‘76 I25 : 6.6"
Mu ‘2. gr”:
T0r3e+ Mn; {)5 : 3'8“ : 50.7 io'k ‘ 5 l A
 As.“ «(3.9295 ' bot,92$55  O'Hob «Cf M h 5'0. .7
c) What size bars and bar spacing would you select for this location? (use #3 = 0.11in2, #4 =
0.20ln2, or #5 = 0.31 inz). ( Y b 5>
#5 cm“: * ‘y : 7‘qsan ASMM:.5ol8lpln: ,00l8 tz , ' L‘V.’ .(r' 2 4401/3164»
7:5 ‘ . l \\ O ( do 89“.“ #4 [156 1 l2. " (A'S‘V‘ 1g3C? 7’7— ’ .. r’ \
n “ ’4 H o 2 K //,
5m: Sk=5o .r ta [email protected] l4 (" > x/ d) What additional checks would you make? ~ctdvaﬂ Mn
\. gt J gt ) 9b '_ CﬁMﬁad— > M“ 7. For the given beam, if the service load moment is 90 kft , is the spacing of the ﬂexural reinforcement adequate to control crack widths? (1 ea. #8’bar = 0.79 inz) (neutral axis depth,
c = 3.71in.) [10] n—zw—ﬁ T #4 stirrup l8" 2" clear cover over stirrup
3 ea #8 f'c= 5000 psi L fy = 60,000 psi note: inside bend diameter for #4 bar is 2" Cc : 2* pg : 20% in O~$$um¢ L, : 40 task :9 3: Eco _ : 87SU‘ “J 4 AOL
4.0 > na+ 0k 0A 5 ' 5
5M: 26  zaps“) : (as M
» _ , : S" i"\
Chad; w/ MM; (5 a. I82 6 .s I
“M ﬂqo“;  Bai ~
2 ‘ ' I ' o S.
a : Ami— Z3 2.3106  k 8. Given the following beam crosssection and loading diagram: [20] f'C = 6000 psi V A _ 5w24
fy = 60,000 psi m _ 384E] wD =I500 Ib/ft (including selfth
/wL =I500 Ib/ft IIIIIIIIIIIIIIIIIIIIIIIIIlllllllllllllllllllllllIlllfiilllll
IIIIIIIIIIIIIIIIIIllIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 6 ed. #9 bars _ 2 25—0"
As = 6.0 In a) What is the cracked transformed moment of inertia, la? 1
3 7&9)
EC —— 574F760; = 4416 u: 1—“, 51mm”) J, (icyrad 1 2
h = 222:0 : (057 + 5.9.97 [205 7,88)
4 .. . , “x
1%; = aésﬁzos c3 I“ ; 33% m C : 7o Rh
b) What is the cracking moment, Mcr? «
13 : ‘(ZUchqP : (8)432 in £(r : 7,5 Jéooox 1 {’5‘ “\WM/
. ‘2. . : : : 8%2“h"; 6" l1 _.
7
c) For the given loading conditions, what is le? , + . 7' I ‘
Ma= Us lag5 : 234.4 MC" V
* 5 m4 4 _ 74.3 _ 7:32. ) 890 = m ;”
IC ' (254A) .8432 *0 (234.4) 8 . m d) What is the instantaneous midspan deflection caused by dead plus live load? 5( aZlesqu 123 0 kg. ..\
: . W\ ” 2;
384'446 753‘ °ll‘l4‘m"t ,/ w: 9. For the following v.,/¢ diagram (for half’of a beam) and beam cross—section: [20] #— 36“ —~—~1
l00 centerline
of span I
V 80 l95" '2” I2"
TU 60 60k '
kips L
40
20 f'c = 4000 psi
fy : 60,000 psi
2 4 6 8 l0 I2 lec1.#3:0.lin2 \
Distance from support, ft I 90. #4 z 0.20 in? a) What is the maximum Vul¢ for which the beam must be designed? Slope: gig“O : 431$ “4+ p
4 ,
\%= ﬁsE‘altm = élik ‘ b) At the critical section, what is the required steel contribution to shear strength (VsVrequ)? V¢;Z \l4ooo ' l‘i'$'lZ 7 Z‘lo‘ck (22 V
0 Vv/gvb: 87.9—29.4, = 683k c) What is the reguired spacing of #3 bars to provide this strength?
\I ,_ AV 4‘3 a 5 : M45 4 .. L ﬂ/ O,\IK3X(DO‘lQ0$ _ w
5 = 6 8 .3 ’ é' E ‘7‘
d) What is the required, spacing of #4 bars to provide thisstrength? 1 0,7.x3 : IZ’O“ 5 53.5
e) Whatisthe maXimum sPacin allowed at this location? t, W J g ' d’z “ Z4 L3
4 4000 ' H.341 = 64.2.!4 >585L 
C\/ 7s“ I‘"’ ice/Vial}
f) What bar size and spacing would you choose? 2 ; a. #[email protected] 0%" or “46 qﬁuboverm 50.12! '93 : 0,0563: 4 a3? ...
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 Fall '07
 CLRobertsWollmann

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