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Unformatted text preview: CE4401 Fall 2009 Homework #12 Due Mon. Dec. 14 For all problems (for this homework and future homeworks) assume a #3 tie and max agg size of 3/ ” unless otherwise stated. You may make any other assumptions necessary, but be sure to state
them in your solutions 1. Make a list of all the references you have seen in class or doing homework to the AC1 code this
list should include the classes through Wed Dec 9 and this homework assignment. Include the
code section, code page number, the topic, and a brief summary of what the code says in that
section if possible ( i.e. if you can summarize it in a few words otherwise leave this column
blank). Do not copy word for word out of the code. Please hand in your ENTIRE list (i.e back
from HW 8) and indicate what is new on the list. 2. Determine the development length required in the column for the bars shown using both the “short
method” and the “long method.” If the available development length is not sufﬁcient to develop the tensile strength of the steel, design an anchorage using a 1800 hook and check its adequacy. fc’ = 4,000
psi. fy = 60,000 psi. The clear space between the No 10 bars is 3 in. with side cover of 2.5 in. The top cover is 2 in. (as shown) and the cover from the end of the bar is 1.5” (as shown). No.
3 stirrups spaced at 7 inches are present throughout the development length. I 2" clear __ —______ l 1.5"C1631' lén 1" 3 3
l
1 4 3" l 3. For the simply supported beam shown, draw the moment capacity (oMn) diagram. The beam has a total length of 16’ from center—to—center of supports and is symmetric about the midspan.
The effective depth, d=21.5”. Assume that the cutoff location 3 ft from the support satisfies the
requirements of 12.105, and the 1211.3 is satisfied. What is the maximum uniformly
distributed factored load that the beam may carry. Use f,” = 3500 psi. fy = 40,000 psi. 1’14"“ 4. An 18 ft normal—weight concrete cantilever beam is subjected to uniform factored load which
produces a factored Mu = 3500 “k and a factored shear V“ = 32.4 k at the face of the support. Design the top reinforcement and the appropriate embedment of 90° hook into the concrete wall
to sustain the external shear and moment. fc’ = 4,500 psi. fy = 60,000. You do not need to design the shear reinforcement. Include cutoff of reinforcement in your design. Do not worry
about satisfying 12.105. CE 4401 Problem 2. 14 Dec. 2009 HW #12 KEY 2.1 Determine the development length required in the column for the bars shown
using the short method and the long method. if id is not sufficient to develop the tensile strength of the steel, design an anchorage using a 180° hook and check
its adequacy. fc = 4 ksi and fy = 60 ksi. Clear space between number 10 bars is 3 inches, side cover is 2.5 inches, top cover is 2 inches and end cover is 1.5
inches. No. 3 stirrups spaced at 7 inches are present throughout ld. Development length using short method 0 No. 10 bars No. 10 bar diameter db := 1.270in Cover values amp 2: 2in aside := 2.5in
Clear spacing ._ 3 , between No. 10 bars 561" 'm Clear cover Ccl : min(Ctop ’ Cside) Ccl : Z'in Determine short
method equation to Equation 2: "Use 1/20 Equation” if SCI 2 2db /\ Cd 2 db use for NO_ 7 bars "To Be Determined" otherwise
and lar er
9 Equation = "Use 1/20 Equation" Matieral strengths fy 2: 60‘ksi fc := 4ksi
> 12" concrete below \Iit :.= 1.3
Assume uncoated bars \Ire :: 1.0
Assume normal . >\ = 1.0
weight concrete
Not more than 1.7 Factorl := min(1.7,\IItII‘6) Factorl = 1.3 Not more than 100 FactorZ : 63.2psi 1cc
Factor2z= psi~min 100, ——.
p51 fyFaCtorl
Development length 1 := max ———————~d ,12~in 1 = 78.3~i
dshort ZOAFactorz b dshort
Development length using long method
No. 7 and larger bars \IIS z: 1.0
On center spacing coc := sc1+ db coc = 4.27~in
No. 10 bars
Minimum cover ﬂ cover := mjn(ct0p,cside) + cover = 2.635‘in distance to bar center 2 CE 4401 14 Dec. 2009 HW #12 KEY 2.2 Problem 2: (Continued) Factor Cb := nﬁn(cover,£§£) ch = 2.135in 80 K U will be determined by a horizontal splitting plane Area of transverse _ ( . 2) . 2
reinforcement Atr '= 2' 011'111 Atr 2 0.22m
Number of bars ._ 3 being developed n '— Spacing of transverse ._ 7 . reinforcement S " '1“ Transverse K '_ 40A“ _ 041 .
reinforcement factor tr " Sin ' Ktr — . 9~1n
Cb + Kn. Not more than 2.5 Factor3 z: mn[__.é_ ’25) Factoﬁ : 201
b 3fyFactor1II's
Development length ldlong I: max mdb, 12111 ldlong = 58.41 Available distance L := 4.1“: + 3~in — 1.5‘in L = 49.5in
Determine if available distance is sufficient to develop full tensile strength of bars Check 2: "Length Not Sufficient" if L < ldlomy
5 "Length Sufﬁcient" otherwise Check = "Length Not Sufﬁcient" We see the available distance is not sufficient to develop the full
tens/e strength of the No. 10 bars so a 1800 hook will be designed Hook development length Hook development 002me'fy .
length ldh:= Web 1dh=24.11n
Cover factor Factor4 z: if(cside 2 2.5in,0.7,1.0) Factor4 = 0.7
Spacing factor FactorS := if(s > 3db,1.0,0.8) FactorS = 1
Minimum Fact0r6 := max(8db,6.in) Factor6 = 10.16in Modified hook _ .
development length ldhook .= max(ldh~Factor4Factor5 ,Fact0r6) ldhook = 16.91 11 8
424362 1
42389 2 ﬁaé‘iamﬂMwﬁ s3 [04% mm. E‘wyzéa CE 4401 V Problem 3. Development length using short method and #8 bars only which is conservative 14 Dec. 2009 HW #12 KEY 3.1 Draw the chn diagram for the simply supported beam. The beam is 16 feet from center to center of supports and is symmetric about its midspan. d = 21.5".
Assume cutoff locations 3 ft from the supports satisfy 12.105 and 1211.3. What is the maximum uniformly distributed factored load that the beam may
carry. f0 = 3.5 ksi fy = 40 ksi. No. 8 bar diameter db := lin Assume conidtions are sufficient to use 1/20 equation for No. 7 bars and larger Matieral strengths < 12" concrete below q] _
and assume uncoated t' Assume normal
weight concrete Not more than 1.7 Not more than 100 Development length 1d := max[ fy := 4O~k51
= 1.0
>\:= 1.0 Factorl := min(1.7,\Pt\Ile) fc
Factor2 := psirmin 100, —7
1381 f Factor1 Y Determine capacity of (2) #8 bars and (1) #7 bar Area of individual bars Area of all bars Beam width and
effective depth Depth of compression
block assuming f5 = f Concrete strength
factor and
ultimate strain Depth to neutral axis Steel modulus of
elasticity A7 := 0.60in2 Aloft i: 2A8 + A7 b := 14111
a ._ Aleft'fy
15f" 0.85fcb
pl 1: 0.85 C ._ aloft
left 61 E := 2900mm db, 12in] 20 X Factor2 fC :2 3.5ksi
‘l'e := 1.0
Factorl = 1 FactorZ = 59.2 psi 1d = 33.811 A8 := 0.794112
. 2
Aleft = 111 d := 21.5in aleﬁ = in ecu :2 0.003 Cleft = in CE 4401 Problem 3: (Continued) Steel yield strain Steel strain Check assumptions
of steel yielding and
tension controlled
section Nominal moment
capacity Strength
reduction factor Design moment
capacity Area of all bars Depth of compression block assuming is = fy Depth to neutral axis Steel strain Check assumptions
of steel yielding and
tension controlled
section Nominal moment
capacity Strength
reduction factor Design moment
capacity 14 Dec. 2009 HW #12 KEY 3.2
fy
5y 2: E 8y 2 0.00138
_ ecu'(d _ Cleft)
eleft .= ———— sleﬁ = 0.023 Cleft AssumpCheck :2 if (Eleft < 8y v Eleft < 0.005, "Not OK" , "OK" ) AssumpCheck = "OK" ‘ a1ch .
Mnleft ~= Atefrfy' ‘1 “ T Mnleft = 149 klp‘ft (heft := if (AssumpCheck = "OK" , 0.9, "TBD") ¢left = 0'9 ¢1eern1eﬁ = 1338 kiP' Determine capacity of (4) #8 bars and (1 i #7 bar . 2 := + = 3.76111 Aﬁght' fy , F . =3.61in
aﬁght 0.85_fc.b aﬁght aright
Cri ht :
g 51 Cﬁght = in Eright 1: —_ eﬁght = 0.012 right < Ey V Eri AssumpCheckZ = "OK" . aﬁght _ .
Mnright .2 d — — AssumpCheck2 := if (E 0ht < 0.005, "Not OK" , "OK")
D dpﬁoht := if (AssumpCheckZ = "OK" ,0.9, "TBD") (bright = 0.9
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This note was uploaded on 06/09/2010 for the course CIVIL 4401 taught by Professor Shield during the Fall '09 term at University of Minnesota Crookston.
 Fall '09
 Shield

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