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Unformatted text preview: CE4401 Fall 2009 Homework #8 Due Wed. Nov. 11 For all problems (for this homework and future homeworks) assume a #3 tie and max agg size of
3/4” 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 ACI code
this list should include the classes through Wed Nov. 4 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. For Example code code Topic I summary
section ae ' must be deformed '
10.2.4 stressstrain assumtions for steel elasticerfectl lastic 2. A 20 ft long simply supported rectangular concrete beam supports a uniform service dead load
consisting of its own weight plus 0.5 kips/ft and a uniform service live
load of 2.3 kips/ft. The width of the cross section is 14 inches. The
overall depth of the cross section is 22 inches. The effective depth of the
cross section (distance from extreme compression fiber to the centroid of 196"
the steel) is 19.6 inches. The beam is reinforced with 3 No. 9 bars. The
concrete strength is 4000 psi and the yield strength of the reinforcement is
60,000 psi. The concrete is normalweight concrete. Compute the design 9 C .
ﬂexural capacity of the beam (oMn). Is this beam safe? On a sketch of
the cross section at midspan, show the location of the compression zone. What is the strain in
the steel at ultimate? 3. a) Compare ¢Mn for singly reinforced rectangular beams having the following properties: No 22 3 No. 8 b) Taking beam 1 as the reference point, discuss the effects of changing As,fy,fc', b, and d on ¢Mn,
What is the most effective way of increasing ¢Mn? What is the least effective way? Look at the %
change in the variable being changed and the resulting % change in ¢Mn. 4. A 30 ft span simply supported beam has a rectangular cross section with b = l4 inches, and h =
32 inches. The cover is 1.5 inches and the beam uses a No. 3 tie. The beam is made from normal—
Weight 4000 psi concrete and has 9 No. 8 Grade 60 bars (in three rows). This beam supports its
own dead load plus a uniform service superimposed dead load of 0.5 k/ft. Compute the maximum
service live load that the beam can support. CE 4401 Problem 2: Given 11/11/2009 HW #8 KEY 2.1 A 20 ft long simply supported normal weight rectangular concrete beam supports a
uniform service dead load consisting of its own weight plus 0.5 k/ft and a uniform
service live load of 2.3k/ft. The width is 14 inches, the height is 22 inches, and the
effective depth is 19.6 inches. The 4000 psi beam is reinforced with 3 grade 60 #9
bars. Compute the design flexural capacity ¢Mn. ls this beam safe? Sketch the location of the compression zone on the cross section at midspan. What is the strain
in the steel at ultimate? Beam length L :2 20ft
Beam width b 2: 14111
Beam height h := 22~in
Dead load WD 2: 0.5E9
ft
Live load wL :: 2.3—ij
ft
Effective depth d 2: 19.6in
Concrete strength fc z: 4000psi
Rebar yield stress fy :: 60ksi Total steel area AS :2 31.00in2 = 3~in2 Determine factored moment Self weight of concrete «(C :: iSOpcf
. kip
Self weight w := «b~h=0.321——
S “{C kip
Factored load Wu := 1.2(wD + wS) + 1.6wL = 4.665?
t
wuL2
Factored moment Mu : = 233‘kipft
8
Determine moment desi n stren th A f
Depth of compression block a ;= A 2 378411
0.85fcb
Assume strength (in 2: 0.9 reduction factor Nominal moment strength
7 Mn := ASfy[d — 3) = 266~kip~ft
Design moment strength ¢~Mn = 239kipft
Check moment capacity (13an Mu Check := if [ < 1 , "This beam is not safe for moment" ,"This beam is safe for moment") CE 4401 11/11/2009 HW #8 KEY 2.2 Problem 2: (Continued) Check = "This beam is safe for moment" Compute strain in steel at ultimate Maximum strain in concrete ecu := 0003
Factor 01 := 0.85
Depth to neutral axis 0 := —a— : 4.45in
B1
. . . Ecu‘m — 0
Strain in steel at ultimate at z: —————— = 0.010 The strain in the steel at ultimate (>0.005) validates the assumption of it) = 0.9 The design moment strength ¢Mn is 239 kipft which is safe for our factored moment of 233 kipft. The strain in steel at ultimate is 0.010
and a sketch of the copression block at midspan is shown below. CE 4401 11/11/2009 HW #8 KEY 3.1 Problem 3: a) Compute ¢Mn for singly reinforced rectangular beams having the following properties b) Taking beam 1 as the reference point, discuss the effects of changing AS, fy, fC, b
and d on ¢Mn. What is the most effective way of increasing (Wm, what is the least?
Look at percent change in the variable versus the percent change in (Mi/in. Beam 1
Width b1 2: 14in
Effective depth d1 :: 20in
. 2 . 2
Steel area A81 := 30.79~1n = 2.37m
Concrete strength f01 :2 4000psi
Steel yield strength fyl 2: 60000~psi
Factor 011 := 0.85
Ultimate concrete strain 5011:: 0.003
. Asl‘ y1
Depth of compressron block a1 := ————————— = 2.99 in
0.85f b
C] 1
Design moment strength at _
assuming q; = 0.9 ¢Mn1'" O‘g'Asl'fy1'(d1 ‘ ‘3 _ 197191”:
a1
Depth to neutral axis c1 := —— = 3.51 in
611
Strain in steel at ultimate, ._ €cu‘(d1 _ C1) _ 0 014
¢=0.9 if greater than 0.005 Etl ’_ c1 ‘ '
Beam 2
A f
2 1
A52 2: 34.004112 = 3m2 S y = 3.78 in a ‘= ——
2 ‘
0.85fC1b1 a2 CE 4401
Problem 3: (Continued)
a
2
c2 2: — = 4.45 in
511
Percent change in variabie
Percent change in ¢Mn
Ratio to determine efficiency
Beam 3
fy3 2: 4000Opsi
a
C3 — —3' = 1n
511
2:  d — c
1 3
EB = C“( ) = 0023
C3
Beam 4
fc4 := SOOOpsi I: 0 a4
C4 = — = m
.814
M 4
Rati04 := m = 0.065 %Variab1e4 1 1/1 1/2009 HW #8 KEY 5:2 := ~————— : 0.010 A — A
2 1
%Variab162 := (5—5) 2 27 %
A51 43M — ¢M
%M0mcnt2 := L—EL—n—l) = 24%
¢Mn1 M
RatioZ z: W = 0.898 %Variab162 Asl‘fy3 = ——— =1.99in
0.852.14:1 a3: a3 f _.
%Variable3 := M = —33 %
fyl ¢Mn3 " ¢Mn1 %Moment3 :2 ————— : —32 %
¢Mn1 M t '
Ratio3 := W = 0.946 %Variable3 Asl‘ y1 .
a4 := —— = 2.39 In
0.85fc4~b1 a 4 . d _
em := M = 0.017 C4 f  f
%Variable4 := (—C4—C—1) = 25 %
fcl 43M 4" (MM
%M0ment4 := M = 2 %
Cme 3.2 CE 4401 11/11/2009 HW #8 KEY 3.3 Problem 3: (Continued)
Beam 5
A .
b5 := 16111 215:: ——S—1—¥l— = 2.61111
0.851fC1b5
a5
211 b —b
05 :=  = 3.51 in %Variab165 := M = 14%
b
811 1
5 . d — c ¢M — 111M
1 5 1
s6 := ﬂL—w—Z = 0.014 %M0ment5 :2 LEE—IL) = 1%
C1 ¢Mn1
7M 15
RatioS 2: 1—93163— 2 0.071
%Variab1€5
Beam 6
d = 22111 a1
6 ¢Mn6 2: 0.9Asl yl d6 — —2— = 219kipft
a1 a  d6 — c
116:: — = 3.51111 em := ill—~32 = 0.016
311 ‘36
d —d ¢M —¢M
1 1
%Variable6 := £6—1 = 10% %Momcnt6 :2 (—né—n) = 11%
d1 ch111
M
Ratio6 := %°—mem6 = 1.081
%Va11‘able6 This is the answer for
part a with beams listed
in numerical order with
beam 1 at the top and
beam 6 at the bottom: Answer for part b ‘ ll of the variables in this parametric study are directly related to ¢Mn meaning that an
ncrease in the parameter results in an increase in ¢Mn. AS, fy and d are all relatively ffective in increasing (11Mn while 1C and b are relatively ineffective at increasing 0M”. l
rder from most effective to least effective in changing ¢Mn: d, fy, AS, b and fc. CE 4401 11/11/2009 HW #8 KEY 4.1 Problem 4: A 30 ft span simply supported beam has a rectangular cross section with b = 14 inches
and h = 32 inches. Cover is 1.5 inches and #3 tie is used. Beam is normal weight
concrete and grade 60 rebar is 9 #8 bars in 3 rows. The beam supports its own dead
load plus a uniform dead load of 0.5 k/tt. Compute the service maximum live load that
the beam can support. General
Beam length L = 30ft
Beam width b = 14in
Beam height h = 32m
Dead load wD 2: 0 53“—2
ft
Tie diameter due — 0375111
Rebar diameter db = LOOin
Clear cover cc = 1.5in
Self weight of concrete «(c :: 150pcf
Concrete strength fC := 4000~psi
Rebar yield stress f := 60ksi Total steel area Vertical clear distance y
AS := 90.79in2 = 7.11in2 between rows of rebar S ;= Lin
Steel modulus of elasticity E :2 29000.ij
fy
Yield strain of steel 5y := —— = 000207
E
Maximum strain in concrete gcu ;= 0003
Factor 61 := 0.85
Determine if steel has ielded g l rm?)
ASfy V
Depth of compression block a = = 8.96.1n
0.85fCb
Depth to neutral axis 0 z: i = 10.54in
51
Effective depth _
for bottom row .= h _ CC — dtie " — 111 Effective depth
for middle row Effective depth
for top row (12:: (11 —db ~s=27.6in I: "db —s=25.6in CE 4401 11/11/2009 HW #8 KEY 42 Problem 4: (Continued) Steel strain for Eculdl — c)
bottom row 51 1: w = 0.00543 C Steel strain for €cu'(d2 ' C) middle row 52  = 0.00486 C Steel strain for 6cu'(d3 _ C)
bottom row ' E3 ' = 0.00429
0 We see that all of the steel has yielded so our value for a using the steel yield stress is correct.
We can compute the moment capacity using the average of the three cl values which is d2. Determine moment design strength Assume strength (1):: 0.9
reduction factor . a _
Nominal moment strength Mn 2: ASfy[dz — = 823hpft
Design moment strength ' (taMn = 740~kip~ft The strain in the extreme tension steel at ultimate (51>0. 005) validates the assumption of q) = 0.9 Determine maximum service live load Maximum factored h 8¢Mn _ 658 kip
uniform load Wumax — 2 — . L
. kip
Self weight WS :: ﬁdbh 2 0.467.__ Maximum service ‘ .
live load and round w z: —————————— = 3.39.—  L 1 6 ft
down conservatively  ...
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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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