Q3__Bar_Cutoff_3.pdf
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231 page CIVL 4135 Bar Cutoff Example 2. The simply supported beam shown below is subjected to a uniform load. Cut the longitudinal reinforcing bars where they are not required for flexure. Satisfy the provisions given in Chapter 12 of ACI 318--89. 21” 18.5” 24” 14” f c ′= 5 ksi f y = 60 ksi 24’--0” 10” 10” 12” 0 20 40 60 80 100 0 2 4 6 8 10 12 Distance from support, ft V u (kips) φV n 0 100 200 300 400 500 0 2 4 6 8 10 12 M u (ft--kips) Distance from support, ft φM n 5 φM n 4 φM n 3 φM n 2 x
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Transcript of Q3__Bar_Cutoff_3.pdf
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231page
CIVL 4135 Bar Cutoff
Example 2.Thesimply supportedbeamshownbelow is subjected toauniformload. Cut the longitudinal reinforcingbarswhere they
are not required for flexure. Satisfy the provisions given in Chapter 12 of ACI 318--89.
2118.524
14
fc = 5 ksify = 60 ksi
24--010 10
12
0
20
40
60
80
100
0 2 4 6 8 10 12
Distance from support, ft
Vu (kips)
Vn
0
100
200
300
400
500
0 2 4 6 8 10 12
Mu (ft--kips)
Distance from support, ft
Mn5Mn4Mn3Mn2
x
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232page
CIVL 4135 Bar Cutoff
2118.5
14
5 -- #10 bars 4 -- #10 bars 3 -- #10 bars 2 -- #10 bars
Mn5 = 480 kips ft Mn4 = 393 kips ft Mn3 = 327 kips ft Mn2 = 225 kips ftd = 20 in d = 19.75 in d = 21 in d = 21 in
Vc = 2 fc bwd = 2 5, 000 (14)(20)
1, 000= 39.6 kips
Determine the permited shear force
Vs =Avfyds =
(0.4)(60)(20)8
= 60 kips
Vn = Vc + Vs = 99.6 kips
d varies with cross-sec-tionbut assume d = 20 in forshear calculations.
Vn = 84.7 kips23Vn = 56.5 kips
if x is measured in feet from the beam centerline:
Mu = wuL2
8 wux
2
2 Vu = wux
NumberofBars
5 480 -- --
4 393 4.8 57.7
3 327 6.6 79.0
2 225 8.6 103.7
Mnkip--ft
Mu with MnIntersection of
x (ft) x(in)
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233page
CIVL 4135 Bar Cutoff
OPTION 1: (d varies with cross-section, use d = 21 in.)Cut two upper bars first
ld from centerline OR d, 12db from where nolonger needed for flexure
79 in + 21 in
43.1 in 100 in
Vu(x = 100in) = 6.5 10012 = 54.2 kips
