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ESR-1917 ( Most Widely Accepted and Trusted Page 13 of 13
<br /> A4 Y tr. 771 A
<br /> Given: ,
<br /> Two 1/2-Inch carbon steel KB-TZ anchors under static tension ANI
<br /> load as shown. s----- 1 c'.•- 1- --------- ,
<br /> 1,5 herf
<br /> her=3,25 in. '>:: ':,:
<br /> Normal weight concrete,f,=3,000 psi
<br /> No supplementary reinforcement(Condition B per ACI 318-11
<br /> Assume cracked concrete since no other information Is available, = : .
<br /> I 1,5hei
<br /> . Needed:Using Allowable Stress Design(ASO)calculate the I
<br /> allowable tension load for this configuration. L1,.5 her o'°4".,-..i
<br /> AA
<br /> Calculation per ACI 318-11 Appendix D and this report. Code Ref. Report Ref.
<br /> Step 1.Calculate steel capacity: ON = nAf�=0.75 x 2 x 0.101 x 106,000=16,O59Ib D.5.1.2 §4.1.2
<br /> Check whether fete is not greater than 1.9fy,and 125,000 psi. D.4.3 a Table 3
<br /> Step 2.Calculate concrete breakout strength of anchor in tension:
<br /> N = c N
<br /> cbg ec,N ed,NYc,NWep,N b 0.5.2.1 §4.1.3
<br /> A
<br /> Step 2a.Verify minimum member thickness,spacing and edge distance:
<br /> h,,,,=6 in.5 6 In. .'.o]C SmIn 2.375,5.75
<br /> 2.375-5.75 D.8 Table 3
<br /> slope= =-3.0 Fig.4
<br /> 3.5-2.375
<br /> For cnin=4in=
<br /> 2.375 controls 3.5,2.375
<br /> San =5'75-{(2.375-4.0)(-3,0)]=0..875<2.375 in<.6 in.'.ok 0.875 .w. min..
<br /> 4 Cmin
<br /> Step 2b.For AN check 1.5he =1,5(3.25)=4.88 in>C 3.0h,=3(325)=9.75 in>s D.5.2.1 Table 3
<br /> Step 2c.Calculate AN,,.and AN,for the anchorage:
<br /> AN„=91.4r=9 x(3.25)2=95,1in.2
<br /> D.5.2.1 Table 3
<br /> AN,_.(1,5/tee+c)(3he f+s)=[1.5 x(3.25)+4113 x(3.25)+6]=139.8in.z<2AN„ .:bk
<br /> Step.2d.Determine W,,., N: eN=0,.+VecN=1.0 0,5.2.4 -
<br /> Step 2e.Calculate Nb:Nb =k,,.Aamief=17 x 1.0 x afro x 3:2515=5,456 lb 0.5.2.2 Table 3
<br /> Step 2f.Calculate modification factor for edge distance: V"ed,N=0.7+0.3 4 =0.95 D:5.2.5 Table 3
<br /> 1.5(3.25)
<br /> Step 2g.Calculate modification factor for cracked concrete: Vr,,N=1.00(cracked concrete) D.5.2.6 Table 3
<br /> Step 2h.Calculate modification factor for splitting: , N=1.00(cracked concrete) §4.1.10
<br /> Table 3
<br /> Step 21.Calculate 0.Nebg:0 Nag=0.65 x 139,e x 1.00 x.0.95 x 1.00 x 5,456=4,952 lb D.5.2.1 §4.1,3
<br /> 95.1 D.4.3 c) Table 3
<br /> Step 3.Check pullout strength:Table 3,OM/we=0.65 x 2 x 5,515 lb x 13,000=7,862 lb>4,952 .'.OK D,5.3,2 §.4..1.4
<br /> V 2s0o D,4.3 c) Table 3
<br /> Step 4.Controlling,strength:0 N,b,=4.,952 lb<OnN,.<ON, ;, 0N,,g,controls D.4.1.2 Table 3
<br /> Step5.To convert'to.ASD,assume U=1.2D+1.6L: T 'l'.952_
<br /> au,w=--3,346 lb. _
<br /> 1.48 §4.2
<br /> FIGURE 7-EXAMPLE CALCULATION
<br /> /6 /v
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