4.2- Allowable Strength Design

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4.2- Allowable Strength Design
Steel reached Fs
fbfc/3
Fs24,000 or 20,000
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Example 4
Assuming a dead load of of 800 lb/ft and a live
load of 400 lb/ft, applied at the top level of
the wall shown below. Design the beam assuming
fully grouted CMU with a nominal thickness of 8
in., and a specified design strength of 1500
lb/in.2.
4 ft
12 ft
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1- The span length 12 x 12 ½ x 8 x 2 152
in. 2- w ( 400 800 )/12 100 lb/in 3-
Mdesign wl2/8 100 x 23104 /8 296,131
lb-in. 4- Determine depth d the minimum masonry
cover is 1.5 in. (MSJC 1.13.4(b)) assume the
half bar diameter (unknown yet) is 0.5 in. Then,
d 4 x 12 -2 46 in 5- 6- As 296,131
/(7/8)/46/24,000 0.31 in2 7- Check on masonry
compressive strength AsFs ½ x fb x b x kd fb
2 x 24000 x 0.31/7.63/(0.375 x46) 112 psi lt
1500/3 8- No minimum or maximum reinforcement
ratio
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5- Shear Design of RM Beams
Sum of the horizontal forces gives
(masonry)
5
1- the shear stresses 2- The allowable shear
stresses is the minimum of 50 psi or (fm)0.5
(MSJC 2.3.5.2.2 (a)) and if fv gt Fv ADD
reinforcement (MSJC 2.3.5.2.3). 3-Ensure that fv
lt Fv 3(fm)0.5 4- If steel stirrups are used,
the min. reinf. area Av Vs/(Fs x d) 5- max.
spacing is d/2 or 48 in
1- the nominal shear force 2- if Vu gt 0.8 Vm
(MSJC 3.3.4.2.2.2) ADD shear Reinforcement and
Vn Vs Vm 3- Vn lt 4An (fm)0.5 4- Av
Vs/(fy d) 5- max. spacing is d/2 or 48 in 6-If
steel stirrups are used, the minimum
reinforcement area 0.0007 bs
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Example 5
Now, we will complete the design of the lintel in
examples 3 and 4. First using Allowable stress
design method V (12 4/12 x 2)/2 x 1200
7600 lb
4 ft
12 ft
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1- the shear stresses 2- The allowable shear
stresses is the minimum of 50 psi or (fm)0.5
(MSJC 2.3.5.2.2 (a)) 3- if fv gt Fv ADD
reinforcement Or increase depth (MSJC 2.3.5.2.3).
1- fv 7600/7.63/46 21.65 lb/in2 2- Fv
minimum (50 or 38.7) psi 3- fv lt Fv no
reinforcement
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Second using Strength design method Vu (12
4/12 x 2)/2 x (800 x 1.2 400 x 1.6) 10.1 kips
1- P 0, approx. (M/Vdv 1), Vm 2.25 x (7.63
x 4 x 12) x 38.7 31.9 kips 2- 10.1 kips lt
0.8 x 31.9 No shear reinforcement and Vn 31.9
1- the nominal shear force 2- if Vu gt 0.8 Vm
(MSJC 3.3.4.2.2.2) ADD shear Reinf. and Vn Vs
Vm
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Example 6
Assume that the shear forces in example 5 is
increased by 100. Design the lintel shear
reinforcement using allowable stresses design
method. V 7600 x 2 15,200 lb
1- the shear stresses 2- The allow. shear
stresses Min. of 50 psi or (fm)0.5 and if
fv gt Fv ADD reinforcement. 3- fv lt Fv
3(fm)0.5 4- Minimum reinforcement area Av
Vs/(Fs d) 5- max. spacing is d/2 or 48 in
1- fv 15200/7.63/46 43.3 lb/in2 2- Fv 38.7
psi 3- fv gt Fv Add reinforcement 4- 43.3 lt 3 x
38.7 116.1 psi, the Beam dimensions are ok. 5-
Try a 4 single leg stirrup with Av 0.2 in2 6-
s Av (Fs d)/V 0.2x24,000x46/1520014.5 in.
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Example 7
Assume that the shear forces in example 5 is
increased by 200. Design the lintel shear
reinforcement using strength design method.
1- P 0, approx. (M/Vdv 1), Vm 2.25 x (7.63
x 4 x 12) x 38.7 31.9 kips 2- 30.3 kips gt
0.8 x 31.9 Add shear reinforcement
and Vult0.8(VmVs) Vs 6 kips 3- OK 4- Vs
Avdfy/s Using 4 Av 0.2 s 0.2x46x60/6 92
in. use d/2 23 in Avmin 0.0007 bs 0.12 in2
1- the nominal shear force 2- if Vu gt 0.8 Vm
ADD shear Reinforcement and Vn Vs Vm 3- Vn
lt 4An (fm)0.5 4- Av Vs/(fy d) 5- max.
spacing is d/2 or 48 in 6-If steel stirrups are
used, the minimum reinforcement area 0.0007 bs