Homework #8 (Due Monday 11/26/2007)

— David Wagner 2007/11/24 11:21, CE3253 Fall 2007

Problem 1.

Using the same sequence of steps as in Lecture #19, design the center pier of the shear wall with openings at the end of that lecture. If you find that flexural reinforcement is required, say so, but do not continue further.

V_pier=V_total/3 = 3600/3 = 1200 lb
M = V_pierL/2 = 1200*10/2 = 6000 ft-lb

Assume the roof is bearing on the other walls, as in the examples. Worst shear and bending are at the top of the windows.

a) Use the allowable-stress design provisions of the 2005MSJC Code.

Try face-shell bedding with critical shear section near top of window.

$N_v = (3.33+{{16.67-10}/2}) * 48 =6.67*48=$ 320 lb/ft
$f_v = {VQ}/{I_n b} = {3}/{2} {V}/{A} = {3*1200}/{2*2*1.25*12}=$ 60 lb/in²

Fv is the smallest of:

$1.5 sqrt{f prime_m}=1.5 sqrt{1500}=$ 58.1 lb/in²
120 psi
$v+0.45{{N_v}/{A_v}} = 37+0.45*{{320}/{12*2*1.25}}=37+4.8$ =41.8 lb/in² < 60 lb/in²✘

Try fully-bedded, fully grouted.

$f_v = {VQ}/{I_n b} = {3}/{2} {V}/{A} = {3*1200}/{2*7.675*12}=$ 19.5 lb/in²

Fv is the smallest of:

$1.5 sqrt{f prime_m}=1.5 sqrt{1500}=$ 58.1 lb/in²
120 psi
$v+0.45{{N_v}/{A_v}} = 37+0.45*{320/{12*7.625}}=37+1.6$ =38.6 lb/in² > 19.5 lb/in²✔

The critical bending section is near window mid-height, with 3.33+(16.67/2)=11.67' masonry bearing on it, fully grouted and fully bedded.

$N_f = (3.33+{16.67/2})*{7.625/12} * 120 =11.67*0.635*120=$ 890 lb/ft = 74.15 lb/in
c=4*12/2
$I={7.625*(4*12)^3}/{12} =$ 70272 in4
$f_t={{Mc}/{I}}-{{P}/{A}}={{Vhc}/{I}}-{{P}/{A}} le F_t$ ≡ $f_t={{6000*12*4*12}/{2*70272}}-{{74.15}/{7.625}} =$ 14.9 < 65 (fully-grouted, M or S)✔

Standard CMU, Fully Bedded Running Bond Type M or S PCL, Fully Grouted

b) Use the strength design provisions of the 2005MSJC Code.

Assume f'm=1500, Fully-bedded Running Bond, Fully Grouted

V_u=1.6*1200= 1920 lb; N_u=0.9*320= 288 lb

Vn least of

$3.8 sqrt{f prime_m} A_n=3.8*sqrt{1500}*7.625*4*12=$ 53865
33070 lb (Fully Grouted Running Bond)

phi V_n = 0.8*33070=26456 » 1920✔

f_t =1.6*14.9= 23.8 lb/in²; phi f_r=0.6*163= 97.8 > 23.8✔

Standard CMU, Fully Bedded Running Bond Type M or S PCL, Fully Grouted

Problem 2.

A simply supported masonry lintel spans across a 20 ft opening, and supports a uniformly distributed load of 1400 lbs/ft, including self-weight.

a) Using the allowable-stress provisions of the 2005 MSJC Code, design the masonry beam using 8-inch, fully grouted CMU. As your initial assumptions, use k = 3/8, and j = 7/8.

Assume simple support:

14470 lb
$M={wl^2}/8={1400*20.67^2}/8=$ 74770 ft-lb = 897000 in-lb

f'm=1500 (Type M or S, 1900 psi units)

$F_v= min (sqrt{f prime_m},50)=min(38.7,50)=$ 50 psi

$f_v=V/{bd} right d ge V/{b F_v} = 14470/{7.63*50} right right d ge$ 37.9 in → Need at least 5 courses to resist shear. d=8*5-1.5-0.5= 38 in > 37.9 in✔

$A_s ge M/{F_s j d} approx {8*89700}/{24000* 7*40}=$ 1.07 in² ⇐ 1.2 (2#7)

Use at least 5 courses above the lintel with 2 #7 bars near the bottom

b) Repeat (a) using the strength provisions of the 2005 MSJC Code. Assume that two-thirds of the unfactored load is due to dead load, and one-third to live load.