Masonry Magazine August 1992 Page. 31
f =
Mpx
Ajd
(203 ft-lb)(12 in./ft)
(0.026 in.2)(0.963)(3 in.)
32.4 ksi
Step 5: Check the compressive stress in masonry and the tensile stress in steel.
F4/3(1/3f'm)
1555 psi
422 psi OK
F, 4/3(30 ksi) = 40 ksi
32.4 ksi OK
Step 6: Calculate shear stress.
f=
10 (9 ft)
bjd (12 in.)(0.963)(3 in.)
= 3 psi
Step 7: Check shear stress.
F4/3f79 psi, 4/3(50 psi) 67 psi maximum
Thus, F, 67 psi > 3 psi OK
Step 8: Design the reinforced brick masonry pier. Repeat steps 3 through 7 with the following design values:
M1000L 1000 (9 ft)
9000 ft-lb
V200 L200 (9 ft)
1800 lb
Step 3a: Calculate required pier reinforcement.
As req'd=
M
=
= 0.259 in.2
(9000 ft-lb)(12 in./ft)
Fajd (4/3)(24 ksi)(0.9)(14.5 in.)
Try 2 #5 bars, A, 0.307 in.2 per bar (Table 3). Ignore compression steel.
pn =
An
=
(0.307 in.2)(9.7)
bd (15.63 in.)(14.5 in.)
-0.013
k(pn)2 + 2pn
j
pn
=(0.013)2 + 2(0.013) (0.013) - 0.149
1-k/30.95
Step 4a: Calculate the masonry compressive stress and the steel tensile stress.
f=
2M
=
jkbd2
= 458 psi
P
f-
2(9000 ft-lb)(12 in./ft)
(0.95)(0.149)(15.63 in.)(14.5 in.)2
2557 lb
bkd (15.63 in.)(0.149)(14.5 in.)
= 75 psi
f=
M
-
(9000 ft-lb) (12 in./ft)
Ajd (0.307 in.2)(0.95)(14.5 in.)
= 25.5 ksi
Step 5a: Check compressive stress in masonry and the tensile stress in steel.
F4/3(1/3fm) - 1555 psi
fmfaf75 psi + 458 psi
533 psi 1555 psi OK
F₁ = 4/3(24 ksi) 32 ksi 25.5 ksi OK
Step 6a: Calculate shear stress.
f=
V
=
1800 lb
bjd (15.63 in.)(0.95)(14.5 in.)
8 psi
Step 7a: Check shear stress.
Fv4/3fm79 psi, 4/3(50 psi) 67 psi maximum
Thus, F, 67 psi > 8 psi OK
With pier designed, check the induced vertical cantilever deflection in the panel. Tensile stresses must be within those given in Table 2 for unreinforced masonry. Determine if the pier is cracked. Assume that the modulus of rupture, fer for solid brick masonry is equal to
2√f'm 118 psi (Ref. UBC 91).
=(1/12)(b)(tpier)3= (1/12)(15.63 in.)(19.63 in.)3
= 9850 in.4
Mer=
ferlg
c
=
(118 psi)(9850 in.4)
(0.5)(19.63 in.)(12 in./ft)
= 9870 ft-lb
M9000 ft-lb < 9870 ft-lb. Thus, the pilaster is not cracked under service loads. Calculate the deflection of the pier based on uncracked section properties:
Δ=
2Vpxh4
8Emlg
2(90 lb/ft)(10 ft)4(1728 in.3/ft³)
8(3 x 106 psi)(9850 in.4)
- 0.013 in.
Calculate the induced moment on the panel assuming the panel must maintain the curvature of the pier:
Wpanel
8Eml panel
h
8(3×106 psi)[(1/12)(9 ft)(3.63 in.)3](0.013 in.)
= 7.9 lb/ft
(10 ft)4(144 in.2/ft²)
Mbase (0.5)(Wpanel)(h)2 (0.5)(7.9 lb/ft)(10 ft)2
= 395 ft-lb
Calculate the flexural tensile stress in the panel at the bottom of the wall.
f₁=
MbaseC
Ipanel
(395 ft-lb)(3.63 in./2)
(1/12)(9 ft)(3.63 in.)3
20 psi
Check the allowable tensile stress value given in Table 2:
F, 4/3(40 psi) = 53 psi 20 psi OK
Finally, the combination of Mpy and the moment due to vertical cantilever deflection at the middle of the panel must be checked.
Mpy (1/10)(h/L)(Mpx)= (1/10) (10 ft/9 ft)(203 ft-lb)
- 23 ft-lb
MMpy+ (1/8)(Wpanel)(h)2
23 ft-lb+ (1/8)(7.9 lb/ft)(10 ft)2 = 122 ft-lb
Mbase 395 ft-lb 122 ft-lb, therefore, OK. If M is greater than Mbase, use M to calculate f, and check the allowable tensile stress, F₁.
Step 9: Sliding and overturning resistance of the caisson is a function of the lateral soil pressure and is beyond the scope of this Technical Notes.