CE 3253, Fall 2007

2007)

— David Wagner 2007/11/17 21:07

1. Explain the meaning of nominal dimensions, specified dimensions and actual dimensions.

Nominal dimensions include the width of the mortar (half the mortar thickness added to all four sides).
Specified dimensions are the required dimensions of the masonry unit itself.
Actual dimensions are the sizes of masonry units delivered and vary from the specified dimensions within the acceptable tolerance.

2. What is a barrier wall? Give an example.

A barrier wall attempts to stop all water at the face of the wall. The walls of a swimming pool are barrier walls, as are the tiled walls of many mosques.

Photo from The chambers of the Sacred Relics , http://www.ee.bilkent.edu.tr/~history/religious.html.

3. What is a curtain wall? Give an example

A curtain wall is a non-loadbearing wall, and the only vertical load it is expected to bear results from its own weight and it need not be supported at every story. The Circular Rose window in Chartes Cathedral is set in a masonry curtain wall, as is the less traditional wall shown below using cordwood and wine bottles as masonry units.

Chartres photo from the blog Things that Inspire; cordwood masonry photo from http://www.kathymoser.com/naturalnew.htm by Kathy Moser.

4. What methods are commonly used to design masonry components?

The masonry code is currently in transition between allowable stress design (ASD) and load reduction factor design (LRFD). In addition, some empirical design methods may be used. (However, I would hazard to guess the most common design technique used is to simply copy an existing design, making minor changes as necessary then checking the new version for obvious blunders. If your firm already has plans for a 20'x30' masonry structure, why start from scratch on a 20'x33' structure for another job?)

5. What is the difference between mortar and grout?

Mortar is used to stick masonry units together. Grout is used to fill the voids in masonry structures.

6. Name each of the masonry units shown assuming that they are viewed perpendicular to the plane of the wall

L = length, H = height, T = thickness, L > T > H

Soldier	Stretcher	Sailor	Shiner

7. What are the most common types of masonry mortar usually specified?

The most common types of masonry mortar are Type N for general masonry walls above grade, Type S for structural masonry applications, and Type M for below grade applications.

8. What is the difference between a Masonry Cement Mortar and a Mortar-Cement Mortar?

Masonry cement mortar is standard mortar made (mostly) from sand, lime, and portland cement. Mortar-cement Mortar is (usually) a proprietary ready-mix made of heaven-knows-what and generally mixed with just sand and water onsite.

9. Design a cavity wall for the top story of a 4-story building in San Antonio, Texas. The outer wythe has a vertical span of 11 feet and a horizontal span (simply supported) of 24 feet. The inner wythe has a vertical span of 10 feet and a horizontal span (simply supported) of 22.5 feet. Assume a uniform story height of 11 feet. The wind pressure is 22 psf. Use an outer wythe of standard modular brick, and an inner wythe of hollow or grouted CMU. Use Type S cement-lime mortar.

[Design caveat: it is usually smarter to make upper stories of lighter materials than lower ones. The Romans understood this two millenia ago.]

As it is the top story, bearing is unlikely to control. Also, not enough information is given to design for shear or bending loads parallel to this wall, so bending due to wind loads perpendicular to the wall face is assumed to control. Also, assume for this problem that additional bond beams and pilasters are not allowed to be added to help resist this bending. Since roof information is not given to calculate uplift, the vertical axial load in the wall will be assumed to be zero. The masonry elements will be conservatively designed as one-way diaphragms, with the wind load transferring in the short (vertical) dimension. The allowable-stress load combination is D+W.

Load transfer in the vertical direction causes bending stresses perpendicular to the bed joints. ~~The wall is designed for composite action with the collar joints and such specified in 2.1.5.2, so the~~ Most of the wind load is transferred to the more rigid wythe of CMU. (If all the load is conservatively estimated to transfer to the inner wythe, the outer layer of bricks would properly be considered as a veneer, not an element in a double-wythe cavity wall as implied by the problem statement.)

Wind load on a 1” strip:

$w=22 {{lb}/{ft^2}} * {1 {ft^2}/{144 i n^2}} * 1 i n =$ 0.153 lb/in.

Inner Wythe: Hollow 8x8x16 CMUs with 6” square voids, Type S cement-lime mortar, Fb = 25 psi ungrouted, 65 psi fully-grouted, f'm: 1000 to 3000 psi, 120 lb/cf.

$M={wl^2}/8 = {0.153*120^2}/8=$ 275.4 in-lb
Bedding area of one nominal unit (including mortar): 8×16 -2x6x6=</m>56 in²
$A_{net}=56/16=$ 3.5 in²/in
$I = {16*8^3-2*6*6^3}/{12*16}={8192-2592}/{12*16} = 467/16$ 29.17 in^4/in
=2.89 in
<99

If the inner wythe is conservatively designed to carry all of the wind load, the outer layer of bricks would be a veneer, not an element in a double-wythe wall. Also, because the outer wythe is longer than this inner one, the outer wythe as a whole is somewhat less stiff than is indicated here, but this difference is neglected.

Assuming roughly equal f'm:

M= $275.4 {{E_1 I_1}/{E_1 I_1 + E_2 I_2}}=$ 275.4*{{900*29.17}/{900*29.17+700*5.33}}= 275.4*{{26253}/{26253+3731}} = 275.4*{26253/29984}= 275.4*0.88= 241 in-lb

The bending stress is $f_b={Mr}/{I}={241*2.89}/{29.17}=$ 23.9 psi.

The axial load at mid-height is about 10*120=1200 lb/sf = fa=8.33 psi.

Try minimum fb=25 and f'm=1000 psi

$F_a={1/4}f prime_m(1-(h/{140r})^2)={1/4}*1000*(1-(120/{140*2.89})^2)=250*(1-0.088)=$ 228 psi.
${{f_a}/{F_a}} + {{f_b}/{F_b}} = {{8.33}/{228}} + {{23.9}/{25}} =0.036+0.955=$ 0.991 ≤ 1✔

∴Any standard CMU compliant with standards is sufficient to resist the wind bending load.

Outer Wythe: Solid clay units, 4×2 2⁄3×8, Type S cement-lime mortar, Fb = 40 psi, f'm: 1000 to 4000 psi, 120 lb/cf.

$M={wl^2}/8 = {0.153*132^2}/8=$ 333.2 in-lb
$A_{net}=A_{gross}={4*8}/8=$ 4 in²/in
$I={8*4^3}/{12*8} = {42.67}/{8}=$ 5.33 in^4/in
=1.15 in
>99
- $F_a={1/4}f prime_m({70r}/{h})^2$
- $F_b={1/3}f prime_m$

M= $275.4 {{E_2 I_2}/{E_1 I_1 + E_2 I_2}}=333.2 *(1-0.88)=275.4*0.08=$ 33.0 in-lb

The axial load at mid-height is about 11*120=1320 lb/sf = 9.17 psi.

Since h/r > 99, forget about the outer layer of bricks carrying any load and design it as a veneer, with the inner CMU carrying all the load.

The bending stress is $f_b={Mr}/{I}={275.4*2.89}/{29.17}=$ 27.3 psi.

${{f_a}/{F_a}} + {{f_b}/{F_b}} = {{8.33}/{228}} + {{27.3}/{F_b}} =0.036+{27.3/F_b} le$ 1
→→grout one of every 12 cores, one every 96”
$F_B=25+40grout=25+{40/12}=28.33$
${{f_a}/{F_a}} + {{f_b}/{F_b}} = {{8.33}/{228}} + {{27.3}/{28.33}} =0.036+0.963=$ 0.9995 ≤ 1✔