= =

MORGAN STATE UNIVERSITY

SCHOOL OF ARCHITECTURE AND PLANNING

LECTURE IX

Dr. Jason E. Charalambides

Unreinforced Masonry Wall Design - Axial

Compression and Flexure (ASD Method)

Proportions and Loading

Many masonry walls are designed for both axial compression and flexure. Examples include: load-bearing exterior walls (subject to wind and/or seismic loads), tall non load-bearing exterior walls (where the self-weight of the wall decreases wall capacity), and walls with an eccentric vertical load.

The figure shows the bending moment and transverse loading distributions assumed in the design of such walls.

The Dom-ino frame design by Le Corbusier during the early 20th century became a paradigm ,a datum in the history of architecture and a reference to the modernist movement.

PROPORTIONS AND LOADING

Axial capacity decreases with increasing wall height. For very tall slender walls, elastic buckling is the limiting factor. For most masonry walls, however, the main effect of slenderness is the development of additional bending due to deflection (known as the P-Δ effect). The MSJC Code uses h/r to evaluate the performance of hollow and partially grouted masonry.

Walls subject only to compressive loads may be in pure compression or may be subject to flexure.

PROPORTIONS AND LOADING

For unreinforced walls where the compressive load is within a distance S/An (i.e., t/6 for solid sections) from the centroid of the wall, there will be no net tensile stress due to load eccentricity.

The critical section for walls subject to eccentric load and out-of-plane loads (such as wind or seismic) may be either at the top of the wall or at mid-height, depending on the magnitudes of the loads. Both locations should be checked.

BASIC EQUATIONS

Unreinforced walls subjected to axial compression, flexure, or both must satisfy Eq. 1 through Eq. 6 [MSJC Sec. 2.2.3].

In Eq. 3, e should be the actual or estimated eccentricity. It should not include virtual eccentricity due to lateral load.

1fa=Applied Axial Compressive StressFa=Allowable Compressive Stressfb=Applied Flexural StressFb=Allowable flexural Stress 2P=Applied Axial LoadPe=Euler's Buckling Load 3Em=Modulus of Elasticity of Masonry in CompressionIn=Moment of Inertia in the Net Cross Section of the member

BASIC EQUATIONS CONT

Unreinforced walls subjected to axial compression, flexure, or both must satisfy Eq. 1 through Eq. 6 [MSJC Sec. 2.2.3].

4For h/r < 99

5For h/r > 99

6

rkobfkclo`ÉÇ=j^plkov=t^ii=abpfdk=Determine the maximumallowable axial load at a giveneccentricity ofon an 8" hollowunreinforced concrete masonrywall.

r 2.84in:= Em 1800ksi:=An 30

in2

ft:= f'm 2ksi:=

S 81in3

ft:= h 12ft:=

In 334in4

ft:=

ecc 1.5in:=

Determining h/r relation:hr

50.7=

Determining the AllowableStress due to Axial Load:

Fa14f'm 1

h140 r⋅

2−

⋅ 434.4 psi=:=

Determining the Allowable Stress due toFlexural Load:

Fb13f'm 666.7 psi=:=

Solving for Applied Axial Compressive Stress and Applied Flexural Stress:

faPAn

= fbMS

= P ecc⋅

S=

faFa

fbFb

+ 1≤

Therefore:

P

An

Fa

P ecc⋅

S

Fb+ 1≤ Solving for P: P

An Fa⋅ Fb⋅ S⋅

Fb S⋅ An Fa⋅ ecc⋅+9.569

kipft

=:=

Checking for Buckling using Euler's equation:

Peπ2 Em⋅ In⋅

h2

1 0.577

eccr

⋅−

3⋅ 96.163

kipft

=:= P14Pe⋅≤

PeP

10= Approved

Checking Tension:

F.t is given as 19lbf/in^2 for N type mortarPAn

P ecc⋅

S− Ft< Ft 19psi:=

StressPAn

P ecc⋅

S− 141.8 psi=:= Stress is positive thus the section is in Compression.

= =

MORGAN STATE UNIVERSITY

SCHOOL OF ARCHITECTURE AND PLANNING

LECTURE IX

Dr. Jason E. Charalambides

Reinforced Masonry Wall Design - Axial Compression and Flexure (ASD Method)

Proportions and Loading

Reinforced masonry walls, like their unreinforced siblings, are designed for both axial compression and flexure. Examples include: load bearing exterior walls (subject to wind and/or seismic loads), tall non load-bearing exterior walls (where the self-weight of the wall decreases wall capacity), and walls with an eccentric vertical load.

The figure shows the bending moment and transverse loading distributions assumed in the design of such walls, just like the case of unreinforced masonry

The Dom-ino frame design by Le Corbusier during the early 20th century became a paradigm ,a datum in the history of architecture and a reference to the modernist movement.

PROPORTIONS AND LOADING

There are many methods applicable for the design of reinforced masonry walls subjected to compression and/or flexure. The most common are iterative methods for which computer programs are necessary to perform loop processes.

Another type of methods is that of graphic use of interaction diagrams which we shall use in this session.

BASIC EQUATIONS

Reinforced walls subjected to axial compression, flexure, or both must satisfy Eq. 1 through Eq. 3 [MSJC Sec. 2.2.3.2.2].

Eq. 2, is actually a combination of two equations and an embedded condition of the h/r>99 or h/r≤99. The following determines the maximum compressive force due to axial loading

1f'm=Specified Compressive Stress of MasonryFb=Allowable flexural Stress 2P=Applied Axial LoadPe=Euler's Buckling LoadAn=Net Cross Sectional Area of MasonryAs=Cross Sectional Area of laterally tied Longitudinal steel reinforcement in MasonryFs=Allowable Reinforcement in Tensile Stressh=Wall heightr=Radius of gyration

BASIC EQUATIONS CONT

The following interaction diagrams are used to determine the ρt based on the relations we can establish. The interaction diagram to be used depends on the condition of stress (Tension or Compression)

4k=ratio of the distance between the compression face of a wall and the neutral axis to the effective depthn=modular ratio of E1/E2Fs=Allowable reinforcement tensile stress

Interaction Diagram for reinforced Masonry Wall (Compression Controls)

Interaction Diagram for reinforced Masonry Wall (Tension Controls)

Reading

Reading: Required: This set of slides! Recommended: This set of slides!!!

obfkclo`ÉÇ=j^plkov=t^ii=abpfdk=cçê=bÅÅÉåíêáÅ^ñá~ä=iç~Ç=Determine the amount of steel needed toadequately reinforce an 8" fully grouted concretemasonry wall subjected to a compressive load P withan eccentricity at midheight as determined in thefollowing data:.

An 91.5in2

ft:= P 1000

lbfft

:= Fs 24000lbf

in2:= S 116.3

in3

ft:=

ecc 6in:= h 15ft:= b 12in:= Em 1350ksi:=In 443.3

in4

ft:= f'm 1.5ksi:= r 2.20in:= t 7.62in:=

Determining h/r relation and n value:

hr

81.8= n29000ksi

Em21.5=:=

Determining the kb value and chosing diagram:

kbf'm

3f'm3

Fsn

+

0.309=:=

Determining the relevant parameters for the ρInteraction Diagrams and determining t:

3Pf'm b⋅ t⋅

0.0221ft

=3P ecc⋅

f'm b⋅ t2⋅0.017

1ft

=

Based on interaction diagrams

nρt 0.014:= ρtnρt f'm⋅

3 Fs⋅0.00029=:=

Determining the amount of steel to be used:

As_ini ρt b⋅ t⋅ 0.0267 in2⋅=:= Chose #3 rebars at 48 in O.C. As .028

in2

ft:=

Calculating allowable compressive force:

Pa .25 f'm⋅ An⋅ 0.65 As⋅ Fs⋅+( ) 1h

140 r⋅

2−

.5

hr

99−

2hr

99−

2−

⋅70 r⋅

h

2

.5

hr

99−

2hr

99−

2+

⋅+

⋅ 22.9kipft

⋅=:=

Condition if Pa P> "OK", "No Go", ( ) "OK"=:=