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Section B: Masonry Design.

Part A) Vertical Loading.

Note: The design of masonry has been broken down into two comparisons: A Clay brick wall design

and a concrete block wall design.

Clay Brick Wall Design:

Assumed a wall thickness of 90mm. Manufacture and Construction: Normal, with ties. Mortar

Designation 3. 7%-12% water absorption.

Loading: Roof and wall.

D.L: Wall - P= Density x 1m x Height x Thickness

=18 x 1 x 0.3 x 0.09

= 4.86KN/m

D.L: Roof - P = 1m x Height x D.L

= 1m x 3 x 3.5

=10.5KN/m

L.L: Roof - P= 1m x Height x L.L

=1m x 3 x 2

=6KN/m

L.L: Roof with safety Factor: 1.6 x 6 = 9.6KN/m

D.L: Roof with Safety Factor: 1.2 x 10.5 = 12.6KN/m

Total Roof Load: (9.6 + 12.6)/2 = 11.1KN/m Note: Divided by 2 as half roof load taken by wall.

Total D.L: Wall with safety factor: 1.2 x 4.86 = 5.83KN/m

Ultimate Load: 5.83 + 11.1 = 16.93KN/m

Therefore: Ultimate loading on wall: P =16.93KN/m

Now work out Q: Load Capacity. But before need: Slenderness Ratio (S.R) and thus Teff (Effective

thickness).

Teff =2/3(t1+t2)

=2/3(90+90)

=120mm

S.R =Heff/Teff

=3/0.12

= 25 Check: OK: 25 < 27

From Table: β = 0.34

fk: From table: Mortar desig. 3 use interpolation.

15-18/5-x = 15-20/5-5.8

-3/5-x=-5/-0.8

2.4=-25+5x

X=5.48

Therefore fk=5.48.

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Now Solve Q

Q=β.Teff.fk/γm

=(0.34)(120)(5. 48)/3

=74.53KN/m

P < Q therefore: Ok.

Concrete Block Wall Design:

Assumed a wall thickness of 190mm. Mortar desig. 3. Construction and Manufacture: normal, with

ties.

Total loading on Roof: P= 11.1KN/m

D.L: Wall: P=Density x 1m x Height x Thickness

=20 x 1m 3 x 190

= 11.4KN/m

D.L. with Safety factor: 1.2 x 11.4 = 13.68KN/m

Therefore Total loading: P = 13.68 + 11.1

=24.78KN/m

Now work out Q: Load Capacity

Slenderness Ratio (S.R.) and Teff.

Teff=2/3(t1+t2)

=2/3(190+190)

=253.33mm

S.R.=Heff/Teff

=3/0.253

=11.84

Approximately 12

Therefore ok as 12 < 27

From Table: β = 0.73, from table: fk = 5.8

Now Solve for Q.

Q=β.Teff.fk/γm

=(0.73)(253.33)(5.8)/3

=357.533KN/m

Therefore Ok, as P < Q

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Part B: Lateral Loading.

Part 1: Clay Brick Wall.

Note: Roof pinned, foundation pinned. Fixed both sides.

Assumed 7%-12% water absorption. Mortar desig. 3.

Support Condition G.

Step 1: Aspect Ratio (A.R.); Orthogonal Ratio (µ)

A.R. = H/L

=3000/7000

= 0.43.

Orthogonal Ratio = fkb/fkp

From table: fkb=0.4

fkp=1.1

µ=0.36

On table 0.36 lies between 0.35 and 0.40

Use Interpolation to find α

@ 0.3

0.35 -0.014

0.36 – x

0.40 – 0.013

0.35 – 0.36/0.014 – x = 0.35 – 0.40/0.014 – 0.013

-0.01/0.014 – x=-0.05/0.001

-0.00001=-0.0007 + 0.05x

X = 0.0138 @ µ: 0.3

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@ 0.5

0.35 - 0.025

0.36 - x

0.40 – 0.023

0.35 – 0.36/0.025 – x=0.35 – 0.40/0.025-0.023

-0.01/0.025-x =-0.05/0.002

-0.00002=-0.00125 + 0.005x

X=0.0246 @ µ:0.5

Now Find α for 0.36

α - 0.3 - 0.0138

α - 0.43 – x

α - 0.5 – 0.0246

0.35 – 0.43/0.0138 – x=0.30-0.50/0.0138-0.0246

-0.13/0.0138-x=-0.2/-0.0108

0.0001404=-0.00276+0.2x

X=0.021

Therefore α @0.36µ = 0.021

Step 2: Applied Bending Moment (m)

Given: Wk =0.8KN/m

yf =1.2

γm = 3

m = α.Wk.yf. L²

=(0.021)(0.8)(1.2)(7)²

=0.988KN.m

Step 3: Moment Resistance (M)

M = fkb.b.t²/ γm.6

=(1.1)(1)(120)² (1x10¯³)/(3)(6)

=0.88KN.m

For Clay brick wall to be acceptable: m<M.

This is not the case. Therefore clay brick wall of 90mm thickness will not be acceptable.

Therefore not OK.

Part 2: Concrete block: Complete concrete block wall through the same steps as clay brick wall.

Step 1: Aspect Ratio (A.R.); Orthogonal Ratio (µ)

A.R. =H/L

=3000/7000

=0.43

Orthogonal Ratio: µ=fkb/fkp

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Through Interpolation

fkb: From tables

100-0.25

250-0.15

190-x

100-190/0.25-x=100-250/0.25-0.15

-90/0.25-x=-150/0.1

-9=-37.5+150x

X=0.19

Therefore fkb=0.19

fkp: From Tables

@ 100

3.5-0.45

5.8-x

7.0-0.60

3.5-5.8/0.45-x=3.5-7.0/0.45-0.60

-2.3/0.45-x=-3.5/-0.15

0.345=-1.575+3.5x

X=0.55

@250

3.5-0.25

5.8-x

7.0-0.35

3.5-5.8/0.25-x=3.5-7/0.25-0.35

-2.3/0.25-x=-3.5/-0.1

0.23=-0.875+3.5x

X=0.32

@5.8

100-0.55

190-x

250-0.32

100-190/0.55-x=100-250/0.55-0.32

-90/0.55-x=-150/0.23

-20.7=-82.5+150x

X=0.412

Therefore fkp=0.412

Orthogonal Ratio (µ)=fkb/fkp

=0.19/0.412

=0.46

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Further interpolation:

From table: 0.46 lies between 0.40 and 0.50

@0.4

0.40-0.013

0.46-x

0.50-0.011

0.40-0.46/0.013-x=0.40-0.50/0.013-0.011

-0.06/0.013-x=-0.1/0.002

-0.00012=-0.0013+0.1x

X=0.012

@0.5

0.50-0.021

0.40-0.023

0.46-x

0.40-0.46/0.023-x=0.40-0.50/0.023-0.021

-0.06/0.023-x=-0.1/0.002

-0.00012=-0.0023+0.1x

X=0.0218

@0.46

α-0.30-0.012

α-0.43-x

α-0.50-0.0218

0.30-0.43/0.012-x=0.30-0.50/0.012-0.0218

-0.13/0.012-x=-0.2/-0.0098

0.001274=-0.0024+0.2x

X=0.019

Therefore α = 0.019

Step 2: Applied Moment (m)

m = α.Wk.yf. L²

=(0.019)(0.8)(1.2)(7) ²

=0.89KN.m

Step 3: Moment Resistance (M)

M = fkb.b.t²/ γm.6

=(0.4120(1)(253.33) ² (1x10exp-3)/(3)(6)

=1.462KN.m

Therefore m is < then M: OK. This tells us that the concrete block wall is a workable design.

Thus we can now assume to use the concrete block wall.

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We now use the concrete block wall’s units in further design.

Step 4: Slenderness Limits.

For a panel that’s supported on four edges:

Height x Length < and/or = 2050(teff) ²

3000 x 7000 = 21 x 10⁶

2050(253.33)²= 129.96 x 10⁶

Therefore ok as 21 x 10⁶ < 129.96 x 10⁶

Step 5: Shear Design

Total load to support = γf.Wk.Area loaded

=(1.2)(0.8)(0.5 x 7 x 3.5)

= 11.76KN

11.76KN becomes an assumed UDL

Design Shear Force (S.F.) per meter

11.76/7 = 1.68KN/m

Design for shear strength (fv)

Vh = vf.Wk.Aw/Ay

=1.68 x 10³/253.33 x 1000

=0.0066N/mm²

Characteristic Shear strength (fv)

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0.35 + 0.6gd

Also ignore self weight.

Vh=Fv/ γmv

=0.35/2.5

=0.14N/mm²

Therefore ok along the base.

Part C: Comments: After reviewing the above work, one can see that during the design for a clay brick (90mm thickness)

wall the applied moment was greater than the moment resistance. This would cause a failure. (A

greater thickness of a clay brick could possibly work -230mm thick). We then chose to look at a

concrete block of 190mm thickness. Here we see that the applied moment was less than the

moment resistance -this case wouldn’t cause a failure. We thus carried on further design with the

concrete block of 190mm thickness.

Part D: Retaining Wall Design:

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Assumed: From rules of thumb: min footing 300.

Cover: 40

Fcu:30Mpa

High yield Steel: 450Mpa.

h=H/12

=3000/12

=250mm

B=2/3H

=2/3(3000)

=2000

Force: KN/m

Horizontal Load (Rh) Distance(m) Moment(M)(KN.m)

Surcharge:

(0.3)(6)(3.5)=5.94KN/m 1.65m 9.801KN.m

Earth:

(0.5)(0.3)(18)(3.3)²= 29.403KN/m 1.1m 32.343KN.m

Total Rh: 35.343KN/m

Vertical Load (Rv)

Surcharge:

(6)(1.083)=6.498KN/m 0.542m 3.522KN.m

Soil:

(3)(1.083))(18)=58.482KN/m 0.542m 31.697KN.m

Stem:

(3)(0.250)(24)=18KN/m 1.208m 21.744KN.m

Base:

(2)(0.3)(24)=14.4KN/m 1.000m 14.4KN.m

Total Rv: 97.38KN/m Total M: 113.507KN.m

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Checks:

Safety against sliding:

Rh < Rv tan30°

35.343 < (97.38)tan30°

35.343 < 56.22

Therefore OK.

Check against Overturning.

e =∑M/∑Rv

=113.507/97.38

=1.16m Therefore OK: stable: lies in middle segment (2)

Factor of Safety:

=Rv tan30°/Rh

=56.22/35.343

=1.59

Therefore ok: greater than 1.5.

Reinforcement:

Cover 40

Fcu: 30Mpa

High yield steel: 450Mpa

M=113.507KN.m

A.S.=M/0.87.fy.Z

Need K

K=M/fcu.b.d²

=113.507x10⁶/(30)(3000)(250²)

=0.020

Therefore ok: K<0.156: No compression reinforcement needed.

Z=d(0.5+√(0.25-(k/0.9))

=d(0.5+0.477)

=d(0.977)

0.977d>0.95d

Use 0.95d

A.S=M/0.87.fy.z

=113.507x10⁶/(0.87)(450)(0.95)(250)

=1220.75mm²

From Table 14:

Use : 1256mm²

Therefore Rebars: 1-Y 40

Spacing at every 1m centers.

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