structural calculations

8/11/2019 Structural Calculations

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PROPOSED RESIDENTIAL

HOUSE IN KYANJA

PLOT 3318 - BLOCK 195, KYANJA

STRUCTURAL CALCULATIONS

CLIENT STRUCTURAL DESIGNER

MR. TONNY SAKAAZA

P.O.BOX 10109, KAMPALA

UGANDA

SOLOMON N. BALEMEZI

BSc. CIV. ENG. MUK; M.U.I.P.E

P.O. BOX 8493 KAMPALA

EMAIL: [email protected]

PHONE: 0779-918-103


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STRUCTURAL SUMMARY SHEET

Client : MR TONNY SAKAAZA

Project: PROPOSED RESIDENTIAL HOUSE IN KYANJA

Regulations and Design 1. Kampala Capital City Authority RegulationsCodes: 2. BS 6399, Part 1 1985 Design Loadings

3. CP3 Chapter V, Part 2 1978 Wind Loadings4. BS 8110, Parts 1 & 3 1991 Structural Use of Concrete

Loading Conditions: 1. Roof Imposed load = 0.6KN/m 2

Dead load (Tiles, timber, ceiling, e.t.c) = 1.0KN/m 2

2. Floor Imposed load (office or bedroom) = 1.5KN/m 2

3 Wind Basic SpeedS 1=1.0 S 2 = 0.9 S 3 =1.0

Foundations: 1. Reinforced concrete Strip Footings

Walls: 200mm Thick Load Bearing walls

Roof: IT4 sheets on Timber purlins, Timber trusses

Sub-Soil Conditions: Evaluated soil bearing pressures = 180k N /m 2

Materials: CONCRETE1. Blinding Concrete Grade 15 (1:3:6) Max 20mm aggregate.

2. Structural Concrete

Grade 20 (1:2:4)

Max 20mm aggregate2. Structural Concrete Grade 25 (1:1.5:3) Max 20mm aggregate

REINFORCEMENT1. T & Y high yield (f y) Grade 460 N/mm 2 2. R Low yield Grade 250N/mm 2

MASONRY WALLSolid concrete blocks of compressive strength 5N/mm 2

TIMBER

Pine - strength class C14

Fire Resistance: 1 hour fire resistance for all elements

Purpose of Building: Residential


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REFERENCE PARTICULARS OUTPUT

Eurocode 5(ENV 1995-1-1)


PROPOSED SAKAZA'S RESIDENCE IN KUNGU

DESIGN OF THE TIMBER PURLINS

Design Assumptions and Details;

Angle of inclination of the roof to horizontal is 4

75x50mm Pine Timber Purlins Iron Sheets as Roof Covering (weight = 4.8kg/m 2) No wind load considered because roof is surrounded by parapet wall

Note: The Design output is to determine the span of purlins that shall preventultimate failure of purlins and to limit excessive deflection.

Timber properties Pine; Strength Class C14/ SG4

Grade Stresses

Loading;

Permanent Load Due to;

1. Corrugated Iron Sheets, F sheets :

2. Purlins, F purlins :

Total Permanent Load

Variable Load;

For Roof without access, Live load Hence, Design Load, F d;


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Resolving the LoadNormal to the roof;

Tangential to the roof;

Determine Maximum Span of Purlins using Bending Criterion

Moments Due to applied Loads ;

Ignore bending about y-y axis because it is negligible

Moment Capacity of a 75x50mm Pine Timber Section ;

f b.allow - Allowable Bending Stress

Ixx - Moment of Inertia about the x-x axis

y - Distance to neutral axis

Allowable Bending Stress f b.allow :

Where:

f m,k - Characteristic Bending Stress kmod - Modification factor for Load Duration and Service Class

kh - Modification factor for Member Size Effect

k sys - Modification factor for System Strength = 1.1

m - Material Safety Factor = 1.3

Moment of Inertia about x-x axis I xx:

Distance to Neutral Axis y: ;

Hence;


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REFERENCE PARTICULARS OUTPUTEquating moment due to applied load to moment capacity of the timber section

L = 1389mm

Determine Maximum Span of Purlins using Limiting Deflection CriterionDeflection Criterion:

L = 1600mm

Bending Criterion governs the Maximum Span for Purlins.

Therefore place Trusses at 1300mm c/c

L = 1389mm Due toBending Criterion

L = 1600mm Due toDeflection Criterion

Bending governs


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DESIGN OF THE MAIN TRUSS

Loading;

Permanent Load Due to;

1. Purlins, F purlins :

2. Corrugated Iron Sheets, F sheets :

3. Ceiling, F ceiling (Assume 0.15kN/m2

):

4. Services, F services (Assume 0.1kN/m 2):

Total Permanent Load

Variable Load;

For Roof without access, Live load Hence, Design Load, F d;


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Eurocode 5

(ENV 1995-1-1)


Design of the Tie Beam

Timber properties Pine; Strength Class C14/ SG4Grade Stresses

Results of Truss Analysis for the Tie Beam;From the Analysis of the main truss for the Ultimate Limit State using PROKONStructural Analysis software, the following maximum forces have been obtainedfrom PROKON Structural Analysis software.

Maximum Compressive Force = 4.87kN

Design of Tie Beam to Resist Maximum Compressive Force

Member Geometric Properties

Try a 50x75mm Timber Section

Since the slenderness ratio is greater than 0.3, the member will fail by bucklingand hence clause 6.3.2(3) applies.

Buckling Resistance Condition:


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Clause 6.3.2 (3)



Factor k; EC5, equation (6.28))

Instability factor, k c; (EC5, equation (6.26))

Design Buckling Strength f c.design :

Where:

f c,design - Design Buckling Strength

f c,k - Characteristic Compression strength parallel to grain

kmod - Modification factor for Load Duration and Service Class

kc - Instability Factor


Applied Compression Stress, f c.applied

Section is Adequate.Hence provide 50x75mm Section for the Tie Beam


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Results of Truss Analysis for the Rafter;

From the Analysis of the main truss for the Ultimate Limit State using PROKONStructural Analysis software, the following maximum forces have been obtainedfrom PROKON Structural Analysis software.

Maximum Force = 4.93kN Compressive

Design of Rafter to Resist Maximum Compressive Force



Since the slenderness ratio is greater than 0.3, the member will fail by buckling

and hence clause 6.3.2(3) applies.Buckling Resistance Condition:



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Where:


f c,k - Characteristic Compression strength parallel to grain kmod - Modification factor for Load Duration and Service Class

kc - Instability Factor



Section is Adequate.

Hence provide 50x75mm Section for the Rafter


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Results of Truss Analysis for the Struts & Ties;

From the Analysis of the main truss for the Ultimate Limit State using PROKONStructural Analysis software, the following maximum forces have been obtainedfrom PROKON Structural Analysis software.

Maximum Force = 6.69kN Compressive

Design of Rafter to Resist Maximum Compressive Force



Since the slenderness ratio is greater than 0.3, the member will fail by bucklingand hence clause 6.3.2(3) applies.

Buckling Resistance Condition:



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Where:


f c,k - Characteristic Compression strength parallel to grain

kmod - Modification factor for Load Duration and Service Class kc - Instability Factor



Section is Adequate.

Hence provide 50x75mm Section for the Ties & Struts


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DESIGN OF PLYWOOD WEBBED BEAM 01REFERENCE CALCULATIONS OUPUT

Illustration

Section PropertiesFirst Moment of Area, QQ = Q flange + Q web

= A + a = ((100mm*50mm)*(200-25))+(200*12*2)= 875,000 + 4,800

Q = 879,900mm 3

Second Moment of Area, QI = (1/12)*(BH 3-bh 3))

= (1/12)*((100*400 3) - (100*300 3))= (1/12)*((6,400,000,000)-( 2,700,000,000))

I = 308,333,333mm 4

Zziwa, A.

Ziraba, Y.K.& Mwakali,J.A (2010)15 th NTCMay 20-21,2010

BS5268-2:2002 Table8

Timber properties Local/Trade name; Flanges from Pine

Strength Class C14/ SG4

Grade Stresses

Bending parallel to grain; m = 14 N/mm 2

Tension parallel to grain; t = 8 N/mm 2

Compression parallel to grain; c = 16 N/mm 2

Compression perpendicular to grain (no wane); cp1 = 2 N/mm 2

Shear parallel to grain; = 1.7 N/mm 2

Mean modulus of elasticity; Emean = 7000 N/mm 2

Characteristic density; char = 290 kg/m 3

Average density; = 350 kg/m 3

3200mm


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Loading Characteristic Permanent Loads

Ceramic Tiles - 0.5kN/m 2

Screed - 1.2kN/m 2

Plywood Decking - 0.2kN/m 2

Plaster - 0.5kN/m 2 Self Weight - 0.7kN/m 2

Miscellaneous - 0.5kN/m 2

Ceiling - 0.5kN/m 2

Services - 1.0kN/m 2

Partitions Walls - 2.5kN/m 2 Total 7.6kN/m 2

Characteristic Variable Loads

Category A; Areas for Domestic and Residential Activities

Imposed Loads = 2.0kN/m 2

Design LoadingFd = 1.35F k.perm + 1.5F k.varFd = (1.35*7.6) + (1.5*2.0)kN/m 2

F d = 13.26kN/m2

Fk.perm = 7.6kN/m 2

F k.var = 2.0kN/m 2

F d = 13.26kN/m 2


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Determine Joist Spacing using Bending Criterion

Mallow Mapplied

Mallow = (1/y)*( f b.allow * I): y = 200mm; I = 308,333,333mm 4

f b.allow = (1/ m) * kmod * kh* ksys * f m,k

Where:

f b.allow - Allowable Bending Stress f m,k - Characteristic Bending Stress kmod - Modification factor for Load Duration and Service Class kh - Modification factor for Member Size Effect ksys - Modification factor for System Strength = 1 .1


f b.allow = (0.769)*0.6 *1.1*14N/mm2

= 7.106N/mm 2

Mallow = 0.005mm-1 x 7.106N/mm 2 x 308,333,333mm 4

M allow = 10,955,083Nmm (or 10.955kNm)

Mapplied = (1/8)* *L2; L=3.2m Assume M applied = M allow;

allow = Mallow x 8/L 2 allow = 10.955kNm x 8/(3.2m) 2

allow = 8.559kN/m

Let z = joist spacing centre to centre

Fd (kN/m2)* z = allow (kN/m)

13.26kN/m 2* z = 8.559kN/mz = 8.559/13.26

z = 0.645m

K mod = 0.6 for serviceclass 1, long termduration

m = 1.3 for Timberand Timber products

M al low = 9.959kNm

allow = 7.7805kN/m

To limit failure due tobending, placePlywebbed BoxJoists at a distanceless than 645mm c/capart.


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Determine Joist Spacing using Shear Criterion

Vallow Vapplied

Vallow = (1/Q)*(2 * t * I * f v.allow ):

Q = 879,800mm 3; I = 308,333,333mm 4; t = 12mm

f v.allow = (1/ m) * kmod * ksys * f v,k

Where:

f b.allow - Allowable Bending Stress f m,k - Characteristic Bending Stress kmod - Modification factor for Load Duration and Service Class ksys - Modification factor for System Strength = 1 .1


f v.allow = 0.769* 0.6 *1.1 * 5N/mm 2 = 2.438N/mm 2

Vallow = (1.13710-6)*(2 * 12 * 308,333,333 * 2.538)

V allow = 21,347N (or 21.347kN)

Vapplied = L/2; L=3.2m

Assume V applied = V allow;

Hence,

allow = V allow * 2/L

allow = 21.347kN * 2/3.2 allow = 13.342kN/m



13.26kN/m 2* z = 13.342kN/mz = 13.342/13.26

z = 1.000m

V al low = 21.347kN

allow = 13.342kN/m

To limit failure due toPanel shear, placePlywebbed BoxJoists at a distanceless than 1000mmc/c apart.


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Determine Joist Spacing using Deflection Criterion

allow loads

allow = L/200

allow = 3200mm/200

allow = 16mm

loads = [((5L4(1+k def,f )/384EI) + (L 2/8)((1+k def,w )/G w Aw))]

Where:

kdef,f - Deformation Factor for flange material kdef,w - Deformation Factor for web material E - Modulus of Elasticity for Transformed Section

I - Second Moment of Area for transformed section

Aw - Total Cross Section Area of the web Gw - Mean Shear Modulus of the web

loads = [((5*32004(1+0.6)/384*7000*453,942,476) +

(3200 2/8)((1+0.8)/578*12*400*2))]

loads = (0.687 + 0.415) = 16

= 14.519kN/m



13.26kN/m 2* z = 14.519kN/m

z = 14.519/13.26

z = 1.095mTo satisfyserviceabilityrequirements due todeflection, placePlywebbed BoxJoists at a distanceless than 1095mmc/c apart.

Bending cr i te r ion go verns . Therefore , p lace Plywebbed Box Beams @ 600mm c/c


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BEAM 02


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Project RESIDENTIAL HOUSE IN KYANJJA REINFORCED CONCRETE COUNCIL

Client TONY SAKAZA Made by Date Page Location FIRST FLOOR, from grid 1 to 3 SNB 8-Sep-14 3

CONTINUOUS BEAM (Analysis & Design) to BS 8110:1997 Checked Revision Job No

Originated from RCC41.xls on CD 1999 BCA for RCC SNB - 1

SPAN 1 LEFT CENTRE RIGHT ACTIONS M kNm 82.3 113.0 82.3

b 0.70 1.60 0.70DESIGN d mm 327.0 343.0 327.0

As mm 673 945 673 As' mm 351 164 351

TOP STEEL Layer 1 3 T 16 2 T 12 3 T 16Layer 2 1 T 16 1 T 16

As prov mm 804 As' prov 226 As prov 804BTM STEEL Layer 1 2 T 16 3 T 16 2 T 16

Layer 2 2 T 16 As' prov mm 402 As prov 1005 As' prov 402

DEFLECTION L/d 19.067 Allowed 28.109 SHEAR V kN 108.6 Link 108.6

v N/mm 1.661 8 1.661

vc N/mm 0.565 Nominal 0.565LINKS R8 @ 105 for 1575 R8 @ 125 R8 @ 105 for 1575

legs No 2 2 2

CHECKS % As ok ok okCover ok ok okmin S ok ok ok

max S ok ok okLinks

Main barsmax V ok ok

Deflection ok

SPAN 2 LEFT CENTRE RIGHT ACTIONS M kNm

b 1.00 ~ 1.00DESIGN d mm -81.0 -45.5 -57.0

As mm #DIV/0! #DIV/0! #DIV/0! As' mm #DIV/0! #DIV/0! #DIV/0!

TOP STEEL Layer 1 #DIV/0! T 32 #DIV/0! T 25 #DIV/0! T 16#DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0!

As prov mm #DIV/0! As' prov #DIV/0! As prov #DIV/0!BTM STEEL Layer 1 #DIV/0! T 25 #DIV/0! T 25 #DIV/0! T 16

#DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0! As' prov mm #DIV/0! As prov #DIV/0! As' prov #DIV/0!

DEFLECTION L/d Allowed #DIV/0!SHEAR V kN #VALUE! Link #VALUE!

v N/mm #VALUE! 8 #VALUE!vc N/mm #DIV/0! Nominal #DIV/0!

LINKS #VALUE! #DIV/0! #VALUE!legs No 2 2 2

CHECKS % As #DIV/0! #DIV/0! #DIV/0!Cover ok ok okmin S #DIV/0! #DIV/0! #DIV/0!

max S #DIV/0! #DIV/0! #DIV/0!Links

Main barsmax V #VALUE! #VALUE!

Deflection #DIV/0!

okok

#DIV/0!#DIV/0!

#DIV/0!#DIV/0!

okok

okok

#DIV/0!#DIV/0!


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BEAM 05


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Project RESIDENTIAL HOUSE IN KYANJJA

Client TONY SAKAZA Made by Date Sheet No Location FIRST FLOOR from grid 1 to 3 SNB 8-Sep-14 1

CONTINUOUS BEAM (Analysis & Design) to BS 8110:1997 Checked Revision Job NoOriginated from RCC41.xls on CD 1999 BCA for RCC SNB - 1

LOCATION Supports from grid 1 to grid 3

MATERIALS COVERS (to all steel)fcu 25 N/mm h agg 20 mm Top cover 25 mmfyl 460 N/mm gs 1.05 Btm cover 25 mm

fyv 250 N/mm gc 1.50 Side cover 25 mm

SPANS L (m) H (mm) bw (mm) hf (mm) Type bf (mm) LOADING PATTERNSPAN 1 5.01 400 200 R 200 min maxSPAN 2 DEAD 1 1.4SPAN 3 IMPOSED 0 1.6

SPAN 4 REBAR LAYERINGSPAN 5 Support steelSPAN 6 in alt layer ? Y

SUPPORTS ABOVE (m) H (mm) B (mm) End Cond BELOW (m) H (mm) B (mm) End CondSupport 1 0.40 400 4550 F 0.40 400 4550 F

Support 2 0.40 400 4550 F 0.40 400 4550 FSupport 3Support 4Support 5Support 6Support 7

LOADING UDLs (kN/m) PLs (kN) Position (m)Dead Imposed Position Loaded Dead Imposed Position Loaded

Span 1 Load Load from left Length Span 4 Load Load from left LengthUDL 20.2 2.0 ~~~~~ ~~~~~ UDL ~~~~~ ~~~~~PL 1 ~~~~~ PL 1 ~~~~~PL 2 ~~~~~ PL 2 ~~~~~

Part UDL Part UDL

Span 2 Span 5UDL ~~~~~ ~~~~~ UDL ~~~~~ ~~~~~PL 1 ~~~~~ PL 1 ~~~~~PL 2 ~~~~~ PL 2 ~~~~~

Part UDL Part UDLSpan 3 Span 6

UDL ~~~~~ ~~~~~ UDL ~~~~~ ~~~~~PL 1 ~~~~~ PL 1 ~~~~~PL 2 ~~~~~ PL 2 ~~~~~

Part UDL Part UDL

LOADING DIAGRAM

1 3

REACTIONS (kN)SUPPORT 1 2

ALL SPANS LOADED 79.0 79.0ODD SPANS LOADED 79.0 79.0

EVEN SPANS LOADED 50.7 50.7Characteristic Dead 50.7 50.7

Characteristic Imposed 5.0 5.0

REINFORCED CONCRETE

COUNCIL


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BENDING MOMENT DIAGRAMS (kNm)

1 Elastic Moments 3 1 Redistributed Envelope 3

SUPPORT No 1 2Elastic M 65.9 65.9 ~ ~ ~ ~ kNm/m

Redistributed M 56.0 56.0 ~ ~ ~ ~ kNm/mb 0.850 0.850 ~ ~ ~ ~ ~Redistribution 15.0% 15.0%

SPAN No 1Elastic M 33.04 ~ ~ ~ ~ ~

Redistributed M 42.93 ~ ~ ~ ~ ~b 1.299 ~ ~ ~ ~ ~

SHEARS (kN)

1 Elastic Shears 3 1 Redistributed Shears 3

SPAN No 1Elastic V 79.0 79.0 ~ ~ ~ ~

Redistributed V 79.0 79.0 ~ ~ ~ ~

Elastic V ~ ~ ~ ~ ~ ~

Redistributed V ~ ~ ~ ~ ~ ~

1 2 ALL SPANS Above 33.0 -33.0

LOADED Below 33.0 -33.0ODD SPANS Above 33.0 -33.0

LOADED Below 33.0 -33.0EVEN SPANS Above 21.1 -21.1

LOADED Below 21.1 -21.1

COLUMN MOMENTS(kNm)

-100

-80

-60

-40

-200

20

40

60

80

100

0 1 2 3 4 5 6-100

-80

-60

-40

-200

20

40

60

80

100

0 1 2 3 4 5 6

-40

-20

0

20

40

60

80

0 1 2 3 4 5 6-60

-40

-20

0

20

40

60

80

0 1 2 3 4 5 6


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b 0.85 1.30 0.85DESIGN d mm 349.0 361.0 349.0

As mm 291 295 291 As' mm

TOP STEEL Layer 1 3 T 12 2 T 12 3 T 12Layer 2


Layer 2 As' prov mm 226 As prov 339 As' prov 226


v N/mm 0.884 8 0.884


legs No 2 2 2


max S ok ok okLinks


Deflection ok


b 1.00 ~ 1.00DESIGN d mm -81.0 -45.5 -57.0











Deflection #DIV/0!

okok

#DIV/0!#DIV/0!

okok

#DIV/0!#DIV/0!

#DIV/0!#DIV/0!

okok


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BEAM 07


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Part UDL Part UDL



UDL ~~~~~ ~~~~~ UDL ~~~~~ ~~~~~PL 1 ~~~~~ PL 1 ~~~~~PL 2 ~~~~~ PL 2 ~~~~~

Part UDL Part UDL

LOADING DIAGRAM

1 3





REINFORCED CONCRETE

COUNCIL


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b 0.85 1.30 0.85DESIGN d mm 349.0 361.0 349.0

As mm 228 196 228 As' mm





v N/mm 0.575 8 0.575


legs No 2 2 2


max S ok ok okLinks


Deflection ok


b 1.00 ~ 1.00DESIGN d mm -81.0 -45.5 -57.0











Deflection #DIV/0!

okok

#DIV/0!#DIV/0!

okok

#DIV/0!#DIV/0!

#DIV/0!#DIV/0!

okok


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BEAM 09


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SPANS L (m) H (mm) bw (mm) hf (mm) Type bf (mm) LOADING PATTERNSPAN 1 3.35 400 200 200 R 200 min maxSPAN 2 4.95 400 200 200 R 200 DEAD 1 1.4SPAN 3 4.55 400 200 200 R 200 IMPOSED 0 1


SUPPORTS ABOVE (m) H (mm) B (mm) End Cond BELOW (m) H (mm) B (mm) End CondSupport 1 1.00 950 200 K

Support 2 1.00 950 200 KSupport 3 1.00 950 200 KSupport 4 1.00 950 200 KSupport 5Support 6Support 7


Span 1 Load Load from left Length Span 4 Load Load from left LengthUDL 2.8 ~~~~~ ~~~~~ UDL ~~~~~ ~~~~~PL 1 130.0 2.98 ~~~~~ PL 1 ~~~~~PL 2 ~~~~~ PL 2 ~~~~~

Part UDL Part UDL

Span 2 Span 5UDL 2.8 ~~~~~ ~~~~~ UDL ~~~~~ ~~~~~PL 1 130.0 3.08 ~~~~~ PL 1 ~~~~~PL 2 ~~~~~ PL 2 ~~~~~


UDL 2.8 ~~~~~ ~~~~~ UDL ~~~~~ ~~~~~PL 1 130.0 1.53 ~~~~~ PL 1 ~~~~~PL 2 ~~~~~ PL 2 ~~~~~

Part UDL Part UDL

LOADING DIAGRAM

1 3

REACTIONS (kN)SUPPORT 1 2 3 4

ALL SPANS LOADED 6.7 188.9 200.1 44.7ODD SPANS LOADED 6.7 141.9 111.6 44.7

EVEN SPANS LOADED 0.3 61.5 106.9 2.8Characteristic Dead 4.6 11.7 13.4 6.3

Characteristic Imposed 0.2 172.5 181.4 35.9

REINFORCED CONCRETE

COUNCIL


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b 0.85 0.70 1.30DESIGN d mm 349.0 361.0 349.0

As mm 52 36 125 As' mm





v N/mm 0.049 8 1.909


legs No 2 2 2


max S ok ok okLinks


Deflection ok


b 0.85 1.16 0.85DESIGN d mm 349.0 343.0 349.0

As mm 213 738 294 As' mm





v N/mm 0.704 8 1.343vc N/mm 0.356 Nominal 0.408

LINKS R8 @ 250 for 750 R8 @ 250 R8 @ 125 for 1500legs No 2 2 2


max S ok ok okLinks


Deflection ok

okok

okok

okok

okok

okok

okok


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Client TONY SAKAZA Made by Date Page Location FIRST FLOOR, from grid 1 to 3 SNB Sep-2014 4




b 0.91 1.07 1.00DESIGN d mm 349.0 359.0 349.0

As mm 271 485 210 As' mm




DEFLECTION 12.674 Allowed 32.448 SHEAR V kN 99.9 Link V 41.5

v N/mm 1.431 8 v 0.595vc N/mm 0.408 Nominal vc 0.356



max S ok ok okLinks


Deflection ok


b 1.00 ~ 1.00DESIGN d mm -53.0 -43.0 -59.0





DEFLECTION Allowed #DIV/0!SHEAR V kN #VALUE! Link #VALUE!






Deflection #DIV/0!

okok

#DIV/0!#DIV/0!

#DIV/0!#DIV/0!

okok

okok

#DIV/0!#DIV/0!


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BEAM 10


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Part UDL Part UDL



UDL ~~~~~ ~~~~~ UDL ~~~~~ ~~~~~PL 1 ~~~~~ PL 1 ~~~~~PL 2 ~~~~~ PL 2 ~~~~~

Part UDL Part UDL

LOADING DIAGRAM

1 3





REINFORCED CONCRETE

COUNCIL


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b 0.70 1.60 0.70DESIGN d mm 327.0 343.0 327.0

As mm 673 945 673 As' mm 351 164 351

TOP STEEL Layer 1 3 T 16 2 T 12 3 T 16Layer 2 1 T 16 1 T 16




v N/mm 1.661 8 1.661


legs No 2 2 2


max S ok ok okLinks


Deflection ok


b 1.00 ~ 1.00DESIGN d mm -81.0 -45.5 -57.0











Deflection #DIV/0!

okok

#DIV/0!#DIV/0!

#DIV/0!#DIV/0!

okok

okok

#DIV/0!#DIV/0!


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BEAM 11


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b 0.85 1.00 1.15DESIGN d mm 349.0 361.0 349.0

As mm 7 12 14 As' mm





v N/mm 0.051 8 0.078


legs No 2 2 2


max S ok ok okLinks


Deflection ok


b 0.85 1.30 0.85DESIGN d mm 349.0 361.0 349.0

As mm 277 271 278 As' mm





v N/mm 0.758 8 0.758vc N/mm 0.408 Nominal 0.408



max S ok ok okLinks


Deflection ok

okok

okok

okok

okok

okok

okok


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Client TONY SAKAZA Made by Date Page Location FIRST FLOOR, from grid 1 to 3 SNB Sep-2014 4




b 1.15 0.85 1.00DESIGN d mm 349.0 361.0 349.0

As mm 193 88 162 As' mm




DEFLECTION 13.019 Allowed 56.917 SHEAR V kN 31.3 Link V 29.2

v N/mm 0.448 8 v 0.419

vc N/mm 0.356 Nominal vc 0.356LINKS R8 @ 250 for 500 R8 @ 250 R8 @ 250 for 500

legs No 2 2 2


max S ok ok okLinks


Deflection ok

okok

okok

okok


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BS 5628; Part 11978


DESIGN OF LOAD BEARING MASONRY WALLS

Assume a 200mm thick wall


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Loading (per m-run of wall);

Load from: DLkN/m

LLkN/m

1. Roof Trusses 2.2 4.3

2. Ring Beams 5.63. Wall from ringbeam to 1 s floor 11.44. Concrete Beams on 1 st Floor 9.35. Timber Beams on 1 s Floor 11.4 36. Wall from 1 st floor to GL 11.4

Total 51.3 7.3

Slenderness ratio;

Since there are no intersecting walls, the effective height of the wall will govern theslenderness ratio.

Vertical Design Strength f d ;

Where;

Critically loaded wall is 600mm Length

Design Philosophy;

Thus;

From which


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DESIGN OF THE STRIP FOOTINGS

Loading (kN/m run) Characteristic Permanent Loads

Slab : 7.6kN/m 2 * 3.2m = 24.32kN/m

Concrete Beam: 25kN/m 3 * 0.2m*0.4m = 2.00kN/m

Masonry Walling: 22kN/m 3 * 8.2m*0.2m = 36.08kN/m

Roofing (Assumed) 2.5kN/m 2 * 3.2m = 8.00kN/m

Total Dead Load = 70.4kN/m

Characteristic Variable Load

Single Family Dwellings : = 2.00kN/m2

Maximum TributaryLength = 3.2m

UDL variable load 2kN/m2 * 3.2m = 6.4kN/m

Sizing of Strip Width (B)

From Geotechnical Investigation Report, at 900mm depthTest Pit 1 Test pit 2 Average

Cohesion, C (kN/mm 2) 39 36 37.5Angle of Friction, (degrees) 19 24 22

Bulk Density, (kN/m3

)17.99 18.15 18.07

Using Terzaghi's Bearing Capacity Equations for strip Footings;

B


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54.4kN

54.4kN

108.8kN

0.6m

181.333kN/m


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REFERENCE PARTICULARS OUTPUT Assume R6 bars at a spacing S

For a 1-m run;

Hence Provide R8 at 200mm centre to centre

structural calculations

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