Al- Quds Open University -Nablus

Al-Quds Open University -

Nablus

Prepared by: Rana Adli Ramahi

Supervised by: Eng. Imad Al-

Qasim

An-Najah National UniversityFaculty of EngineeringCivil Engineering Department

Outline:

Introduction Slabs preliminary design Beams preliminary design Three-dimensional structural modeling Columns design Walls design Footings design Stairs design

2 An-Najah National University

An-Najah National University

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Chapter One:

Introduction

Project Description

The structure that designed is the building of Al-Quds Open University, Nablus-Palestine.

the land area is 3208 m2 It consists of two Buildings, Academic & Administrative

Buildings. They consists of five floors over ground & join together in

two basement floors. The ground floor elevation is 4.5 m, all other floors are

3.5m elevated.



Project Description

The academic building will be designed in this project

It’s divided by structural joints with thicknesses equal 10cm roughly into four parts: Part A is the central entrance for the academic

building Parts B & C consists of garages in the second basement

floor, lectures rooms in floors (B1-F3), & offices in the fourth floor.

Part D is an auditorium (it’s design not completed)


7An-Najah National University

Philosophy of analysis & design The structure was analyzed & designed by one & two

dimensional structural models for the elements manually & using SAP, then using three dimensional structural model


SAP2000 14.2.4

Design code & load combinations

The structural design was according to American concrete institute code (ACI318-08)

Depending on the used code the load combinations are: U1 = 1.4D U2 = 1.2D+1.6L+1.6H U3 = 0.9D+1.6H



Materials Structural materials used in the design are:

concrete:f’c=24 MPa for horizontal elements f’c=28 MPa for vertical elements Steel(both longitudinal & transverse): fy=420 MPa

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Soil Bearing Bearing capacity of the soil is 400 KN/ m2

Loads: Gravity loads: Live loads:

for all floors = 4 KN/m2 except the second basement floor which equal 5KN/m2

Dead Load:In addition to own weight of the structural elements, Superimposed dead load =4.36 KN/m2 Perimeter walls load

=20.6 KN/m ………………. for h=3.5m=26.5 KN/m …………..…... for h=4.5m

Lateral loads: Only soil pressure was taken into consider.


Structural Systems: Part B & Part C are designed as one way ribbed slabs in X-

direction, with hidden beams generally

Part A & Part D are designed as two way solid slabs with drop beams.



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Chapter Two:

Slabs Preliminary Design

Thickness: For Part (B) & Part (C):One way ribbed slabL max = 5.8 mHmin=L / 18.5

=5.8 / 18.5 = 0.314 m→Use h = 0.32 m

For Part (A) & Part (D):Two way solid slabDirect design method does not satisfied for this slab, so

many trials for thicknesses done in the 3-d model to determine the suitable thickness.


Loads estimation: As a sample calculation Take the ribbed slab (rib1.1):

Own weight = 3.15 (KN/m) /0.55 = 5.73 KN/m2 Superimposed dead load =4.36 KN/m2 Total dead load = 5.73+4.36 = 10.2 KN/m2 Live load = 4 KN/m2 →Wu (for rib) = 10.2 KN/m/rib


Analysis using SAP:


Reinforcement: φVc = φ*0.1667*√24*150*290 = 26.67 KN > Vu No need for shear reinforcement As min = 0.0033*150*290 = 143 mm2



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Chapter Three:

Beams Preliminary Design

Thickness: Take Beam B2 as a sample calculation:All spans of this beam have a length 5.6 m

Hmin=L / 18.5 = 5.1 / 18.5 = 0.28 m

→Use hidden beams (h=32cm)


Loads estimation: Wu from slab (5.7) =60.7 KN/m Wu from slab (3.2) =29.6 KN/m Factored O.W. of beam = 0.32 * 1 * 25 *1.2 = 9.6 KN/m Total ultimate load on beam = 60.7 + 29.6 + 9.6 = 100 KN/m

An-Najah National University20

Using ACI-coefficient the moment diagram(KN.m),shear diagram (KN):

.

As =ρ*b*dAn-Najah National University21

Torsion

From ribs models in SAP, the moments in the supports of the slab act on the beams as torsion ……..

Tm = 16.96 KN.m/m Tv(due to shear on the face of the beam from both

sides)= 15.53 KN.m/m→Total factored torsion = 73.42 KN.m/m (64.58 at d from

face of support) Reduction for compatibility torsion to be 47.5 KN.m/m No equilibrium torsion occur because the center of the

beam same as the center of columns carry it


Torsion Check section adequacy for beam

OK, Good section

Av/s +AT/s = 2.09 mm2/mm Use 1φ12/100mm Longitudinal steel at each support = 510 mm


Total reinforcement (manual design)


Model in SAP The beam was modeled in SAP with fixed ends firstly &

pin ends also, then the average of loaded was used for the design


Pin ended

fix ended

Model in SAP The average moment (KN.m):

The total (torsion & flexure) longitudinal reinforcement:


Section in beam B2



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Chapter Four:

Three Dimensional Structural Modeling

General


A three dimensional structural modeling done using SAP2000(v14.2.4)

Many verification checks must be achieved, serviceability, equilibrium, compatibility, & stress-strain relation ship.

ServiceabilityLong Term Deflection Check Take Part (C) & Part (B) as sample calculation, the check

occurs for many spans, the critical one is the span between H-G&15-17 SAP information: Mu=10.3 KN.m ∆D = 0.0094, ∆D+L = 0.0114 Calculations summery: ∆allowable = L/480 = 2.9/480 = 0.00604 m ∆LT = ∆L +α ∆D+α T ∆Ls = 0.0226 >∆all ,find Ig/Icr


Ig = 7.03*104 cm4

Icr = 8277 cm4

.

.

∆L = ∆D - ∆D+L = 0.125 mm ∆LT = ∆L +α ∆D+α T ∆Ls = 0.001 < ∆all=0.006 OK


Check Compatibility


From start animation in SAP, the compatibility(structure work as one unit) was verified

Check Equilibrium


Take Part (B) as sample calculation Dead load (own weight)


Dead load (Superimposed dead load)

Live load


Comparing results: Error (own weight) = 2.4% OK Error in SID = 2.64% OK Error in live load = 0.33% OK


Take Part (B) as sample calculation, check for span 5-6 in beam B4 (1m*0.32m) & Wu =92 KN/m

+ve M = Wu*Ln2/16 = 92*3.92/16 = 87.46 KN.m

-ve M =127.2 KN.m at each support & avg is 127.2KN.m 1-D model: (+ve M) + [(–ve MR + -ve ML)/2] = 214.67 KN.m 3-D model: M = 233 KN.m

Error = 8.7………………….. OK

Check stress-strain relationship


Take Part (B) as sample calculation, design slab thickness 32cm (ribbed slab R1) & cover 3cm.

Bending moment diagram for rib 1 (KN.m/m)

Slab Design



Take Part (B) as sample calculation, design beam on grid (F)

Beam Design



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Chapter Five:

Columns Design

Preliminary Design

Take group G2 as sample Pu = 2000 KN , assume ρ = 0.01 Pu = 0.65 * 0.8 * [0.85* f’

c *(0.99Ag) + Fy*(0.01Ag)]

→ Ag = 0.1523 m2 ,

use section b*h = 0.5 m *0.5 m As = * Aρ g = 0.01 *0.25 = 2.5*10-3 m2 = 2500 cm2

So, use 8 20 mmφ


(3-D)Design Take group G2 as sample( Pu = 1727.8KN & M2=47 KN.m

&M1=29.66 KN.m), design for axial, flexure & shear Check buckling, depend on stiffness's of column & related beams:

K factor =0.82 (Non-sway)


K*Lu/r = 0.82*4.2/0.15 = 22.96 Check slenderness: 34 – 12(M1/M2) >22.96 (neglect slenderness) δns = 0.92, use δns = 1 →Md=Mu = 47 KN.m Using interaction diagram(f’c=28, fy=420, =0.75)ƴ →ρ=0.015, As=30 cm2 → Use 8 22 mmφ Use 10/200 mmφ

Vn<Vc/2 …………… no shear reinforcement



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Chapter Six:

Walls Design

Bearing wall design


Take wall 6as sample, design for axial loads.

average axial load=337 KN


Thickness:

Section used has a thickness =25 cm = 1952KN > Pu OK Reinforcement (minimum) Vertical reinforcement = 0.0012*Ag (Use 1 10/300 mm)φ Horizontal reinforcement = 0.002*Ag (Use 1 10/300 mm)φ

Basement wall design


Take wall 11 as sample, design for axial loads & soil load. Vu (from sap) = 122 KN Vc = 127KN>Vu OKφ From SAP: Vertical Moments: M +ve max = 55 KN.m/m (ρ=0.004) → (1 14/250mm)φ M-ve max = 30 KN.m/m (ρ=0.0033) → (1 14/250mm)φ Horizontal Moments: M +ve max = 33 KN.m/m (ρ=0.003) → (1 12/300mm)φ


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Chapter Seven:

Footings Design

Footing system

Footings that used in this project can be classified into three types:

Wall footing Single footing Combined footing No need for using mat foundation because the soil has

high bearing capacity. The footings was grouped into 27 group, depending on

the column load, dimension, shape, adjacency.


Single footing (G22)


Footing area=Pa/Qall →B= 2.5m & L= 2.6m Footing thickness designed to

resist the wide beam & Punching shear: φVc > Vu &

Vcp > Vupφ Flexure design: Mu = qu*l2/2 In B direction: M= 411 KN.m → 7 20/mφ In L direction: M= 197 KN.m →5 18/mφ

Single footing (G22)



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Chapter Eight:

Stairs Design

The stair was designed manually & using SAP:



Thank You

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Al- Quds Open University -Nablus