lectures notes of advanced concrete design 1 ceng 306

Advanced Concrete 1

Modern University

For Information and Technology

Civil Engineering Department

Lectures Notes of

Advanced Concrete Design 1

CENG 306

Prepared By

Dr : Yasmin Hefni

(First Edition 2021)

Advanced Concrete 1

Vision

The vision of the Faculty of Engineering at MTI university is to be a center

of excellence in engineering education and scientific research in national

and global regions. The Faculty of Engineering aims to prepare graduates

meet the needs of society and contribute to sustainable development.

Mission

The Faculty of Engineering MTI university aims to develop distinguished

graduates that can enhance in the scientific and professional status,

through the various programs which fulfill the needs of local and regional

markets. The Faculty of Engineering hopes to provide the graduates a

highly academic level to keep up the global developments.

Advanced Concrete 1

Introduction

This content has been prepared as a lecture note, which is a compilation of

the books of the professors mentioned in the references.

The content has been prepared with clarification of all technical and

engineering terms, as well as some solving examples, in order to achieve for

the student, the ability to understand and comprehend, as well as familiarity

with technical and engineering terms.

تجميع من كتب الاساتذة الافاضل المذكورين بالمراجع. يوه كمذكرة محاضراتتم اعداد هذا المحتوي

والهندسية وكذلك بعض الامثلة المحلولة وبما يحقق ولقد تم اعداد المحتوي مع توضيح جميع المصطلحات الفنية

للطالب القدرة علي الفهم والاستيعاب وايضا الإلمام بالمصطلحات الفنية والهندسية. اعداد

د/ ياسمين حفن

Advanced Concrete 1

Content Title Pages

Chapter 1: Design of Stairs 1-18

Chapter 2 : Design of Eccentric Sections 19-45

Chapter 3 : Design of Frames 46-92

Chapter 4 : Design of Arched Slabs 93-107

Chapter 5 : Design of Arch Girders

108-121

References

• ECP code for design.

• Design of Reinforced Concrete Structure II, D. Mashour A. Ghoneim.

• Lectures in Reinforced Concrete, Prof. Sherief Elwan.

• Lectures in Reinforced Concrete, Eng. Yasser Elleathy.

• Design of Reinforced concrete, Dr. Magdy A. Tayel.

• Reinforced Concrete, Dr. Nasr Younis.

• Lectures in Reinforced Concrete, Prof. Ahmed Ghallab.

• Lectures of Reinforced concrete, Tanta University, Facluty of Engineering

• Lectures of Reinforced concrete, Shubra University, Facluty of Engineering

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Chapter 1

Design of Stairs

1.1 Introduction

A Stair is a system of steps consists of risers and treads (going), stairs are used for accessing

different floors. due to the discontinuity of stair slab stairs are considered a critical slab and

should have a perfect design.

The height of the riser (r) is about 150-200 mm, the width of the going (tr) is about 250-

300 mm A good design of the stair should comply with the following role of thumb

2r+tr=600 to 620

Ɵ=tan -1 [𝑟𝑖𝑠𝑒𝑟

𝑔𝑜𝑖𝑛𝑔]

Geometric of stairs

1.2 Classification of stairs

1.2.1 According to shape

➢ Straight stairs

Straight stairs consist of several steps leading to the same direction. Usually,

straight stair consists of one flight.

➢ Dog legged stairs

Succeeding flight rise are in opposite directions.

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➢ Open newel stairs

A rectangular opening or well is exist between flights.

➢ Geometrical stairs

A curved well exists between flights.

1.2.2 According to spanning direction

➢ Stairs spanning horizontal

These stairs are support at both sides by sides by walls or beams. The stairs may

be acts as a cantilever from a supporting beam or an RC wall

➢ Stairs spanning longitudinally

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These stairs span between supports at the bottom and the top flight and remain

unsupported at the sides

1.2.3 According Structural Systems of Stairs

The main differences among these types are the way through which the flights and the

landings transfer their loads to the supporting beams and columns.

➢ Cantilever type.

➢ Slab type.

➢ Slab-beam type.

➢ Spiral type

➢ Free-stranding type

1.3 Loads on Stairs

Calculate the horizontal load Wsh (for landing) and Wsi for inclined slabs (flights).

Wsh = 1.4 (t s * ˠ c +F.C) + 1.6 L.L.

Wsi = 1.4 (t av * ˠ c +F.C) + 1.6 L.L. cos Ѳ

Live loads

= 3.0kN/m2 (Residential)

= 4.0kN/m2 (Office building)

= 4.0kN/m2 (Hospitals)

= 4.0kN/m2 (School)

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= 5.0kN/m2 (Cinemas, theaters)

1.4 Steps of design

❑ Take Strips in load direction.

❑ Calculate BM, design strips,.

❑ Draw reinforcement of slabs in plan taking inclination into consideration.

❑ Calculate BM, reactions for beams.

❑ Design of stair beams.

1.5 Design of sectors subject to torsional moments

1.5.1 Introduction to Torsion moments

❑ Some structural elements are subjected to torsion moments accompanied by shear

forces and bending moments.

❑ When the beam is subjected to a torsion moment, it results in shear stresses that are

of greater value at upper and lower faces.

❑ The stresses (principal stresses) in compression and tension are equal to the shear

stresses resulting from torsional moments, and the main tensile stresses cause helical

annular cracks around the beam as shown in the drawing.

❑ These cracks must be resisted with a longitudinal reinforcement in the beam added

to the original reinforcement for bending moments in addition, external stirrups

added to original shear stirrups beams and a beam added to the shear stresses

resistant to the beam

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عزم و القص لقوى مصاحبة تكون و الاحيان بعض فى الالتواء لعزوم الانشائية العناصر بعض تتعرض ❑

. العنصر لها يتعرض التى الانحناء

الكمرة اوجهه عند لها قيمة اكبر تكون قص اجهادات عنها ينتج فانه التواء لعزم الكمرة عرضتت عندما ❑

الاجهادات وتكون( فرضا فقط التواء لعزوم الكمرة تعرض حالة فى و والسفلية العلوية و الجانبية

(principal stresses ) وتسبب الالتواء عزوم عن الناتجة القص لاجهادات مساوية الشد و الضغط فى

. بالرسم موضح هو كما الكمرة حول حلزونية حلقية شروخا الرئيسية الشد اجهادات

الاصلى التسليح ألى يضاف و( طولى تسليح ) الكمرة فى طولية سياخاب مقاومتها يجب الشروخ هذه و ❑

المقاومة الكانات إلى تضاف ولكمرة محيطة خارجية كانات الى بالاضافة الانحئاه لعزوم المقاوم بالكمرة

ألكمرة لها تتعرض الى القص لاجهدات

1.5.2 Some cases where beams are exposed to torsion moments:

بعض الحالات التي تتعرض فيها الكمرة لعزوم التواء

❑ Case No. (1): Cantilever slab resting on a beam

كمرة على مرتكزة كابولية بلاطة( : 1) رقم حالة

❑ Case No. (2): A cantilevered slab that has no extension within the adjacent slab

and is analyzed in the same manner as in Case No ( .1 )

فى المتبع الاسلوب بنفس تحليلها ويتم المجاورة البلاطة داخل امتداد لها يوجد لا كابولية بلاطة( : 2) رقم حالة

( 1) رقم حالة

❑ Case No. (3): a cantilevered slab protruding on both sides of the beam with

different lengths on both sides of the beam .

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.الكمرة جانبى على الطول اختلاف مع الكمرة جهتى على بارزة كابولية بلاطة( : 3) رقم حالة

❑ Case No. (4): a cantilever beam rested on a beam, the cantilever beam loaded by

a concentrated or distributed load .

.طرفى مركز حمل أو موزع بحمل محمولة و كمرة من بارزة كابولية كمرة ( : 4) رقم حالة

❑ Case No. (5): L shape beam loaded with distributed or concentrated edge load .

كمرة علي شكل حرف ومحملة بحمل موزع او حمل مركز طرفي

1.5.3 Design of Sections subjected to torsion moments

Steps

❑ Calculate Qu and Mu at the critical section of the beam

❑ Calculate shear stress due to both shear and torsion

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q u= 𝑸𝒖

𝒃𝒅 q t=

𝑴𝒕𝒖

𝟐 𝑨𝒐∗ 𝒕𝒆

q u actual shear stress due to shear

q t actual shear stress due to torsional moment

Ph = perimeter of the stirrup which resists torsion

الكانة الخارجيةمحيط المقاومة لزم الالتواء مقاسة من محورها

Ao= 0.85 Aoh = X1*Y1*0.85.

Where Aoh =مساحة القطاع المحصور داخل الكانة

te= 𝐴𝑜ℎ

𝑃ℎ=

𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑠 𝑠𝑡𝑖𝑟𝑟𝑢𝑝

𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑖𝑟𝑟𝑢𝑝

te= 𝑥1∗𝑦1

2(𝑥1+𝑦1)

qt= 𝑀𝑡𝑢(𝑋1+𝑌1)

0.85∗(𝑋1∗𝑋1∗𝑌1∗𝑌1)

Tاو Lللتسهيل يمكن اهمال الجزء الفعال من البلاطة في حالة قطاعات

❑ Calculate the following limit stresses

qcu-uncracked = 0.16 √ (fcu / γc)

qcu-cracked = 0.12 √ (fcu / γc)

qu max = 0.7 √ (fcu / γc )

qt min = 0.06 √ (fcu / γc )

❑ Calculate the resultant stress due to shear and torsion =√ 𝑞 𝑢 + q 𝑡

Case a. √ 𝑞 𝑢 + q 𝑡 > qu max increase the concrete dimensions

Case b. √ 𝑞 𝑢 + q 𝑡 ≤ qu max concrete dimensions are safe and check the

following :

✓ qu ≤ q cu uncr q tu ≤ qt min use stirrups 5 Ǿ 8/m

✓ qu > q cu uncr q tu ≤ qt min use stirrups to resist (qu - q cu cracked )

✓ qu ≤ q cu uncr q tu > qt min use stirrups to resist (q tu)

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✓ qu > q cu uncr q tu > qt min use stirrups to resist (qu - q cracked)+(q tu)

❑ reinforcement to resist torsion

❑ torsion is resisted by either closed stirrups or longitudinal reinforcement

Closed stirrups

Astr = 𝑀𝑡𝑢∗𝑆𝑡

1.7∗𝐴𝑜ℎ∗(𝐹𝑦𝑠𝑡

ˠ𝑠)

Astr min = 0.2∗𝑏∗𝑠

𝑓𝑦𝑠𝑡

Smax the smaller : 𝑃ℎ

8 or 200mm

Longitudinal reinforcement

As L= 𝐴𝑠𝑡𝑟∗𝑃ℎ

𝑆𝑡 (

𝐹𝑦𝑠𝑡

𝐹𝑦)

Where

Fy yield strength of longitudinal reinforcement

Fyst yield strength of stirrups

Astr Area of stirrups

AsL Area of longitudinal reinforcement

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Example

For the shown plan of a one-story-building, the following data are given: fcu = 25 MPa fy

= 360 MPa fyst = 240 MPa Flooring cover = 2.0 KN/m2 Live load on H.P. = 2.0 KN/m2

(assume permanent load) Slab thickness = 140 mm Wall load/m3 = 16 KN/m3 (positions

as shown in figure) Girder section 400*800 mm Column section 400*400 It is required to:

- Calculate the external loads and internal forces.

- Design the girder critical sections for flexure, shear, and torsion (if any).

- Draw the girder details of reinforcement in elevation and cross sections.

- Design & draw a column with uniform distributed steel to carry Pu= 800 kN and Mu=

160 KN.m

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Solution

g slab = ts * c + F.C = 0.14 * 25 + 2 = 5.5 kN/m2

Lslab = 2 kN/m2 (H.P) Lslab = 2 cos = 1.9 kN/m2 (R.P)

Uslab max = 1.5 (g slab + L slab ) = 1.5 (5.5 + 2.0 ) = 11.25 kN/m2 (H.P)

= 1.5(gslab+Lslab)= 1.5 ( 5.5 + 1.9 ) = 11.1 kN/m2 (R.P)

Uslab min = 0.9 * gslab = 0.9 * 1.5 = 4.95 kN/m2

o.w of girder = b* ( t - ts )* c = 0.4 (0.8 - 0.14) (25) = 6.6 kN/m

parapet wall load = w * hp * tw = 16 * 1.0 * 0.12 = 1.92 kN/m

Case of max. shear

R1=11.25*3 + 2*1.5*1.92 = 39.51 kN/m

Mo1 = 0.0

Umax=1.5(o.w + wall load ) + R1 =1.5(6.6+0.0) +

39.51 = 49.41 kN/m

-49.41(1.5) + 274.59 - 48.96 * X = 0

Then X = 4.09 m

R1=11.1*3 + 2*1.5*1.92 = 39.06 kN/m

Mo2 = 0.0

Umax = 1.5 (o.w + wall load) + R2

= 1.5 (6.6 + 0.0) + 39.06 = 48.96 kN/m

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Case of max torsion

R3 = 0.9*1.92 + 1.5*1.92 + 4.95*1.5

Mo3 = {1.5*1.92(1.5) + 11.25*1.5(0.75)}-{0.9*1.92(1.5) + 4.95(1.5)(0.75)} = 8.8

kN.m/m

Umax = 1.5 (o.w + wall load) + R3 = 1.5(6.6+0.0) + 28.9 = 38.8 kN/m

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R4 = 0.9*1.92+1.5*1.92 + 4.95*1.5 + 11.25*1.5 = 28.9 kN/m + 11.1*1.5 = 28.68 kN/m

Mo4 = {1.5*1.92(1.5) + 11.1*1.5(0.75)} - {0.9*1.92(1.5) + 4.95(1.5)(0.75)} = = 8.65

kN.m/m

Umax = 1.5 (o.w + wall load) + R4 == 1.5(6.6+0.0) + 28.68 = 38.58 kN/m

Qu = 149.6 kN & Mtu = 53.34 kN.m

Aoh = (400 - 2*35)(800 - 2*35) + 2*420(140 - 2*15)= 333300 mm2

Ph = 2{(400-2*35)+(800 - 2*35)} + 4*420 = 3800 mm

te = Aoh/Ph = 87.71 mm Ao= 0.85*Aoh = 283305 mm2

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SHEAR

qu = Qu /( b*d ) = 149.6*10³/(400*750) = 0.5 N/mm2

qu max = 0.7* √(fcu/c) = 1.2 > qu

qsu = qu - 0.5*qcu = 0.291N/mm2

Astr/s = (Astr/s)shear + (Astr/s) torsion = 0.139 + 0.45 = 0.589 mm

use 2*8/m`

TORSION

qtu = Mtu/( 2*Ao*te ) = 53.34*106 /(2*283305*87.71) = 1.07 N/mm2

qtu max = 0.7√(fcu/c) = 2.57 >qtu

(Astr/s)torsion = Mtu/ ( 2*Ao*fystr/s ) = 53.34*106 *1.15/(2*283305*240 ) = 0.45 mm

Asl = (Astr/s)torsion* Ph * fyst/fy = 0.45*3800*(240/360) = 1140 mm2

Acp = 400*800+2*420*140 = 437600 mm2

Asl min = 0.46√(fcu/c)*Acp/fy - (Astr/s)torsion* Ph* fyst/fy

= 0.46*√(25/1.50*437600/3600 - 1140 = 1142.74 > Asl

Asl = 1142.7 mm2 = 4 +

For -ve moment Design as R- Sec. Mu-ve = 55.85 kN.m

Mu max = Rmax b d² fcu /c = 0.194 * 400 * (750)² * 25 / 1.5 * 10-6 = 727.5 kN.m

Mu max > Mu -ve → single R . F . T

Ru = Mu / ( b.d² ) = 55.85 * 10 6 / ( 400 * (750)²) = 0.247 → = min As =

min*b*d=1.1/360*400*750 = 916.6 mm2 use 418

For + ve moment T-Sec M+ve = 334.33 kN.m

B is the min of :

• C.L to C.L = 3 m

• 16 * ts + b =16*(0.14) + 0.4 = 2.65 m

• Lz/5 + b = 0.8 (1.5 + 6.3)/5 + 0.4 = 1.648m

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Then B = 1.648 m

Mf = 0.67*fcu/c*B*ts*(d - ts/2) = 0.67*(25/1.5)*1648*140*( 750 - 70)*10-6 =

1751.93 kN.m > Mu+ve

Ru = Mu /( B*d²) = 334.33*106 /(1648 * (750)²) = 0.3606

= 0.00125 As= B*d = 0.00125*1648*750 = 1545 mm2

As min = 1.1/360*400*750 = 916.67 mm2 > As

As+ve = 1545 mm2 = 7

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Qu = 800 kN & Mu = 160 kN.m

Pu/( fcu*b*t ) = 800 * 10³/(25*400*400) = 0.2

Mu/( fcu*b*t² ) = 160 * 106 /(25*400*400²) = 0.1

= 4.5

= fcu*10-4= 4.5*25*10-4 = 0.01125

As total = *b* t = 0.01125*400*400 = 1800 mm2 use 8

R = 333 KN

Pu = R * 1.1 * 6.0 = 2197.8 KN

Mu = 49.8 KN.m

Pu / (fcu b t ) = 2197.8 * 103 / (25 * 400 * 400) = 0.55

Mu / (fcu b t2 ) = 49.8 * 106 / (25 * 400 * 4002 ) = 0.03

ρ = 8.1

R = 273.71 KN

Pu = R * 1.1 * 6.0 = 1806.48 KN

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Mu = 116.15 KN.m

Pu / (fcu b t ) = 1806.48* 103 / (25 * 400 * 400) = 0.45

Mu / (fcu b t2 ) = 116.15 * 106 / (25 * 400 * 4002 ) = 0.07

ρ = 9.0

μ = ρ fcu * 10-4 = 9 * 25 * 10-4 = 0.0225

Astotal = μ b t = 0.0225 * 400 * 400 = 3600 mm2 = 12 Ф 20

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1.5 Problems

Problem 1

Figure (1) shows a structural lay out of stair (each step is 300x150mm), if columns are

allowed only on axis x-x, it is required to:

.

Figure (1)

a) With reasonably assumed dimensions choose two statical systems for the stair case.

b) design one statical systems and draw the reinforcement on the plan and sections for the

select statical systems.

Data: - F.C = 1.5 kN/m2 LL = 4.0 kN/m2 –

- Fcu =25 N /mm2 Fy=350N/mm2

Problem 2

Figure (2) shows a structural lay out of different stairs, it is required to:

a) With reasonably assumed dimensions choose statical systems for each stair.

b) design all stair slabs and draw the reinforcement on the plan.

c) design beam B1 for each stair and draw the reinforcement on the plan for the beam.

Data: - F.C = 2.0 kN/m2 LL = 4.0 kN/m2 –

- Fcu =30 N /mm2 Fy=400N/mm2

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Figure (2)

Problem 3

Figure (3) shows a structural lay out of different stairs, it is required to:

a) With reasonably assumed dimensions choose statical systems for each stair.

b) Design all stair slabs and draw the reinforcement on the plan.

c) Design beam B1 for each stair and draw the reinforcement on the plan for the beam.

Data: - F.C = 2.0 kN/m2 LL = 3.0 kN/m2

- Fcu =35 N /mm2 Fy=420N/mm2

Figure (3)

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Chapter 2

Design of Eccentric Sections

2.1 Introduction

Reinforced Concrete Sections in most construction systems are subjected to loads that are

not concentrated in the Sections center, those loads are compressive or tensile loads. These

loads are equivalent to concentrated loads accompanied by bending moments. The load can

be uncentered in one direction (uniaxial) or in two directions (biaxial). In these cases, the

section is designed as a section subjected to concentrated loads accompanied by bending

moments in one or two directions.

تتعرض القطاعات الخرسانية في معظم الانظمة الانشائية الي احمال غير متمركزة في مركز ثقل القطاع سواء كانت

متمركزة مصحوبة بعزوم انحناء. ويمكن ان يكون الحمل تلك الاحمال احمال ضغط او شد. وهذه الاحمال تكافئ احمال

(. وفي هذه الحالات يتم تصميم القطاع كقطاع معرض BiaxiaI( أو اتجاهين )Uniaxialغير متمركز في اتجاه واحد )

لاحمال متمركزة مصحوبة بعزوم انحناء في اتجاه واحد او اتجاهين.

2.2 Sections subjected to compression loads and bending moments

القطاعات المعرضة لاحمال ضغط مصحوبة بعزوم انحناء

The behavior of the concrete section subjected to concentrated compression loads

accompanied by bending moments depends on the value of the load and the value of the

effective moments. If the value of the moments (M) is small compared to the value of the

load (N), then the value of (e = M/N) is relatively small. This means that That a part of the

section ; far from the center of the effect of the force; is subject to weak tensile stresses,

and therefore the failure is often the result of the strain in concrete reaching the maximum

strain before the strain in the steel reaches the yield strain value. This behavior applies to

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most of the cases if the ratio t ≤ 0.5/e, which is the closest to the behavior of columns

subjected to concentrated vertical loads.

ويعتمد سلوك القطاع الخرساني المعرض لاحمال ضغط متمركزة مصحوبة بعزوم انحناء على قيمة الحمل وقيمة العزم

( تكون صغيرة نسبيا وفي e=M/N( فإن قيمة )Nصغيرة بالمقارنة بقيمة الحمل )( Mالمؤثرين. فاذا كانت قيمة العزم )

معرض لاجهادات ضغط يكون القطاع اغلب فان وبالتالي القطاع داخل الضغط قوة تاثير مركز يكون الحالة هذه

(Eccentricity Small معرض لاجها )القوة تاثير )البعيد عن مركز منه يكون جزء أن المحتمل دات شد ( ومن

قبل وصول الاقصي الانفعال الي الخرسانة في الانفعال نتيجة وصول مايكون غالبا الانهيـار فان وبالتالي ضعيفة

وهو t ≤ 0.5/eالانفعـال فـي الحديد الي قيمة انفعال الخضوع. وهذا السلوك ينطبق علي معظم الحالات اذا كانت نسبة

ال راسية متمركزة. اقرب مايكون لسلوك الاعمدة المعرضة لاحم

if the value of the moment (M) is large compared to the value of the load, then the value

of (e=M/N ) is relatively large. Only part of the section (close to the center of the force

effect) is subjected to compressive stresses, and therefore the collapse is often the result of

the strain in steel reaching the maximum strain before the stress in concrete reaches the

maximum strain value. This behavior applies to most of the cases if the ratio (e/t>0.5),

which is the closest to the behavior of beams subjected to bending moments.

( تكون كبيرة نسبيا وفي هذه الحالة =M/N e( كبيرة بالمقارنة بقيمة الحمل فأن قيمة )Mأما اذا كانت قيمة العزم )

Eccentricityيكون مركز تاثير قوة الضغط خارج القطاع وبالتالي فان اغلب القطاع يكون معرض لاجهادات شد )

Bigا فان ( ومن القوة( معرض لاجهادات ضغط وبالتالي تاثير )القريب من مركز لمحتمل ان يكون جزء منه فقط

الانهيار غالبا مايكون نتيجة وصول الانفعال في الحديد الي الانفعال الاقصي قبل وصول الانفعال في الخرسانة الي

(.وهو اقرب مايكون لسلوك e/t 0.5<ئت نسبة )قيمة الانفعال الاقصي. وهذا السلوك ينطبق علي معظم الحالات اذا كا

الكمرات المعرضة لعزوم أانحناء.

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21 Dr/Yasmin Hefni

interaction Diagrams2.2.1 منحنيات التفاعل

Sections subjected to eccentric loads are designed using curves called interaction curves,

ID is the curve connecting a group of points. Each point represents a relationship between

different values of the maximum moments and the corresponding maximum compression

forces. It represents the safety limits of the section. if the section is affected by the value

of moment and compression force, represented by a point, if this point is outside the curve

limits, this means that the section does not bear this load and moment safely, but if the

point is inside or on the curve limits, then the section bears the load and moment safely.

يتم تصميم القطاعات المعرضة لاحمال غير متمركزة باستخدام منحنيات تسمي منحنيات التفاعل وهذا المنحني عبارة

عن المنحنى الواصل بين مجموعة من النقاط تمثل كل نقطة علاقة بين قيم مختلفة من العزوم القصوي وقوي الضغط

و يمثل حدود الامان للقطاع بحيث اذا تم التاثير علي القطاع بقيمة عزم وقوة ضغط تم تمثيلهم القصوي المقابلة لها وه

بنقطة وكانت هذه النقطة خارج حدود المنحني فمعني هذا ان القطاع لايتحمل هذا الحمل والعزم بامان اما اذا كانت

بأم والعزم الحمل يتحمل القطاع فأن المنحني حدود اوعلي داخل قطاع النقطة رسمه لاي يمكن المنحني وهذا ان.

خرساني معلوم ابعاده وتسليحه بحيث ان النقاط الواقعة علي المنحنى تمثل قيم للعزوم القصوي الواقعة علي القطاع وقيم

قوي الضغط القصوي المقابلة لها وبالتالي فان المنحنى يمثل الحدود القصوي للقطاع.

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22 Dr/Yasmin Hefni

As can be seen from the curve, point A represents the case of the section being exposed to

an axial compression force, which means the failure of the section under compression. As

well as for point 'A, which represents the state of the section when it is subjected to an

eccentric load with a ratio of (e/t = 0.05). Point B and all points on the curved part C - 'A

represent the state of the section when the strain on the concrete reaches the maximum

value (0.003) while the steel did not reach the yield and the failure is also a stress collapse.

Point C represents the state of equilibrium where the stress in the concrete reaches the

maximum value (0.003) at the same time that the steel reaches the yield strain, which is the

failure of the equilibrium state (Balanced failure) and this point at which the moment value

reaches the maximum moment that the section can withstand. This point is considered a

demarcation point in terms of the type of failure. If the value of the load exceeds the value

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23 Dr/Yasmin Hefni

of the equilibrium, a failure in compression occurs, and if the value of the load decreases,

a collapse in tension occurs.

تمثل حالة تعرض القطاع لقوة ضغط محورية بما يعني انهيار القطاع في الضغط Aوكما يتضح من المنحني فان نقطة

( Compression failure' للنقطة وكذلك .)A بنسبة متمركز لحمل غير تعرضه عند القطاع حالة تمثل والتي

(0.05 =e/t وتمثل النقطة .)B وجميع النقاط علي الجزء المنحنيC -'A حالة القطاع عندما يصل الانفعال بالخرسانة

حالة Cقطة ( بينما لم يصل الحديد للخضوع ويكون الانهيار انهيار ضغط ايضا. وتمثل الن 0.003الي القيمة القصوي )

( في نفس وقت وصول الحديد لانفعال الخضوع وهو 0.003الاتزان حيث الانفعال بالخرسائة الي القيمة القصوي )

( وهذه النقطة تصل عندها قيمة العزم الي اقصي عزم يمكن للقطاع تحمله. Balanced failureانهيار حالة الاتزان )

النقطة حد فاصل من حيث نوع الانه أنهيار في وتعتبر هذه فانه يحدث قيمة الاتزان الحمل عن قيمة فاذا زادت يار

الضغط واذا قلت قيمة الحمل يحدث انهيار في الشد.

2.2.2 Design of sections subjected to compression loads accompanied by bending

moments using reaction curves

وبة بعزوم انحناء باستخدام منحنيات التفاعل: تصميم القطاعات المعرضة لاحمال ضغط مصح

To design a concrete section subject to eccentric loads (moment accompanied by axial

compression load), initial concrete dimensions of the section are first estimated using one

of the following methods:

لتصميم قطاع خرساني معرض لاحمال غير متمركزة )عزم انحناه مصحوب بحمل ضغط محوري( يتم اولا فرض

ابعاد خرسانية مبدئية للقطاع باستخدام احدي الطرق الاتية:

- If the section is a beam section or the decentralization value is relatively large, the

depth of the section is imposed using equation used to design section subjected to

bending moments

اذا كان القطاع قطاع كمرة اوقيمة اللامركزية كبيرة نسبيا يتم فرض عمق القطاع باستخدام -

المستخدمة فى تصميم القطاعات المعرضة لعزوم انحناء المعادلة

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24 Dr/Yasmin Hefni

- If the section is a column section or the decentralization value is relatively small,

the depth of the section is estimated using the equation used in the design of

columns subjected to axial compression loads

Pu = 0.35 Ac + 0.67 Asfy

Where:

- Ac = b*t

- As = µ*Ac

- µ=1.2 - 1.5%

After estimating the initial dimensions of the section, the ratio (e/t) of the section is

calculated and the design is completed using one of the following methods

- If the ratio (e/t < 0.05), Mu values is neglected and the section is designed as a

section subject to axial compression loads using the following equation:

Pu = 0.35 Ac fcu + 0.67 Asfy

- If the ratio of Nu/(f_cu bt)≤0.04, the value of Nu is neglected and the section is

designed as a section subject to bending moments only taking into account the

type of section (R-sec., L-sec or T-sec)

ونتبع التالي eccentricityلا يمكن اهمال (e/t > 0.05)في حالة

( وهو مايعني ان نسبة اللامركزية عالية اي ان مركز تاثير القوة خارج e/t>0.5 – big eccاذا كانت نسبة ) -1

القطاع ويكون القطاع معرض لعزوم كبيرة نسبيا تسبب اجهادات ضغط وشد فانه يمكن استخدام منحني التفاعل

(interaction diagram ولكنه )لتفاعل للقطاعات ذات في هذه الحالة حيث لايفضل استخدام منحني ا غير مفضل

αنسبة =As

\

As≤ وهي اقصي نسبة مسموح بها في القطاعات المعرضة لعزوم. 0.04

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25 Dr/Yasmin Hefni

وفي هذه الحالة يتم استخدام طريقة تقريبية تعتمد علي تحويل القوة غير المتمركزة المؤثرة علي القطاع الي قوة ضغط

فيمة وعزم مؤثر علي القطاع ب uNتؤثر علي حديد التسليح جهة الشد

(seu= N usM:حيث )

𝑒𝑠 = 𝑒 +𝑡

2− 𝑐 c = 50 mm if t<100 c = 100 mm if t>1000

𝑑𝑜 = 𝑐1√𝑀𝑢

𝐹𝑐𝑢𝑏→ 𝑐1 = − − . 𝑗 = − −

وهذا النوع من القطاعات غالبا مايكون قطاع كمرة وبالتالي يطبق عليه في هذه الحالة نفس اشتراطات تسليح الكمرات

s( وحديد الضغط )sminAنسبة الحديد الدنيا )من حيث \A سواء كان موجود لدواعي التصميم اولتعليق الكانات وكذلك )

بالنسبة لحديد الانكماش.

Mu s approach

( وهو مايعني ان نسبة اللامركزية صغيرة اي ان مركز تاثير القوة داخل .e/t<0.5- small eccاذا كانت نسبة ) -2

القطاع معرض لعزوم صغيرة نسبيا تسبب اجهادات ضغط في الجهتين او ضغط وشد ضعيف )بعيد القطاع ويكون

( وهو مفضل في هذه الحالة interaction diagramعن مركز تأثير القوة( فانه يمكن استخدام منحني التفاعل )

غير متمركزة )قوة ( وهي اقصي نسبة مسموح بها في القطاعات المعرضة لاحمال α=1.0ويفضل استخدام نسبة )

الآتى: بدلالة التصميم فى لإستخدامها تصلح للتصميم كمساعدات مسبقا معدة تفاعل منحنيات وهاك وعزم(

)yType of steel (f

𝑎 =𝐴𝑠

\

𝐴𝑠 (𝑎𝑙𝑚𝑜𝑠𝑡 𝑡𝑎𝑘𝑒𝑛 𝑒𝑞𝑢𝑎𝑙 1.0) 𝜀 =

𝑡−10

𝑡

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26 Dr/Yasmin Hefni

وبعد تحديد المعلومات السابقة يتم اختيار المنحني التصميمي بدلالة هذه القيم ولحدد القيم 𝑁𝑢

fcubtو

𝑀𝑢

fcubt2ثم نحدد

pقيمة

p =1فتؤخذ 1اقل من pاذا كانت

يتم تحديد الاتي:

𝜇 = 𝑝 ∗ 𝑓𝑐𝑢10−5

𝐴𝑠 = 𝜇𝑏𝑡

𝐴𝑠\

= 𝛼𝐴𝑠

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27 Dr/Yasmin Hefni

2.3 Sections subjected to tensile loads accompanied by bending moments

القطاعات المعرضة لاحمال شد مصحوبة بعزوم إنحناء:

behavior of concrete section subjected to concentrated tensile loads accompanied by

bending moments depends on the value of the load and the value of the moment.

Sections subjected to axial tensile loadsالقطاعات المعرضة لاحمال شد محورية:

بالنسبة للقطاعات المعرضة لاجهادات شد فقط يتم فرض ان الخرسانة قد وصلت لحد التشرخ ولاتتحمل اي اجهادات

المؤثرة. ويتم توزيع الحديد الناتج علي كامل القطاع الشد الحديد يتحمل قوة القطاع علي اساس ان شد ويتم تصميم

لحمايته وبالتالي يمكن فرض القطاع مربعا بنفس عرض الخرساني ويكون القطاع الخرساني عبارة عن غطاء للحديد

العنصر الذي يرتبط به الشداد.

𝐴𝑠 =𝑇𝑢

(𝑓𝑦

ˠ𝑠⁄ )

('5φ8/mوبالنسبة لكانة القطاع يتم وضع كانة خارجية فقط )

القطاعات المعرضة لقوى شد غير متمركزة )شد + عزوم إنحناء(:

( ونسبتها الي سمك eيتوقف اسلوب تصميم القطاع المعرض لقوي شد غير متمركزة علي مقدار اللامركزية الحادثة )

( ويمكن تحديد خطوات التعامل مع تلك النوعية من القطاعات كمايلي: e/tالقطاع )

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28 Dr/Yasmin Hefni

Assume 𝑑𝑜 = 𝑐1√𝑀𝑢

𝑓𝑐𝑢𝑏 𝑤ℎ𝑒𝑟𝑒 𝐶1 = 3.5 → 4

.eu= T uM→ If the section is subjected to eccentric tension

o1) d→Take d = (0.9

حيث ان القطاع معرض لقوة شد واحتياجه لحديد التسليح يكون اكبر من odالمستخدمة عن dويلاحظ هنا امكانية تقليل

احتياجه للخرسانة والتي تهمل عند التصميم في القطاعات المعرضة لشد وفي هذه الحالة يتحمل الحديد قوة الشد المؤثرة

بالكامل.

t = d + 50 mm t ≤ 1000 mm

t = d + 100 mm t > 1000 mm

get e/t

e/t < 0.05لة حا

وفى هذه الحالة يتم اهمال العزم وتصميم القطاع كقطاع معرض لقوة شد محورية.

e/t <0.5 and e/t > 0.05حالة

في هذه الحالة تكون القوة المؤثرة لها لامركزية صغيرة تكون داخل القطاع وبالتالي يكون القطاع معرض لقوة شد في

الجهتين وبالتالي فانه يتم تصميم القطاع بحيث يتحمل صلب التسليح قوة الشد المؤثرة بالكامل.

( مساويان في المجموع لقيمة 2T1,Tالمتمركزة المؤثرة الي قوتين )كما هو موضح بالرسم يتم تحليل قوة الشد الغير

uT( ويتم مقاومة القوتين باستخدام حديد شد في الجهتين ,s2, As1A.)

𝑇1 = 𝑇𝑢 (𝑒𝑠2

𝑒𝑠1 + 𝑒𝑠2

) 𝑇2 = 𝑇𝑢 (𝑒𝑠1

𝑒𝑠1 + 𝑒𝑠2

)

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29 Dr/Yasmin Hefni

𝐴𝑠1 =𝑇1

(𝑓𝑦

𝑦𝑠⁄ )

𝐴𝑠2 =𝑇2

(𝑓𝑦

𝑦𝑠⁄ )

es2=e + 𝑡

2-cover

e/t ≥ 0.5حالة

القطاع وبالتالي فان القطاع الخرساني يكون معرض خارجوفي هذه الحالة تكون القوة المؤثرة لها لامركزية كبيرة تكون

لاجهادات شد في الجهة القريبة من القوة واجهادات ضغط في الجهة البعيدة عن القوة.

كما هو موضح بالرسم يتم نقل قوة الشد الغير المتمركزة المؤثرة الي مركز حديد الشد الذي يتحمل هذه القوة وعزم )

usM .)

es=e - 𝑡

2 +cover

Mus = Tt (es)

𝑑 = 𝑐1√𝑀𝑢

𝐹𝑐𝑢𝑏→ 𝑔𝑒𝑡 𝑐1 = ⋯ . . 𝑗 =……..

𝐴𝑠 = 𝑀𝑢𝑠

𝑓𝑦𝑗. 𝑑+

𝑇𝑢

(𝑓𝑦

ˠ𝑠⁄ )

Check Asmin Get (12 – 20 %) As

2.5 Determining the additional moments caused by the slenderness of the section

(Madd) تعيين العزوم الاضافية الناتجة عن نحافة القطات

In plan buckling

buckling occurs in the column as a result of the compression force P, if this buckling occurs

in the plane of the paper on which we draw the elevation, this meaning that we see that

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buckling, the resulting additional moment Madd is parallel to the apparent width of the

column. In this case the buckling is called in plan buckling and the column will design for

P and Madd

يحدث للعمود انبعاج . الانبعاج يحدث في نفس مستوي الورقة التي نرسم Pنتيجة التاثير علي العمود بحمل ضغط

اننا نري ذلك الانبعاج ويكون العزم الاضافي الناتج موازي للعرض الظاهر للعمود ويسمي elevation عليها اي

P and Maddويصمم العمود علي in plan buckling ذه الحالة الانبعاج في ه

Out of plan buckling

buckling occurs in the column as a result of the compression force P, if this buckling occurs

in the perpendicular plane of the paper on which we draw the elevation, this meaning that

we cannot see that buckling, the resulting additional moment Madd is perpendicular to the

apparent width of the column. In this case the buckling is called out of plan buckling and

the column will design for P and Madd

نري اي اننا لا elevation يحدث الانبعاج في هذه الحالة علي مستوي عمودي علي مستوي الورقة التي نرسم عليها

out ذلك الانبعاج ويكون العزم الاضافي الناتج موازي للعرض الغير الظاهر للعمود ويسمي الانبعاج في هذه الحالة

of plan buckling ويصمم العمود عليP and Madd

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31 Dr/Yasmin Hefni

Calculatiom of the additional moment Madd

Madd= P*δ

Madd the maximum buckling moment

δ the maximum displacemnet due to buckling

δ = λ2∗ (b or t)

2000

δ for in plane buckling = λin2∗ (𝐭)

2000 δ for out of plane buckling =

λout2∗ (𝐛)

2000

for circular colums δ = λ2∗ (𝐃)

2000

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32 Dr/Yasmin Hefni

λ = kHo

b

He = kHo

He Effective length of column

K the ratio between the effective length and the original length of the column

Ho the original length of the column

The effective length of the column is defined as the distance between the two inflection

points of the column when buckled, and this length is what is attributed to the width of the

column to determine whether this column is short or long. Also, according to whether this

column is braced.

للعمود عند انبعاجه وهذا الطول هو الذي ينسب الى نقطتي الانقلاببانه المسافة بين eHويعرف الطول الفعال لعمود

من اقلمن او اكبر عرض العمود لتحديد اذا ما كان هذا العمود قصيرا ام طويلا والطول الفعال للعمود ربما يكون

الطول الفعلي للعمود وذلك حسب طريقة تثبيت العمود من طرفيه وكذلك حسب اذا ما كان هذا العمود مقيدا جانبيا

Determination the value of K

There are three types of column-fixation cases: fixed, party fixed, hinged, and free as

illustrated in the following explanation:

( مثبتة للعمود الطرفي للتثبيت حالات ثلاث )Fixedويوجد جزئيا ومثبتة )party fixed( ومفصلية )hinged )

( وفيا يلى شرح حالات تثبيت العمود: Freeحرة )بالاضافة الي حالة عدم التثبيت

: Fixedمثبتة -1

It is the case in which the column is installed with the beam or slab and casted with it at

the same time. It is required that the thickness of the slab or the depth of the beam is not

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less than the width of the column. In this case, the column is completely fixed so that it

cannot rotate.

مرة أو البلاطة ومصبوبا معها في نفس الوقت ويشترط أن يكون وهي الحالة التي يكون فيها العمود مثبت مع الك

سمك البلاطة أو عمق الكمرة لا يقل عن عرض العمود وفى هذه الحالة يتم تثبيت العمود تماما بحيث لا يستطيع

الدوران .

: Party fixedمثبتة جزئيا -2

This is the case in which the beam or slab is not able to fix the column completely, but

can rotate partially. In this case, the depth of the beam is less than the width of the

column.

وهى الحالة التي لا تتمكن فيها الكمرة أو البلاطة من تثبيت العمود تماماً بل يمكنه الدوران جزئيا وفى هذه الحالة

تكون عمق الكمرة أقل من عرض العمود.

: Hingedتثبيت مفصلى -3

In this case, the column can rotate around this end, and there is not the slightest resistance

to this rotation, because the stiffness of the member to which the column is attached is

weak.

وفى هذه الحالة يمكن للعمود الدوران حول هذا الطرف ولا توجد أدنى مقاومة لهذا الدوران وذلك لأن جساءة

العضو المرتبط به العمود تكون ضعيفة.

: Freeتثبت حر -4

In this case, the end of the column is free and not attached to any structural member, in

the case of unbraced columns.

وفى هذه الحالة يكون طرف العمود حر وغير مثبت بأي عضو إنشائي وهذه الحالة لا تحدث إلا في حالة الأعمدة غير

تثبيت أحد طرفي العمود كا يلي: . ويرمز لحالات unbracedالمقيدة جانبي

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The K values can be summed up as followsكما يلي: Kويمكن تلخيص قيم

Upper

End

Condition

Braced Column Unbraced Column

Lower End Condition Lower End Condition

a b c a b c

a 0.75 0.80 0.90 1.20 1.30 1.60

b 0.80 0.85 0.95 1.30 1.50 1.80

c 0.90 0.95 1.0 1.60 1.80 -

d - - - 2.20 - -

The cases mentioned in the table are as follows:

a. The end of the column is fixed (fixed) to the base or supported by a beam having a depth

not less than the width of the column in the same plane. In this case, the column is designed

to withstand a bending moment.

b. Partly fixed end of a column to be supported by beams of less depth than the width of

the column in the same plane.

c. The end of a column connected by a hinged connection, where it is supported by a

concrete member that allows it to rotate and not allow it to move.

d. Free-tip column movement and rotation, as in the case of a free-end cable column.

a. ( طرف عمود مثبتFixedالع لها عمق لا يقل عن عرض القاعدة أو مدعم بكمرة في مود في نفس (

المستوي. وفى هذه الحالة يكون العمود مصمما لمقاومة عزم انحناء.

b. ( طرف عمود مثبت جزئياPartly fixed حيت يكون مدعم بكمرات ذات عمق أقل من عرض العمود )

في نفس المستوي.

c. ( طرف عمود متصل أتصالا مفصلياHinged حيث يكون مدعما بعضو خرسانى يسمح له بالدوران )

يسمح له بالحركة. ولا

d. ( طرف عمود حرfree .الحركة والدوران مثل حالة عمود كابوبي ذو طرف حر )

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2.5 Biaxially Loaded Columns- Biaxial Interaction Diagram Method

a. Using numerical integration, the force and moment in the concrete zone are

determined by integrating the nonlinear idealized stress-strain curve for

concrete in compression. 3-D failure surfaces are drawn for the column. The

biaxial interaction diagrams (curves) are produced by cutting a horizontal

plan Rb (load contour) through the failure surface as shown in the figure.

b. The use of these interaction curves is as follows:

- Determine the curve using the calculated of {fy, Rb, (My /fcu tb2), (Mx / fcu

bt2), ξ}

- From the design ID, get µ, r and calculate As total = µ*b*t

- As total (> Asmin = 0.006 bt) is distributed equally on the four sides.

µ=p*Fcu*10-4

As total = µ*b*t

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Example 2.1

Design the shown rectangular

section if fcu= 25 MPa & fy= 360

MPa, Pu = 2210 kN,

Mxu = 410 kN.m., My = 210 kN.m.

Rb=𝑃

𝐹𝑐𝑢𝑏𝑡=

2210∗1000

25∗400∗600=0.368

At Rb=0.3

𝑀𝑥

𝐹𝑐𝑢𝑏𝑡2=

410∗1000∗1000

25∗400∗600∗600=0.11

𝑀𝑦

𝐹𝑐𝑢𝑏2𝑡=

210∗1000∗1000

25∗400∗400∗600=0.087

p=11.8

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At Rb=0.4

𝑀𝑥

𝐹𝑐𝑢𝑏𝑡2=

410∗1000∗1000

25∗400∗600∗600=0.11

𝑀𝑦

𝐹𝑐𝑢𝑏2𝑡=

210∗1000∗1000

25∗400∗400∗600=0.087

p=15

As, At Rb=0.3 p=11.8

At Rb=0.4 p=15

take p= 13.4

µ=p*Fcu*10-4 =13.4*25*10-

4=0.0335

As total = µ*b*t=0.0335*400*600=8040mm2

Asmin = (0.6/100)*600*400=1440mm2

As=Astotal=8040 mm2 use 20D25

2.6 Biaxially Loaded Columns- ECP Code Method

1- ECP offers a simplified method for the case of a R-section with symmetrical

reinforcement by approximating the curved shape of biaxial I.D. by two

straight lines as shown in the figure. The biaxial state is transferred to a case

of uniaxial eccentric compression.

المصري فريو مبسطة الكود بتسليح R مقطعلل طريقة المنحني المسلح الشكل تقريب طريق عن متماثل

biaxial I.D الحالة نقل يتم الشكل. في موضح هو كما مستقيمين بخطين . biaxial حالة ضغط إلى

uniaxial

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2- If the column section under Pu, Mux & Muy not given, assume the

section as follows: ey = Mux * 106 / (Pu * 103) ex = Muy * 106 / (Pu *

103)

If ex = ey take x = y square sec.

If ex > ey take x > y R-sec.

If ey > ex take y > x R-sec.

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Case ( 2 ) Case ( 1 )

3- In case of rectangular or square columns sections with equal steel on its faces

(uniform steel arrangement), then design moments can be reduced into an

equivalent uniaxial moment as given below. (Consider a` = y – 50 mm, b` = x –

50 mm as shown in the figure)

اختزال ( ، فيمكن تسليح منتظم وجه متساوية )في الأ تسليح في حالة وجود مقاطع أعمدة مستطيلة أو مربعة ذات

a` = y - 50 mm ،b` = x - 50 mm اعتبارمع مكافئ uniaxial moment اختزالالي العزم التصميمي

4- Case (1): If Mux / a` > Muy / b`

then design the section for Pu & M`ux = (Mux + β (a`/b`) Muy) using I.D. of

uniaxial bending

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5- Case (2): If Mux / a` ≤ Muy / b`

then design the section for Pu & M`uy = (Muy + β (b`/a`) Mux) using I.D. of

uniaxial bending.

Table for values of the coefficient β :-

Pu*103/(fcu bt) ≤ 0.2 0.3 0.4 0.5 ≥ 0.6

β 0.80 0.75 0.70 0.65 0.60

6- The calculated steel area from I.D. As total > As min (= 0.006 bt)

Example 2.2

Given: fcu = 25 MPa , fy = 360 MPa , d` = 50 mm

PD.L. = 850 kN PL.L. = 1630 kN

MX D.L. = 350 kN.m MX L.L = 544 kN.m

MY D.L. = 180 kN.m MY L.L = 318 kN.m

Required: Design the column section Biaxial Design Example (Case 1)

Pu = 1.4 PD.L + 1.6 PL.L = 3800 kN where PL.L > 0.75 PD.L

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Mux = 1.4 MX D.L + 1.6 MX L.L = 1360 kN.m where MX L.L > 0.75 MX D.L

Muy = 1.4 MY D.L + 1.6 MY L.L = 760 kN.m where MY L.L > 0.75 MY D.L

ey = Mux * 106 / (Pu * 103) = 357.9 mm

ex = Muy * 106 / (Pu * 103) = 200.0 mm

Assume square or rectangular section

Assume x = 800 mm ,

y = 1000 mm (ey > ex)

a` = t – d` = 1000 – 50 = 950 mm, b` = b – d` = 800 – 50 = 750 mm

Mux / a` = 1.432, Muy / b` = 1.013

Mux / a` > Muy / b`

then Mùx = Mux + β (a`/b`) Muy Pu*103/(fcu bt) = 0.19 ,

then β = 0.8 from the previous table

Mùx = 1360 + 0.8 * (950 / 750) * 760 = 2130.13 kN.m

Then design a column as shown in the Figure

e = Mùx * 106 / (Pu * 103) = 560.6 mm

e / t = 0.561 (1)

K = Pu * 103 / (fcu b t) = 3800 * 103 / (25 * 800 * 1000) = 0.19 (2)

For fy = 360 MPa (uniform steel arrangement)

and = (t – 2d’) / t = (1000 – 2*50) / 1000 = 0.9 then use chart No.

From the chart get = 5.2

µ = fcu * 10-4 = 0.013 = 1.3 %

As = µ b t = 10400 mm2 (use 24 25) Distribute the

steel uniformly over the section

If ex = ey take x =

y

square

sec.

If ex >

ey

take x >

y

R-sec.

If ey >

ex

take y >

x

R-sec.

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2.7 Problems Problem 1

Design a reinforced concrete column knowing that it is subjected to the following

straining actions

Pu = 3000 kN. , Mu= 300 kN.m. Breadth of section = 250 mm.

The material properties are fcu, = 25 N/mm2 360 N/mm2

Problem 2

Design a reinforced concrete section if it is subjected to an eccentric compression

force equal Pu = 150 kN., with eccentricity e= 3.33m. given that: Breadth of

section = 300 mm.

The material properties are fcu=25N/mm2 , 360 N/mm2

Problem 3

Design a reinforced concrete column using interaction diagrarms knowing that it is

subjected to the following straining actions:

Pu = 1200 kN. , Mu = 200 kN.m Breadth of section = 300 mm.

The material properties are fcu=25N/mm2, fy =360N/mm2

Problem 4

design Design the column section using ECP Code Method (Case 1)

P D.L. = 1050 kN P L.L. = 2000 kN

MX D.L. = 390 kN.m MX L.L = 654 kN.m

MY D.L. = 200 kN.m MY L.L = 300 kN.m

The material properties are : fcu = 25 MPa , fy = 360 MPa

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Problem 5

Design the shown rectangular

section if fcu= 35 MPa & fy=

400MPa,

Pu = 2000 kN,

Mxu = 350m.kN, My = 240 m.kN

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Chapter 3

Design of Frames

3.1 Definition of the Frame

A concrete frame is a common form of structure comprising a network of columns and

connecting beams that forms the structural 'skeleton' of a building, the internal forces due

to the loads are resisted by the combined action of the girder and the columns i.e. the

bending moment is distributed among the girder and the columns

If we compare the value of bending moment for beam – column 'skeleton' or frame

'skeleton', a maximum moment will be appeared in frame 'skeleton.

Loads

Deformed Shape

Loads

Deformed Shape

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B.M.D

Simple Girder and column (Hinged roller)

B.M.D

Frame (2 Hinges)

3.2 Types of Frames

The choice of the frame depends to a large extent on the purpose of its use and the loads

on it, as well as the type of soil. Several types of Frames are used in constructions,

increasing the indeterminacy for the frame system reduces bending moment occurs due to

the applied loads.

Therefore, in the case of large spans or weak soils, we resort to use statically independent

frames to reduce the moment resulting from loads. The following are some types of frames

and their sustainability according to; spans and type of soil.

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3.3 Dimensions of frame elements

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Width of frame the greater

• Spacing /20

• 300mm

3.4 Detailing of frames

The following shows some examples of reinforcing details of joint of frame. (Alternative

reinforcing is accepted).

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This frame joint does not resist moment (a joint between a link member and a frame

girder)

Connection between corbel and link member. a frame joint that does resist moment.

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rigid connection between the girder and the column “Closing joint”. The moments of the

girder are resisted by the tension column reinforcements

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rigid connection between the girder and the column “opening joint”. The moments of the

girder will not resist by the column reinforcements to avoid failure of the cover

Corbel and rigid connection between the girder and the column “Closing joint”. The

moments of the girder are resisted by the tension column reinforcements

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a broken part of a girder of a frame creating an “opening joint”. The bottom reinforcement

of the frame girder has to be discontinued at the broken part in order to prevent cover

spalling

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Typical reinforcement for two hinged frame

B.M.D

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3.5.1 Example 3.1

Fig () shows different frames under the given loads. It is required (without any

calculations) to sketch N.F.D, B.M.D. and the corresponding location of main tension

steel.

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3.5.2 Example 3.2

Fig( ) shows different frames under the given loads. It is required (without any

calculations) to sketch the deformed shape, N.F.D, B.M.D., S.F.D and the corresponding

location of main tension steel.

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Loads

Deformed Shape

N.F.D

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B.M.D

S.F.D

Main Reinforcement Location

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Loads

Deformed Shape

N.F.D

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B.M.D

S.F.D


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Loads

Deformed Shape

N.F.D

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B.M.D

S.F.D


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3.5.3 Example 3.3

Calculate loads on the shown frame, spacing between frames = 6m, width of frame =400

mm, slab thickness= 120 mm, Live load (H.P) = 2.5 KN/m2, Floor cover= 2.5 KN/m2

Figure (a)

Figure (b)

Solution

loads on the intermediate frame:

➢ Slab at level +8.000

Spacing between frames = 6.00 m, Ls =4.0 m (short length) < 5.0 m….

Then use solid slab System.

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Slab L

(m) Ls

(m) m ms r=(m L)/(ms Ls) Type

S1 1.0 Cantilever S.S

S2 6.0 4.0 0.87 0.76 1.72 < 2 Two Way S.S

Loads on Slabs (Wus)

D.L= O.W of Slab (ts ɣc) + F.C= 0.12*25+2.5=5.5 kN/m2

. L.L=2. 5 kN/m2.

L.L/D.L=2.5/5.5 =0.45<0.75

then Wus=1.5*(D.L+LL) =1.5*(5.5+2. 5) =12 kN/m2

Loads on beams

Assume b =250 mm and t =600 mm for B1 & B2.

O.W of Beam= b (t-ts) ɣc = 0.25*(0.6-0.12)*25= 3.0 kN/m’;

(O.W) ult. = 1.5*(3.0) = 4.5 kN/m’

Loads on B1

Wus= 1.5* (O.W+W.L) + ∑βWu, slab X

β for triangle = 0.5

Wus= 4.5 + (0.5*12*2*2) = 28.5 kN/m’

RB1 = 0.5Wus L = 0.5*28.5*4 = 57 kN.

Loads on B2


β for trapezoidal = 1-(1/2r)=1-(1/2*1.5)=0.667

Wus= 4.5 + (12*1) + (0.667*12*2.0) = 32.5 kN/m’

RB2 = 1.1 Wus L = 1.1*32.5*6 = 214.5 kN.

Design of post

Assume post dimensions = 250X250 mm

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Rpost = (RB1 + RB2) = 214.5+57=271.5 kN

Pu= 1.1 * Rpost = 1.1*271.5=298.7 kN

298.7*103 = 0.35*35*(250*250-Asc) + 0.67*400*Asc

Asc= -ve value

Use µmin=0.8%

As = (0.8/100)*250*250 = 500mm2 use 4 ∅16

➢ Slab at level +5.000

Reactions of Beams (B3 & B4):

Concrete Dimensions:

Assume b =250 mm and t =600 mm for B3 & B4.

Loads on Beams:

O.W of Beam= b (t-ts) ɣc = 0.25*(0.6-0.12)*25= 3.0 kN/m’;

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(O.W) ult. = 1.5*(3.0) = 4.5 kN/m’

Calculate Straining Action for B3:


Wus= 4.5 + (0.667*12*2) = 20.5 kN/m’

RB3 = 1.1 Wus L = 1.1*20.5*6 = 135.4 kN.

Calculate Straining Action for B4:


Wus= 4.5 + (0.667*12*2*2) = 36.52 kN/m’

RB4 = 1.1 Wus L = 1.1*36.52*6 = 241 kN.

The intermediate frame

Dimensions

For Girder and Column (b=400 mm);

tc1=tg = L/ (12:16) = 20000/ (12:16) =1500 mm.

Then, (b*t) = (400x1500) mm for Girder and Column. tc2 = 0.5 tc1 =750 mm

Loads on Frame

O.W of Frame Girder = b (t-ts) ɣc = 0.4*(1.5-0.12)*25= 13.8 kN/m’;

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Wu= 1.5*(13.8) + (0.5*2*2*12) = 44.7 kN/m

3.5.4 Example 3.4

It is required to design the shown frame, spacing between frames = 6m, width of frame

=400 mm, slab thickness= 120 mm, Fcu= 35.0 MPa The main steel is high tensile steel of

grade 40/60 , The stirrups steel is mild steel of grade 24/35.

Solution

Reactions

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BMD

NFD

SFD

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Design all the critical sections of the frame

Section Mu

(kN.m)

Nu (kN)

Qu (kN)

1 2856.8 -722.9 0.0

2 -3614 -722.9 1097.3

3 -3614 -1232.6 -722.9

4 0.0 -1232.6 -722.9

Section 1

Mu = 3614 kN.m

Nu = 722.9 kN (Comp.)

𝑁𝑢

fcubt=

722.9 ∗ 1000

35 ∗ 400 ∗ 1500= 0.030 ≤ 0.04

Neglect Nu Design the section for Mu only

The section is T sec

B the smaller value of

• 16 ts +b=16*120+400=2320 mm

• CL to CL = 6000 mm

• L/5+b=20000/5+400 =4400mm

Tabe B= 2320 mm

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d = c1 √ 𝑀𝑢

Fcu∗b

1450 = c1 √ 2856.8∗106

35∗2320 → → C1=7.7 J=0.826

As = 2856.8∗106

400∗0.826∗1450 = 5963.09 mm2

µ =As/Ac=5963.09/(400*1450)=0.010

µmax=4.31*10-4*35= 0.015 > µ OK

As = 5963.09 =n*(3.14*252/4 ) use 13D25mm2

As’= (0.1:0.2) As = 940 mm2.

use 4D18mm2

Section 2

Mu (-ve) = 3614.0 kN.m;

Nu = 722.9 kN (Comp.)

𝑁𝑢

fcubt=

722.9∗1000

35∗400∗1500= 0.030 <0.04 Then neglect Nu and design for Mu only.

𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 (R-sec.)

d = c1 √ 𝑀𝑢

Fcu∗b

1450 = c1 √ 3614.0 ∗106

35∗400 → →c1=2.85 J=0.726

As = 3614.0 ∗106

400∗0.726∗1450 = 8582.65 mm2

µ =As/Ac=7543.6 /(400*1450)=0.013

µmax=4.31*10-4*35= 0.015 > µ OK

As = 8582.65 =n*(3.14*252/4 ) use 18D25mm2

Design for Shear section 2:

Critical shear stress:

qu = Qu*103 / ( b . d ) = 1097.3*103 / (400 * 1450) = 1.892 N/mm2

Shear stress limits:

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qcu-uncracked = 0.16 √ (fcu / γc) = 0.77 N/mm2

qcu-cracked = 0.12 √ (fcu / γc) = 0.58 N/mm2

qu max = 0.7 √ (fcu / γc ) = 3.38 N/mm2 < 4 N/mm2

.77 > 1.889 < 3.38

qcu-uncracked < qu < qu max (Design for stirrups)

S = n. AS * (fyst / γs) / [(qu - qcu-cracked) * b]

Take Φ = 8 mm AS = 50.5 mm2 b = 400 mm take n = 4 γs = 1.15

Fyst=240 N/mm2 S=123 mm.

S=80mm use Φ = 10mm AS = 785 mm2

S=124 mm Then Use 9∅ 10/m’-4 Branches

Section 3

Mu = 3614 kN.m Nu = 1232.6 kN (comp).

Check for buckling:

In Plane (t-direction):

t’ = (5/6) t=(5/6)*1500=1250 mm.

Lower case: Hinged (Case (3));

Upper case: Fixed (Case (1)).

From Table , K=1.6.

He = K Ho = 1.6*(5-0.75) = 6.8 m.

λt = (He/t’) = (6.8 /1.25) = 5.44 < 10. Short column in (t-direction)

Out of Plane (b-direction):

Ho=(5000-300 (half depth of upper beam)-600 (depth of wall beam)/2 ) =2050mm

Lower case: Hinged (Case (3)); Upper case: Fixed (Case (1)).

From Table (6.10), K=1.6.

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He = K Ho = 1.6*2.05= 3.28 m.

λb = (He/b) = (3.28/0.4) = 8.2

Short column

𝑁𝑢

fcubt=

1232.6 ∗ 1000

35 ∗ 400 ∗ 1500= 0.058 > 0.04

Design the section for Nu and Mu

The section is R sec

Mu = 3614 kN.m

Nu = 1232.6 kN (comp).

e = Mu/Nu=3614/1232.6=2.9m

e/t = 2.9/1.5>1.9 use Mus approach

es= e-t/2+cover

es= 2.9 -0.75 +0.050=2.2 m

Mus = Nu* es=1232.62.2=2711.7kN.m

d = c1 √

𝑀𝑢

Fcu∗b

1450 = c1 √

2711.7 ∗106

35∗400 → → c1=3.29 J=0.768

As = 2711.7 ∗106

400∗0.768∗1450 +

1232.6∗103

400/1.15= 6087 -+2679.6 = 8766.5mm2 Then Use 18∅25

As min =µmin b d = 0.225(√𝐹𝑐𝑢)𝑏𝑑/𝐹𝑦=1930.1 mm2

Design for section (4-4):

Pu= 0.35 fcu (Ac-Asc) + 0.67 fy*Asc

1232.6*103=0.35*35*(400*750-Asc ) +0.67*400*Asc

Asc = -ve value;

Ascmin = (0.8/100)*400*750=2400mm2

Then Use 7∅25

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3.5.5 Example 3.5

For the following frame it is required to

1- Calculate (loads including own weight of the frame)

2- Design the frame and draw the reinforcement

Data

- Spacing between frames = 5.00 m, slab thickness = 12 cm

- width of frame = 400mm

- Fcu = 25 N/mm2, Fy = 400 N/mm2, Fyst = 240 N/mm2

Solution

1. The straining action on the frame

Reaction (kN)

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B.M.D (kN.m)

N.F.D (kN)

S.F.D (kN)

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2. Design the frame and draw the reinforcement

Straining actions on frame sections

Section Mu( kN.m) Nu (kN) Qu(kN)

1 790.18 0 0

2 1833.75 0 611.25

3 1833.75 -611.25 0

4 1833.75 489.0 -366.75

5 0 489.0 -366.75

Concrete Dimensions

t g = (L/12-14)=(1375 to 1190) take t frame = 1300cm

Design of frame sections

Section 1

Mu= 790.18kN.m Nu=0

The section is T sec

d = 1300-100=1200 mm

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B the smaller value of

• 16 ts +b=16*120+400=2320 mm

• CL to CL = 5000 mm

• L/5+b=16500/5+400 =3700mm

Tabe B= 2320 mm

d = c1 √ 𝑀𝑢

Fcu∗b

1200 = c1 √ 790.18∗106

25∗2320 → →C1=10.3 J=0.826

As = 790.18∗106

400∗0.826∗1200 = 1992 mm2

µ =As/Ac=1910 /(400*1200)=0.010

µmax=4.31*10-4*25= 0.0107 > µ OK

As = 1990 =n*(3.14*252/4 ) use 5D25mm2

As’= (0.1:0.2) As = 300 mm2. Use 3 D 16

Section 2

Mu= 1833.75kN.m Nu=0 kN

The section is R section

d = c1 √ 𝑀𝑢

Fcu∗b

1200 = c1 √ 1833.75 ∗106

25∗400 → → c1=2.8 J=0.72

As = 1833.75∗106

400∗0.720∗1200 = 5305 mm2 use 11D25mm2

Section 3

Mu= 1833.75kN.m Nu=-611.25 kN

𝑁𝑢

fcubt=

611.25∗1000

25∗400∗1300= 0.047 > 0.04

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The section will design for Mu and Nu

Calculation of additional moment due to buckling

He=K*Ho

Ho= H-tg/2=7-1.3/2=6.350m

α1 (hinge)=10 α2=𝐼𝑐/𝐻𝑜

2𝐼𝑐/16.5=1.30

K value the least of

1+0.15 (α1+ α2)=1+0.15 (10 +1.3 )=2.695

2+ 0.3 * αmin=2+0.3*1.3=2.39

K=2.39

He=K*Ho=2.39*6.35=15.17m

Λ=He/t=15.17/1.3=11.6 >10 long column

δ = λ2∗ ( t)

2000

δ= 11.6∗11.6∗ 1300

2000=0.087m

Madd= Nu* δ=611.25*0.087=53.46 kN.m

Mtotal = Mu+ Madd=1833.75+53.46=1886.9 kN.m

e = Mu

Nu e =

1886.9

611.25 =3.08 >

𝑡

2- (

1.3

2-=0.65m) big eccentricity

es=e - 𝑡

2 - cover =3.08-0.65+0.1=2.53m

Mus=Nu * es =611.25*2.53 = 1546.46 kN.m

d = c1 √ 𝑀𝑢

Fcu∗b

1200 = c1 √ 1546.46 ∗106

25∗400 → → C1=3.05 J=0.747

As = 1546.46 ∗106

400∗0.747∗1200 = 4131mm2 use 9D25mm2

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section 4

use 9D25mm2

section 5

Design as short column

Pu = 0.35 fcuAc+0.67As Fy Pu=489.0kN

489*1000=0.35*25*(400*650)+0.67*400*As

As= -ve

Use min reinforcement

As min=( 0.8/100)*b*d =0.8*400*650/100=2080mm2 Use 5 D25

Design for Shear

Maximum shear occurs at section 2:

Qu = 611.25 kN

Critical shear stress:

qu = Qu*103 / ( b . d ) = 611.25*103 / (400 * 1200) = 1.27 N/mm2;

Shear stress limits:

qcu-uncracked = 0.16 √ (fcu / γc) = 0.65 N/mm2

qcu-cracked = 0.12 √ (fcu / γc) = 0.49 N/mm2

qu max = 0.7 √ (fcu / γc ) = 2.86 N/mm2< 4 N/mm2

S = n. AS * (fy / γs) / [(qu - qcu-cracked) * b]

Φ = 8 mm AS = 50.5 mm2 b = 400 mm take n = 4 γs = 1.15

fy =240 MPa S=135 mm.

Use 8∅ 8/m’-4 Branches

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3.6 Problems

Problem 1

Figure 1 shows typical frame section.

a) Choose a structural system for the roof slabs on the frames then make complete

design for the slabs.

b) Draw to reasonable scale the reinforcement roof slabs.

c) Make a complete structural analysis of the frame and draw the N.F.D, S.F.D and

B.M.D.

d) Design all frame elements according.

e) Draw to a reasonable scale the concrete dimensions and complete reinforcement

detail of the frame using scale (1:25) as well as the necessary cross section detail

(1:10)

General Data

- Live load (h.p.) = 1.5 kN/m2

- Floor cover = 3.50 KN/m2

- Frame elements width = 5.00 mm

- Spacing between frames = 7.0 m

- Characteristic strength of concrete used = 35

MPa

- The main steel is high tensile steel of grade

400/600

- The stirrups steel is mild steel of grade 240/350

Figure 1

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Problem 2

The reinforced concrete hall whose cross section is shown in Figure 2 consists of

system of solid slabs and secondary beams supports on the frame

It is required to

a. Make a complete structural analysis of the frame and draw the N.F.D, S.F.D

and B.M.D.

b. Design all frame elements according.

c. Draw to a reasonable scale the concrete dimensions and complete

reinforcement detail of the frame using scale (1:25) as well as the necessary

cross section detail (1:10)

General Data

- Characteristic strength of concrete used = 30 MPa

- The main steel is high tensile steel of grade 400/600


Figure 2

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Problem 3

Figure (3) shows statically determinate frame ABCDEFGHIJ of a series frame

spaced 5 m. The statical system, ultimate loads, and concrete dimensions of an

intermediate frame is shown in the figure. The frame is hinged at A and B and two

intermediate hinges at E and F acting as a link member IJ. The vertical component

reactions at A and B are equal (YA = YB). It is required to make a complete ultimate

strength design one of the intermediate frames having breadth 400mm and the slab

thickness 120mm. Neglect the effect of buckling in out of plane of the frame. For

the given factorized (ultimate) loads, it is required to

a) Make a complete structural analysis of the frame and draw the N.F.D, S.F.D and

B.M.D.

b) Design the critical sections and check shear stresses of the frame.

c) Draw to a convenient scale the intermediate frame shown clearly all concrete

dimensions and reinforcement details in elevation and cross sections.

General Data

- Characteristic strength of concrete = 30 MPa

- The main steel is high tensile steel of grade 400/600


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Figure 3

Problem 4

Figure (4) shows a section in an intermediate frame of a series of frames used

in a stadium. it is required to

a. Choose an appropriate structural system for the RC roof. Then, design

and give detailed reinforcement drawings of an intermediate panel of

the roof.

b. Draw the bending moment, shearing force and normal force diagrams

for the frame.

c. Design all critical sections of the frame.

d. Draw to convenient scale the frame showing clearly all concrete

dimensions and steel reinforcement in elevation and cross sections.

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General Data

- The frames are spaced at 5.0 m.

- Live load on the stadium floor is 6.0 kN/m2 , and on the upper shed is 1.0

kN/m2 .

- Load of the flooring is 2.5 kNm2 .

- Characteristic strength of concrete = 35.0 Mpa

- the main steel is high tensile steel of grade 40/60

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Fig

ure

4

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Chapter 4

Design of Arched Slabs

4.1 Introduction

It should be mentioned that in a simple beam, significant parts of the concrete sections are

not participating in resisting the applied straining actions and constitute additional weight,

as shown in the figure below the hatched part only is participating in resisting the applied

straining the applied straining actions

Loaded beam

If we remove unnecessary concrete from a simple beam, an arch with a tie would result

as shown

Arch slabs

Arched slabs are Type of slab which used to cover large span ranged between (12:25) m.

Main straining actions act on the arch slab are compression force and very small bending

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moment. As concrete can sustain compression force then it is allowing an economical

structural system.

Why arched slabs have very small bending moment?

The arched slab shape is the inverse of the bending moment

Advantages

• Permits covering large spans without colums.

• Low construction period.

• Economical use of construction materials as concrete subjected to compression

only.

4.2 Structural System for arch slab

The Structural System of the arched slab is usually 3-Hinged arch with tie. the thickness of

the slab is reduced at the crown of the arch slab to provide hinge at the crown of the arch.

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On the other side the slab thickness is increased at support to resist the applied

compression force. Tie is provided to resist the thrust force due to the arch action.

DIFFERENT The structural systems of arched slabs

4.3 Component of Arch Slab:

a) Curved Slab:

▪ The span length of the slab (L) is ranged between (12: 25) m. ▪ One way slab with thickness ranged between (80:150) mm. ▪ The rise of the arch slab (f) is about (L/4) to (L/8).

As the reaction of the arched slab is inclined. Accordingly, horizontal and vertical

beams are used to support the slab.

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Vertical and Horizontal beam

▪ Horizontal and vertical beams are used to support the inclined reaction the

arched slab.

▪ The vertical beams is a continuous beams supported by columns spaced with (S).

▪ The horizontal beams are a continuous beam supported with horizontal tie

spaced with spacing (S).

Horizontal Ties

➢ The horizontal ties are used to resist the tension force occurred from the

reaction of the horizontal beams.

To avoid the deflection of the ties, hangers spaced with (2.5 to 3.0) m are used

Hangers

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Hangers spaced with (2.5 to 3.0) m are used to support the horizontal Ties

4.4 Structural analysis of Arch slab

Two types of slabs are used in construction

❑ Parabolic Arches

❑ Circular Arches

Parabolic Arches are more efficient because the arch centerline coincides with the line of

pressure resulting in zero bending moment

The equation of the parapolice arch is 𝑦 =4𝑓𝑥

𝐿2(𝐿 − 𝑥)

Where:

f: the rise of the arch

L: the span of the arch

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Bending Moment in Parabolic Slab

4.5 Dimensions of arched slabs

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4.6 Loads and reaction calculations:

W UDL = 1.4(ˠ* ts+F.C)*1.1 load on horizontal projection

W ULL = 1.6(p)

Wu (Total) = W DL + W LL

Case (1) full load

V ma x= 𝑊𝑢 ∗ 𝐿

2

H max = 𝑊𝑢 ∗ 𝐿2

8ℎ

Case (2) half load

Ya = 𝑊𝑈𝐷𝐿 ∗ 𝐿

2 +

3𝑊𝑈𝐿𝐿 ∗ 𝐿

8

Xa = Xb = 𝑊𝑈𝐷𝐿 ∗ 𝐿2

8ℎ +

3𝑊𝑈𝐿𝐿 ∗ 𝐿2

16ℎ

At X =𝐿

4

Y = 8ℎ𝑥

𝐿2 ( L- X)

Yd = Ya -W*X

Xd = Xa

Ɵ = tan -1 ( 4ℎ

𝐿2 ( L - 2X))

Nd = (Xd cos Ɵ + Yd sin Ɵ) compression

M max = =𝑊𝑈𝐿𝐿 ∗ 𝐿2

64

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Example 4.1

For the following arched slab

Design the arched slab and draw the reinforcement

Design the main supporting element and draw the reinforcement

Fcu=25 N/mm2 Fy=360 N/mm2

FC= 1 kN/m2 LL= 3 kN/m2

Heigh of column is 6 m & Spacing between column 5 m

Solution

Concrete Dimensions

- ts =100 mm (crown), ts=120 mm (quarter), ts =140 mm (support )

- Horizontal beam (250*1000) mm.

- Vertical beam (300*700) mm.

- Column beam (300*1000) mm2.

- Hanger (150*400) mm2.

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Loads

WUDL = 1.4 (ˠ* ts + F.C) * 1.1 load on horizontal projection

= 1.4 * (0.12*25+1) +1.0 ) * 1.1= 6.2 kN/m2

W ULL = 1.6(p)

= 1.6 * 3.0 = 4.8 kN/m2


= 6.2+4.8=11.0 kN/m2

Case (1) full load

V ma x= = =132 kN

H max = = =264 kN

Case (2) half load

Ʃ M @ b=0

6.2 * 24 *12 + 4.8 * 12 * 6 =Ya * 24

Ya = 88.8 kN

Ʃ Y =0 Y b=11.76 kN

Ʃ M @ C (crown)=0

Xa*(3) = 6.2*12*6 + 88.8*12=0

Xa =206.4 kN

Xa = Xb = +

At X= = =6m

Y= (L-X) = (24-4)=2.25m

Yd=Ya-W*X=117.6-11*4=51.6 kN

Xd=Xa=206.4

Ɵ = tan -1 ( ( L - 2X))= tan -1 ( ( 24 – 2*6))=14.04

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Nd = (Xd cos Ɵ + Yd sin Ɵ) =206.4 cos 14.04+51.6 sin 14.04=212.8 kN (compression)

=212.8 kN

Mmax=117.6*6-206.4*2.25-11*6*3=43.2 kN.m.

Or

M max =𝑊𝑈𝐿𝐿 ∗ 𝐿2

64= =

4.8 ∗ 242

64=43.2 kN.m

Design of Arch slab (sec d-d)

M max =43.2 kN.m Nd=212.8 kN ts=120mm

𝑃𝑢

𝐹𝑐𝑢∗𝑏∗𝑡=

212.8 ∗ 103

25∗1000∗120=0.07>0.04 design for Nu, Pu

Emin = 0.05 ts=0.05*12=0.006m

Mmin=N*emin=212.8*0.006=1.2768kN < Mmax consider Mmax

e=Mmax/Pu=43.2/212.8=0.20m >emin big ecc

es= e +𝑡

2 – cover=0.2+0.06-0.15=0.245

Mus=Pu*es=212.8* 0.245=51.1 kN.m

As = 𝑀𝑢

𝐹𝑦∗𝐽∗𝑑-

𝑃𝑢

𝐹𝑦/ˠ𝑠=

52.1∗106

360∗0.781∗105-

212.8∗1000

360/1.15=1084mm2

Use 6D16/m

Design of horizontal beam (25*1000) mm2

Wbeam = Hmax=264.2 kN/m2

Mmax (+ve) = 𝑊𝑏∗ 𝐿2

10 =

264.∗ 52

10=660 kN.m

Mmax (-ve) = 𝑊𝑏∗ 𝐿2

12 =

264.∗ 52

12=550 kN.m

For -ve section 3

d= c1√𝑀𝑢∗106

𝑓𝑐𝑢∗𝑏

2 950= c1√

660∗106

25∗250

2

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c1=2.92 j=0.781

As = 𝑀𝑢

𝐹𝑦∗𝐽∗𝑑 =

660∗106

360∗0.781∗950=2470 mm2 use 8 D20/m

Design of tie

R = Hmax * spacing

264*5=1320 kN (tension)

As =𝑃𝑢

𝐹𝑦/ˠ𝑠 =

1320∗103

360/1.15 =4216 mm2 use 12 D22 /m

Ac = 40 As =40*4216=168640 mm2

Use (450*450) mm2

Design of the vertical beam (300*700) mm2

Wbeam = Vmax +OW of VL beam + OW of H beam

Wbeam =132+1.4*(25*0.30*0.45)+1.4*(25*0.25*1)=145.475 kN/m2

Mmax (+ve) = 𝑊𝑏∗ 𝐿2

10 =

145.47∗ 52

10=363.68 kN.m

Mmax (-ve) = 𝑊𝑏∗ 𝐿2

12 =

145.47∗ 52

12=303 kN.m

Design of sections

= c1√𝑀𝑢∗106

𝑓𝑐𝑢∗𝑏

2 650= c1√

363.68∗106

25∗300

2

c1=2.95 j=0.781

As = 𝑀𝑢

𝐹𝑦∗𝐽∗𝑑 =

363.68∗106

360∗0.781∗650=1989.9 mm2 use 7 D20/m

Design for hanger

Tu= OW of hanger+ OW of tie* hanger spacing

Tu= 1.4*(0.25*0.4*3*25)+1.4*(0.45*0.45*25)*3=27.56 kN

As =𝑇𝑢

𝐹𝑦/ˠ𝑠 =

27.56∗103

360/1.15=88mm2

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Use 6 D10 /m

Design of column (300*1000) mm2

N=Vmax 132 kN

M=Hmax * height of column = 264*6=1584 kN.m

Buckling of column

He = Kho = 6*1.3

λ=𝐻𝑒

𝑡

= 7.8

1=7.8 <10 short column

e= 𝑀

𝑁=

1584

132=12 m >

𝑡

2=0.5 bid ecc

es= e +𝑡

2 – cover=12+0.5-0.1=12.40 m

Mus=Pu*es=132* 12.4=1636.8 kN.m

As = 𝑀𝑢

𝐹𝑦∗𝐽∗𝑑-

𝑃𝑢

𝐹𝑦/ˠ𝑠=

1636.8∗106

360∗0.781∗900-

132∗1000

360/1.15=6046.7mm2

Use 12D25/m

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4.6 Problems

Problems 1

it is required to cover an area with (18 x 28) m, with concrete structure elements.

Columns allowed only in perimeter of long direction every 6.00m.

- Choose a suitable Parabolic Slab system with a tie (y=4hx2/L2) to cover this

area.

- Draw plan and illustrate the main system with full approximate concrete

dimensions.

- Design the arched slab

- Design Vertical &horizontal beams -Design The tie and hungers - design

columns

- Draw full reinforcement details (elevation and sections) are necessary

Data:

- Fcu = 30 N/mm2 , Fy (main steel ) = 360 N/mm2, Fy (stirrups ) = 240

N/mm2

- LL=0.50 KN/m2 , Flooring =2.0 KN/m2

- Clear height =7.

- Drawing scale is 1 : 50 for plans and 1 : 25 for sections Systematic

arrangement of calculations and complete detail drawings are necessary

Problems 2

An Arched slab with a tie of span 18m and rise 3m supported on columns of spacing,

S= 5m. The following data are given: ultimate stiffener load, wu,stif= 7kN/m·, total

ultimate load of a tie, T total, u= 650kN, ultimate own weight of the vertical and

horizontal beams= 12kN/m', Ultimate moment of arched slab at quarter point= 7kN.m.

It is required to

- Determine the maximum dead and live loads carried by the arch.

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- Design the tie and the arch at crown.

- Determine the total loads applied on the column.

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Chapter 5

Design of Arch Girders

5.1 Introduction

Arch girder is used to cover spans up to 40 m, the girder is subjected to high compression

force and low bending moment and shearing force. All section of the girder is

approximately subjected to the same compressive stress which means economic concrete

sections. With this system large spans are coverd with no intermediate columns.

The main idea of the arch girder is shape of the girder is the inverse of the bending

moment, which lead to almost zero moment.

As the number of concentrated loads increased the number of joints or segments increased.

The slab of the arch girder system should be one way slab in the direction of the secondary

beam and the slab load will concentrated load on the girder directly.

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To control the existence of the load to be on joints only the following items should be

followed

- Put beams on the arched joints

- Slabs should be one-way sabs in the directions of the beam

- OW of the arch slabs should be located as concentrated loads on joints

- Hangers or posts should be located at joints

5.2 Location of slabs

The location of roof slab is changes according to the architectural requirement as the

following

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5.3 Dimensions of arched slabs

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✓ t girder =L

20−25

✓ b girder=(0.3-0.4) t girder

✓ h (arch rise) = L

5−8

✓ t column =h

8−9

✓ t slab = 120 mm

✓ spacing between girders =(3-6) m

✓ hanger, tie, secondary beams ( b*t )

5.4 Loads and reaction calculations:

Slabs

gs UDL = 1.4(ˠ* ts+F.C)*1.1 load on horizontal projection

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ps ULL = 1.6(p)

Secondary beams

Wdead = OW+gs* α*𝐿𝑠ℎ

2

WLive =ps* α*𝐿𝑠ℎ

2

Beam RDL =1.1* Wdead*S

Beam RLL =1.1* WLive*S

Post

PDL = OW + Beam RDL

PLL = Beam RLL

Arch girder

WUDL= OW*1.1+Ʃ𝑃𝐷𝐿

𝐿 per horizontal projection

WULL= Ʃ𝑃𝐿𝐿

𝐿 n

Wu (Total) = W UDL + W ULL

Case (1) full load


2


8ℎ

Case (2) half load

Ya = 𝑊𝑈𝐷𝐿 ∗ 𝐿

2 +

3𝑊𝑈𝐿𝐿 ∗ 𝐿

8

Xa = Xb = 𝑊𝑈𝐷𝐿 ∗ 𝐿2

8ℎ +

3𝑊𝑈𝐿𝐿 ∗ 𝐿2

16ℎ

At X =𝐿

4

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Y = 8ℎ𝑥

𝐿2 ( L- X)

Yd = Ya -W*X

Xd = Xa

Ɵ = tan -1 ( 4ℎ

𝐿2 ( L - 2X))

Nd = (Xd cos Ɵ + Yd sin Ɵ) compression

M max = =𝑊𝑈𝐿𝐿 ∗ 𝐿2

64

5.5 Example 5.1

For the following arched girder

- Design the arch girder, tie, posts and hungers and draw the reinforcement

- Design the main supporting element and draw the reinforcement

Fcu=35 N/mm2 Fy=400 N/mm2


Heigh of column is 6 m & spacing between column 6 m

Solution

Concrete Dimensions

- ts=120 mm

- Secondary beam (250*600) mm.

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- Post (400*400) mm2 h post = 21

5−8 h=3m

- t girder =L

20−25=

21

20−25 t girder=1000mm

- b girder=(0.3-0.4) b girder=400mm

Loads

gs = 1.4 (ˠ* ts + F.C)

= 1.4 * (0.12*25+1) = 5.6 kN/m2

ps = 1.6(p)

= 2 * 1.0 = 3.2 kN/m2

Secondary beam

Spacing between secondary beams =3m (one way slab 3*6)

W dead = OW + gs* α*𝐿𝑠ℎ

2 = 1.4(0.25*0.6*25) + 5.6*1*1.5*2 = 22.05kN/m

W live = ps* α*𝐿𝑠ℎ

2 = 3.2*1*1.5*2 = 9.6 kN/m

Beam RDL =1.1* W dead*S=1.1*22.05*6=145.53 kN

Beam RLL =1.1* W Live*S=1.1*9.6*6=63.36 kN

Post (400*400)mm2

PDL = OW + Beam RDL=1.4*(0.4*0.4*3*25) +145.3=162.1 kN

PLL = Beam RLL=63.36 kN

Arch girder

WUDL= OW*1.1+Ʃ𝑃𝐷𝐿

𝐿 =1.4*(0.4*1*25)*1.1+

6∗162.1

21 =15.4+46.3=61.7 kN/m

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WULL= Ʃ𝑃𝐿𝐿

𝐿 =

6∗63.36

21=18.1 kN/m


= 61.7 +18.1 =79.8 kN/m2

Case (1) full load


2=

79.8 ∗ 21

2=837.9 kN


8ℎ=

79.8 ∗ 212

8∗3=1466.3 kN

Case (2) half load

Ʃ M @ b=0

61.7 * 21 *10.5 + 18.1 * 10.5 * 5.25 =Ya * 21

13604.9+997.76=Ya*21

Ya = 695.4 kN

Ʃ Y =0

Y b = 61.7 * 21 + 18.1 * 10.5-695.4 = 790.4

kN

Ʃ M @ C (crown)=0

Xa*(3) = 79.8 *10.5*5.25 -790.4*10.5

Xa =1300 kN

Xa = Xb = +

At X= = =5.25m

Y = (L-X) = (21-5.25) =2.25m

Yd=Ya-W*X=790.4 -79.8*5.25=371.45 kN

Xd=Xa=1300 kN

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116 Dr/Yasmin Hefni

Ɵ = tan -1 ( 4ℎ

𝐿2 ( L - 2X))= tan -1 (

4∗3

212 ( 21 –2*5.25))=15.95o

Nd = (Xd cos Ɵ + Yd sin Ɵ) =1300 cos 19.95+371.45 sin 15.95=1324.06 kN

(compression)

Mmax=790.4*5.25-1300*2.25-79.8*5.25*5.25/2=124.86 kN.m.

Or

M max =𝑊𝑈𝐿𝐿 ∗ 𝐿2

64= =

18.1 ∗ 212

64= 124.7 kN.m

Design of Arch girder (at section d-d)

M max =124.86 kN.m Nd=1324.06 kN tg=1000mm

𝑃𝑢


1324.06∗ 103

35∗1000∗400=0.09>0.04 design for Mu, Pu

e=Mmax/Pu=124.86/1324.06 =0.09m e/t=0.09 <0.5 use ID

𝑀𝑢

𝐹𝑐𝑢∗𝑏∗𝑡2=

124.86 ∗ 106

35∗1000∗1000∗400=0.009

𝑃𝑢


1324.06∗ 103

35∗1000∗400=0.09

p < 1 take p=1

µ=p*Fcu*10-4 =1*35*10-4 = 0.0035

As =As- = µ*b*t=0.0035*400*1000=1400mm2

Asmin = (0.6/100)*400*1000=2400mm2

use 7D16 top and bottom

Design for post

PDL = 162.1 kN

PLL = 63.36 kN

Pu 162.1+63.36=225.46

Ho Height of the post = 3-0.5=2.5m

He = Ho*K=1.2*2.5=3m

λ=𝐻𝑒

𝑡=

3

0.4=7.5 <10 short column

Pu =0.35 Ac Fcu+0.67 As Fy

225.46*1000=0.35*(400*400)*35+0.67*400*As

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117 Dr/Yasmin Hefni

Asmin = (0.6/100)*400*400=960mm2 use 8 D12

Design of tie

R = Hmax * spacing

264*5=1300kN (tension)

As =𝑇𝑢

𝐹𝑦/ˠ𝑠 =

1300∗103

400/1.15 =3737 mm2 use 10 D 22

Ac = 40 As =40*10*380.13=152053 mm 2

Use (400*400) mm2

Design for hanger

Tu= OW of hanger+ OW of tie* hanger spacing

Tu= 1.4*(0.25*0.25*3*25)+1.4*(0.40*0.40*25)*3=24 kN

As =𝑇𝑢

𝐹𝑦/ˠ𝑠 =

24∗103

400/1.15=69mm2

Use 4 D12

Advanced Concrete 1

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118 Dr/Yasmin Hefni

Advanced Concrete 1

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119 Dr/Yasmin Hefni

5.6 Problems

Problem 1

For the following arched girders



Fcu=30N/mm2 Fy=400 N/mm2



Figure 1

Problem 2

For the following arched girders



Fcu=30N/mm2 Fy=400 N/mm2



Advanced Concrete 1

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120 Dr/Yasmin Hefni

Figure 2

lectures notes of advanced concrete design 1 ceng 306

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