Final Structure Coursework handin.docx

THE UNIVERSITY OF GREENWICH

School of engineering

Department of Civil Engineering

STRUCTURE IIICIVI 1015 & CIVI 1009

PLASTIC ANALYSIS OF A PITCHED – ROOF PORTAL FRAME

LABORATORY TESTING OF A REINFORCED CONCERTE SLAB AND YIELD

LINE THEORY

LABORATORY TESTING OF STEEL PORTAL FRAME

BEng (Hons)

Tutor: Dr Ouahid HarirecheLaboratory Technician: Tony Stevens

Group 5

STUDENTS:

Xudong Niu,

Fode Cisse

2011 – 2012

http://www.gre.ac.uk/

Structures III

Contents

Introduction......................................................................................................................................2

I – Plastic Analysis of a Pitched–roof Portal Frame..........................................................................3

1.1 Aim.....................................................................................................................................3

1.2 Objective.............................................................................................................................3

1.3 Description of the Model....................................................................................................3

1.4 Methodology of the Analysis..............................................................................................4

1.5 Result & findings................................................................................................................5

1.5.1 Using SAND programme.........................................................................................5

1.5.2 Manual Plastic Analysis.........................................................................................23

1.5.3 Draw Plastic Bending Diagram..............................................................................27

II–Laboratory Testing of a Reinforced Concrete Slab and Yield Line Theory...............................30

2.1 Aim...................................................................................................................................30

2.2 Objective...........................................................................................................................30

2.3 Description of the Experiment..........................................................................................30

2.4 Procedure for Experiment.................................................................................................31

2.5 Experiment Data & Observations.....................................................................................33

................................................................................................................................................36

2.6 Calculations......................................................................................................................37

2.6.1 Empirical Method..................................................................................................37

2.6.2 Theoretical Method of Calculation (Yield line Method)........................................40

III – Laboratory Testing of Steel Portal Frame...............................................................................42

3.1 Introduction.......................................................................................................................42

3.2 Aim...................................................................................................................................42

3.3 Objective...........................................................................................................................42

3.4 Description of Experiment................................................................................................43

3.5 Procedure for Experiment.................................................................................................45

3.6 Results & Findings............................................................................................................47

3.6.1 Analysis Experimental Results...............................................................................47

3.6.2 Manual Theoretical Calculation.............................................................................51

3.6.3 Interaction Diagram (ID).......................................................................................55

3.7 Discussion &Conclusions.................................................................................................57

4 General Conclusion......................................................................................................................58

5 References....................................................................................................................................59

School of Engineering Page 1

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Introduction

The elastic design method, also termed as allowable stress method (or Working stress

method), is a conventional method of design based on the elastic properties of steel.

This method of design limits the structural usefulness of the material up to a certain

allowable stress, which is well below the elastic limit. The stresses due to working

loads do not exceed the specified allowable stresses, which are obtained by applying

an adequate factor of safety to the yield stress of steel. The elastic design does not

take into account the strength of the material beyond the elastic stress. Therefore the

structure designed according to this method will be heavier than that designed by

plastic methods, but in many cases, elastic design will also require less stability

bracing.

In the method of plastic design of a structure, the ultimate load rather than the yield

stress is regarded as the design criterion. The term plastic has occurred due to the fact

that the ultimate load is found from the strength of steel in the plastic range. This

method is also known as method of load factor design or ultimate load design. The

strength of steel beyond the yield stress is fully utilized in this method. This method is

rapid and provides a rational approach for the analysis of the structure. This method

also provides striking economy as regards the weight of steel since the sections

designed by this method are smaller in size than those designed by the method of

elastic design. Plastic design method has its main application in the analysis and

design of statically indeterminate framed structures.


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I – Plastic Analysis of a Pitched–roof Portal Frame

1.1 Aim

In this part of the report, our aim is to investigate the plastic collapse of a pitched roof

portal frame.

1.2 Objective

1. We would perform an incremental collapse analysis of a pitched-roof portal

frame structure through using an elastic structural analysis of the SAND

software programme;

2. We would interpret output data from the SAND programme;

3. With the calculation process with software programme, we could understand

the theory of incremental collapse method.


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1.3 Description of the Model

Figure 1.1 Model of Pitched-roof Portal Frame

The frame: Span = 20 m

Height to eaves = 10 m

Height to ridge = 15 m

Loading: Horizontal load eaves = 8 kN

Vertical load at mid-points of rafters = 24 kN and 16kN

Section Property: Steel with uniform plastic moment of resistance = 60 kNm.

1.4 Methodology of the Analysis

1. Incremental Elastic Analysis using SAND;

Trace the structural response at all stages of plastic hinge formation

Identify the collapse mechanism and the collapse loads

2. Perform a manual plastic analysis of the structure considering all possible collapse


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mechanisms and using the theorem of virtual work to calculate the collapse loads;

3. Compare the calculation results with the computer output obtained;

4. Draw the plastic bending moment diagram.

1.5 Result & findings

1.5.1 Using SAND programme

In order to trace the structural response at all stages of plastic hinge formation, we use

the SAND programme with elastic analysis method and then hand calculation which

to locate plastic hinge in the structure model. The structure model is shown in figure

1.1; due to the data of load which horizontal and vertical loads contain H 1 is 8kN at

joint No.2, one of vertical load is 24kN at joint No.3 and the other one is 16kN at joint

No.5, therefore, the whole values divided by 8 should get H 2= 1.0, V 3= 3.0, V 5= 2.0.

So we can carry loadsλ, 3λ, 2λ in the figure 1.2. The behaviour of the frame, as λ is

increased, the whole process contains four stages shown below:

Stage 1

Whole structure elastic until first plastic forms at Joint No.7:


THE COURSEWORK OF STRUCTURES III PAGE:ELASTIC STRUCTURAL ANALYSIS MADE BY: XUDONG NXUDONG NIU, FODE CISSE DATE: 10/02/12 REF NO:

STRUCTURE SCALE 1 CM = 1.500 = SUPPORTS

X

Y

1

2

3

4

5

6

7

MOMENT Z SCALE 1 CM = 5.00 LOADING KEY 1

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3λ 2λ

4

3 5

λ 2 6

1 7

Figure 1.2 Stage1 the whole elastic structure with no hinge

The Bending Moment Diagram from elastic analysis of the SAND programme:

0.20 0.20

3.67 1.69 6.81

5.76

3.67 6.81

0.45 7.32


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Figure 1.3 The bending moment diagram with no hinge through SAND programme

In the bending moment diagram from SAND programme, the bending moment value

(kNm) of each major joint is shown above. This situation is a whole structure elastic

without any plastic hinge. The value of maximum bending moment is 7.32 kNm at

joint No.7, so we need to put the uniform plastic moment of resistance value of 60

kNm on joint No.7 which a plastic hinge location, and then we would calculate the

value of load factor using the plastic bending moment diagram programme.


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4

3 1.64 1.64 5

30.08 55.82

30.08 2 47.21 13.85 6

55.82

1 3.69 60 7

Figure 1.4 Stage 1

Load factor λ = 60/7.32 = 8.197

The figure 1.4 is shown that the stage 1 of the plastic bending moment (kNm) diagram

process by SAND and then calculation. Initially the frame structure is elastic

everywhere, and elastic analysis using SAND programme. When λ = 8.197 the largest

bending moment, at Joint No.7 becomes structure apart form joint No.7 is still elastic

and remains so whenλis increased above 8.197. Whenλis increased, Joint No.7

behaves i.e. a hinge in that it can rotate freely, but the bending moment must remain

equal to 60 kNm of the plastic moment.

The value of 8.197 times bending moments from figure 1.3, getting the bending

moments in the stage 1 of the plastic bending moment diagram is shown in figure 1.4.

Stage 2




X

Y

1

2

3

4

5

6

7


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Effective structure contains a frictionless hinge at joint No.7; second plastic hinge

forms at Joint No.6:

2.19 2.19

1.58

2.55 7.32 9.03

2.55 9.03

3.52 0.00

Figure 1.5 the bending moment diagram with hinge at joint No.7


(kNm) of each major joint is shown in figure 1.5. This situation is effective structure

contains a frictionless hinge at joint No.7. The value of maximum bending moment is

55.82 kNm at joint No.6 in figure 1.4, so we need to put the uniform plastic moment

of resistance value of 60 kNm on joint No.6 which a plastic hinge location, and then

we would calculate the value of load factor using the plastic bending moment diagram

programme.


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Structures III


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4 2.65 2.65

3 5 60

31.25 14.58 31.25 2 50.58 6 60

2.07 60 7

1

Figure 1.6 stage 2

Load factorλ = 8.197 + 0.46 = 8.657The figure 1.6 is shown that the stage 2 of the plastic bending moment diagram

process by SAND programme and then calculation. The effective structure model

which resists the loads when λ is greater than 8.197, it is simply the original frame

structure with frictionless hinge at Joint No.7. This frame structure is analysed by the

same elastic method of the SAND programme in stage 1. The result of the analysis is

the change in bending moment diagram in figure 1.5. To get the total moments it is

necessary to add the change in bending moment to the bending moments whenλ=

8.197. The notice: the frictionless hinge at joint No.7 ensures that the change in

bending moment at Joint No.7 is zero so that the total moment remains equal to the

plastic moment.

The maximum moment is the bending moment of 55.82 kNm at joint No.6 shown in

figure 1.4.

M 6 = 55.82 + 9.03λ '

Where λ ' is change inλand M 6 equals the plastic moment (60kNm), 9.03 is the value

of the bending moment (kNm) at joint No.6 in bending moment diagram by SAND

elastic analysis in figure 1.5.

So, λ ' = (60−¿55.82)/9.03 = 0.46




X

Y

1

2

3

4

5

6

7


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Stage 3

Effective structure has frictionless hinges at joint No.6 and 7; third plastic hinge forms

at Joint No.3:

10.83 10.83

10.41

3.34

3.34 11.24 0.00

13.34 0.00

Figure 1.7 the bending moment diagram with hinge at joint No.6 and 7



contains frictionless hinges at joint No.6 and 7. The value of maximum bending

moment is 50.58 kNm at joint No.3 in figure 1.6, so we need to put the uniform

plastic moment of resistance value of 60 kNm on joint No.3 which a plastic hinge

location, and then we would calculate the value of load factor using the plastic

bending moment diagram programme.


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11.72 4 11.72

3 5 60

34.05 23.30

34.05 2 60 6 60

1 79.11 60

Figure 1.8 stage 3

Load factor λ = 8.657 + 0.838 = 9.495

The figure 1.8 is shown that the stage 3 of the plastic bending moment diagram

process by SAND elastic analysis and calculation. The effective which resists the

loads whenλ is greater than 8.657. It is simply the original frame structure with two

frictionless hinges at joint 6 and 7. Elastic analysis method is the same to the stage 2

analysis using SAND programme. The result of the analysis is the change in bending

moments. In order to get the total moments, we should add the change in bending

moments to bending moments when λ is 8.657. The same condition with stage 2,

which changing bending moment at joint No.6 is zero so that the total moment should

equal to the plastic moment. So from now there are two hinges in the effective frame

structure. The maximum moment of 50.58kNm is under the vertical load, joint 3 in

figure 1.6.

M 3 = 50.58 + 11.24λ ' '

Where λ ' 'is change inλ and M 3 equals the plastic moment (60kNm). 11.24 is the value

of the bending moment (kNm) at joint No.3 in bending moment diagram by SAND

elastic analysis in figure 1.7.

So, λ ' ' = (60−¿50.58)/11.24 = 0.838




X

Y

1

2

3

4

5

6

7


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Stage 4

Frictionless hinges at joint No.3, 6 and 7; fourth plastic hinge forms at Joint No.2;

frame structure fails:

3.33 3.33

6.67

18.33 0.00 0.00

18.33

28.33 0.00

Figure 1.9 the bending moment diagram with hinge at joint No.3, 6 and 7



contains frictionless hinges at joint No.3, 6 and 7. The value of maximum bending

moment is 34.05 kNm at joint No.2 in figure 1.8, so we need to put the uniform

plastic moment of resistance value of 60 kNm on joint No.2 which a plastic hinge

location, and then we would calculate the value of load factor using the plastic

bending moment diagram programme.


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Structures III


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16.44 4 16.44

60 3 5 60

60 32.74 6 602 60

49.22 1 60 7

Figure 1.10 stage 4

Load Factor λ = 9.495 + 1.416 = 10.91

It is shown that the stage 4 of the plastic bending moment diagram process by SAND

elastic analysis and calculation. The process can be continued as in stage 4 with three

frictionless hinges, until at λ is 10.91 a fourth plastic hinge forms.

The maximum moment of 34.05 kNm is under the horizontal load, joint 2 in figure

1.8.

M 2 = 34.05 + 18.33λ ' ' '

Where λ ' ' 'is change inλ and M 2 equals the plastic moment (60kNm). 18.33 is the

value of the bending moment (kNm) at joint No.2 in bending moment diagram by

SAND elastic analysis in figure 1.9.

So, λ ' ' ' = (60−¿34.05)/18.33= 1.416

The value of 1.416 times bending moments from figure 1.9; plus the bending

moments in figure 1.8 (the process exclude hinge locations), getting the bending

moments in the final plastic bending moment diagram is shown in figure 1.10.


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Any attempt to continue the process using an effective structure with fourth

frictionless hinges is impossible; the equations become singular and cannot be solved.

In fact, the structure becomes a mechanism and is on the joint of collapse when the

fourth hinge forms.

So, the collapse loads:

H 2 = 10.91 kN

V 3 = 32.73 kN

V 5 = 21.82 kN


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1.5.2 Manual Plastic Analysis

In this part, we would perform manual plastic analysis of the structure in order to

consider all possible collapse mechanisms and calculate the collapse loads using the

theorem of virtual work. Finally, compare the manual plastic analysis result with the

computer output obtained.

3λ 2λD

C E 5mλ B F 15m

10m

A G5m 5m 5m 5m

20m

Steel with uniform plastic moment of resistance, M p = 60 kNm.

Structure:

Redundancy, R = 3

Critical Section (the number of possible plastic hinge locations), N = 7

Mechanism, M = N−¿R = 7−¿3 = 4

Number of plastic hinges, R+1 = 3+1= 4

3λ D Beam mechanism BCD:

C θ 2θ E the virtual work equation is

B θ C ' F 3λ ×5θ = M pθ + M p 2θ + M pθ

15λθ = 4M pθ

A G λ = 4 M p

15 =

4 × 6015

= 16


M1

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D 2λ Beam mechanism DEF

C θ E the virtual work equation is

B 2θ E' θ F 2λ ×5θ = M pθ + M p 2θ + M pθ

10λθ = 4M pθ

λ = 4 M p

10 =

4 × 6010

= 24

A G

D Sway mechanismC D' E' the virtual work equation is

λ B B' C ' E F F ' λ ×10×θ = M pθ+M pθ+M pθ+M pθ

θ θ 10λθ = 4M pθ

θ θ

A G λ = 4 M p

10 =

4 × 6010

= 24


M2

M3

Structures III

tan α = 5

10 =

10IG

IG = 20 m

IF = IG −¿ FG= 20−¿10 = 10 m

D D' = BD× θ = DI × ϕ

F F ' = FG× β = FI × ϕ

θ = DIBD

ϕ

β = FIFG

ϕ

BD = √102+52 = 11.18 m

BI = √202+102 = 22.36 m

DI = BI −¿ BD = 22.36 – 11.18 = 11.18 m

So, θ = 11.1811.18

∅ θ = ϕ

β = 1010

∅ β = ϕ

The virtual work equation is

3λ ×5θ + 2λ ×5ϕ = M pθ + M p(θ+ϕ) + M p(β+ϕ) + M p β

θ = ϕ, β = ϕ

3λ ×5ϕ + 2λ ×5ϕ = M p ϕ + M p(2ϕ) + M p(2 ϕ) + M p ϕ

5λ ×5ϕ = M p×6ϕ

λ = 6

2.5M p =

6 ×602.5

= 14.4


Structures III

I

D

C ϕ hinge cancelled at D

B C '

3ϕ

ϕ=θ2

A G

Combined mechanism

Other combined is not available by trial. If hinge cancelled at D, so the rotation ofθ,

which at D in M1 should be equal to the rotation of 2ϕ, which at D in M4, i.e. θ = 2ϕ.

For M4 (hinge cancelled at D)

5λ ×5ϕ = M p ϕ + M p ϕ + M p ϕ + M p ϕ ¿ = 2ϕ)

2.5λ ×5ϕ = M p×2θ

For M1 (hinge cancelled at D)

3λ ×5ϕ = M p× θ + M p× 2θ

M1 + M4 5.5λ × 5θ = M p× 5θ

λ = M p

5.5 =

605.5

= 10.91

So, the collapse loads:

H 1 = λ = 10.91 kN

V 1 = 3λ = 3×10.91 = 32.73 kN

V 3 = 2λ = 2×10.91 = 21.82 kN

Compare the results with the SAND output obtained; the values are equal (OK).


M1 + M4

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1.5.3 Draw Plastic Bending Diagram

32.73kN 21.82kND

C E

10.91kN B F

A G

M BA = M BC = M CB = M CD =M FE = M FG = M GF = M p = 60 kNm

(Because of having plastic hinges at these locations)

At the horizontal direction, ΣM = 0, so

D

10.91kN C E

HB B F

A G

M AB −¿ M BA + M FG + M GF = HB × 10

M BA = M FG = M GF= M p = 60kNm

M AB = HB ×10−M p−M p+M p = 10.91×10−¿60 = 49.1 kNm

B


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60kNmHB = 10.91kNm

H A ×10 + 49.1−¿60 = 0

10m H A = 1.09 kN

Σ M G = 0

V A ×20−V c ×15−V E ×5 = 0

49.1kNm A H A V A = 32.73× 15+21.82 ×5

20 = 30 kN

V A

32.73kNm

D

B CM BC=M CB = M CD = M p = 60 kNm

H A=1.09 kN M DC = H A ×15 = 1.09×15 = 16.35 kNm A

V A=30 kN

D Σ M D= 0, M DE + M DC = 0

E M DE = -16.35kNm

F M FE = −M p = -60 kNm M FG = M GF = M p = 60 kNm

HG = H A + HB = 1.09 + 10.91 = 12kN

HG= 12kN M ED= 49.1 – 16.35 = 32.75 kNm G

Σ M A = 0 V G=24.5kN −V G ×20+V c ×5+V E ×15 = 0

V G = 32.73× 5+21.82 ×15

20 = 24.5 kN

So, depended by the values of the result process, the bending moment (kNm) diagram at


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collapse load, which is a plastic bending moment diagram shown following as:

D

16.35 16.35

C E

60 60

32.75

60 B 60 F 60

: Fixed Support

A G

49.1 60


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II–Laboratory Testing of a Reinforced Concrete Slab and

Yield Line Theory

2.1 Aim

In this part of the report, our aim is to investigate failure modes of reinforced concrete

slabs.

2.2 Objective

1. Experimental data and observations;

2. Data from Cube Testing and Tensile Testing;

3. Observed deflection readings;

4. Calculation of the Ultimate Moment of Resistance;

5. Experimental crack pattern;

6. Analysis of the Yield Line pattern and calculation of the Ultimate Moment of

Resistance.

2.3 Description of the Experiment


Hydraulic pressure pump

Hydraulic pressure loadSlab Model

Digital display

Point gauge

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Reinforced Concrete Slab Testing System

2.4 Procedure for Experiment

The testing mechanism was set up by laboratory technician “Tony”. For health and

safety raison and for the good conduct of the experiment, the hydraulics pump system

has to be checked to make sure that not damage to any single part of the system is

been neglected which can affect the result of experiment.

A reinforced concrete slab size of 548 mm x 505 mm with depth of 37 mm is set up in

the test rig. The slab was then loaded by hydraulics pressure pump through loading

point which was the centre of the slab


Hydraulic pressure pump Reinforced Concrete Slab Model and Point gauge

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As the load increased, the cracks started to

appears at the opposite face of where the load

was applied to. The first one appears just

horizontally at centre of the slab as shown

marked in dark colour on the face of the slab.

Cracks started to grow from the centre of the slab opposite face to the loading face

and progress toward the edge in horizontal and vertical directions. The cracks in form

of zigzag become larger at the slab surface as loading increased indicating failure in

concrete at flexure.


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2.5 Experiment Data & Observations

Data collected during the experiment are shown in table1.

Table2.1: Record from slab testing.

Graph2.1:


The side which load pressureThe back side of the slab for load pressure

Crack appeared at the side of the slab.Reinforcing steel in both direction, positioned nearer to one edge (tension side) than the other edge.

Loading Increment as follow (100% = 20kN)% Load (kN) Deflection in (mm)0% 0.0 0.03% 0.6 0.96% 1.2 1.19% 1.8 1.312% 2.4 1.415% 3.0 1.518% 3.6 1.721% 4.2 1.825% 5.0 2.230% 6.0 3.235% 7.0 4.340% 8.0 9.845% 9.0 11.5

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0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.010.0

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0Deflection

Deflection

Load (kN)

Def

ecti

on (m

m)

This graph is the interpretation of the direct relationship between Loading and

deflection at the mid span of the slab.

Tensile Testing Result for 1 bar & 3 bars


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12mm

1.1mm

12 mm

Cross-section

Note: Number of steel in 100 mm strip of concrete is 9


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0.815 kN

Graph 2.1 Data from tensile test for 1 bar of reinforcing steel

2.16 kN


The reinforcing steel yielded before splitting into two parts.

Three reinforcing steels yielded before breaking into two

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Graph 2.2 Data from tensile test for 3 bars of reinforcing steel

Concrete Cube Test Result

Sample of concrete cube:

100mm

100mm

Cube (date: 08/09/10) testing in the machine

First sample: 2.105 kg P = 55.39 MPa


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Second sample: 2.115 kg P = 56.54 MPa

2.6 Calculations

2.6.1 Empirical Method

The empirical relationships are:

K1 K3=(27+0.35 f c)/(22+ f c )

K2=0.5−f c /550

Where, f c = the 150 mm cylinder strength = (approx.) 0.78× f cu

Where, f cu = the 100 mm cube strength.


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b = 505 mm

d = 37 mm – (12 mm+1.1/2 mm) = 24.45 mm

Take d = 24.45 mm

From the experiment, the average value of 100 mm cube strength is:

f cu=P1+¿ P2

2=

(55.39+56.54 )2

=55.97 N /mm2 ¿

Therefore, the cylinder strength can be calculated as:

f c=0.78 × f cu=0.78 ×55.97=43.66 N /mm2

K1 K3=27+0.35× 43.6 6

22+43.65=0.64

K2=0.5−( 43.6 6550 )=0.42

The compressive force is given by: F c=K1 K3 × f c × b× x

The tensile force is given by: F t=f y × A s

Where, A s is the area of the tensile reinforcement and f y is the yield stress of the bars.

The reinforcing steel cross-section is:

A s=π d2

4=

π (1.1)2

4=0.95 mm2

A s is the reinforcing steel cross-section area and the diameter d = 1.1 mm.


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The yield stress of one reinforcing steel bars was determinate from yield stress taken

from the plot and the cross-section area of the steel.

f y=PA

=0.815× 103

0.95=857.89 N /mm2

Equating the two forces gives the values of x, the depth to the neutral axis.

x=f y A s

K1 K3 × f c × b

x=857.89 ×37 × 0.950.64 × 43.66 ×505

=2.14 mm

The lever arm of the two forces in then: z=d−K2× x

z=24.45−(0.42 ×2.14 )=23.6mm

Take z = 23.6 mm

The ultimate moment of resistance is M u=Fc × z=Ft × z

M u=F t × z=f y ×37 × A s× z=857.89 ×37 × 0.95×23.6=711654.1 Nmm

Mu = 0.71 KNm

A flexural element will be adequate in bending if its internal moment of resistance is


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not less than the externally applied bending moment. Therefore, the design ultimate

resistance moment M of a concrete slab must be greater than or equal to the ultimate

imposed bending moment.

2.6.2 Theoretical Method of Calculation (Yield line Method)

This picture showed at the right hand side can be represented graphically by:

P

L

Y M

B E C

G ≈550m

I II M


Slab positioned between its two supports: fixed end at the bottom and Simply support at the top end

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A F D

X≈500mm

Calculation of Ultimate Moment of Resistance:

A θ δ θ D

θ F θ

Note : No hinge at support D

For value of displacement in collapse load, δ = 11.5mm

The vertical displacement δ 1 = 11.5

2 = 5.75 mm

The vertical displacement δ 2 = 11.5

2 = 5.75 mm

The External Work W E = q(0.55×0.25)×0.00575 = 7.90625×10−4q

The Internal Work W i = M×0.55×θ + M×0.55×2θ = 3×M×0.55×θ

δ = AF × θ → θ = δ

AF =

0.01150.55

= 0.021

W i = 3×0.55×0.021M =0.035M

From Theorem of Virtual Work:

W E = W i

7.90625×10−4q = 0.035M → M = 7.90625× 10−4

0.035q =

7.90625350

q = 0.0226q kNm

For crack load, Q = 9 kN → q = 9 kN

0.55 ×0.50 = 32.73 kN/m2

M = 0.0226×32.73 =0.74 kNm

As the results from the two methods are very similar, we can understand from there

that the assumption made about support ‘A’ and “D” is about correct. The two results


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converge into one similar answer and the small difference between them might be

from human and system error.

In this particular case, the likely correct result might be from the second method and

the percentage of error might be estimated by:

%err=Theo−expTheo

=0.74−0.710.74

×100=4.05 %

III – Laboratory Testing of Steel Portal Frame

3.1 Introduction

Steel portal frames are commonly used for single-storey construction, particularly for

factory and warehouse buildings. These frames provide support to a steel roof. For

frame collapse load factors research, it has a further influence for health and safety in

practical application. The main collapse load factors contain two direction loads

which horizontal load (i.e. wind load) and vertical load (i.e. self-weight and snow load

etc.). Therefore, it is quite important experiment research for structural engineering to

understand the result of testing of steel portal frame.

3.2 Aim

In this part of the report, we should explore failure modes for a portal frame by model

experimentation. Through learning this experimental process and related knowledge,


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we could understand independent and combined failure modes; learning the method

of drawing and interpret interaction diagrams, in addition to interpret experimental

results as a predictor of failure.

3.3 Objective

1. Collected experimental data and observations;

2. Theoretical calculation of the plastic moment,M p;

3. Observed deflection readings;

4. Theoretical values of V (beam collapse), H (sway collapse) and V & H (combined

collapse);

5. Draw the Interaction Diagram;

6. Analysis the three modes if failure from the experimental results;

7. Draw up a table showing the comparison of theoretical and experimental results,

for the 3 failure modes.

3.4 Description of Experiment


50mm dial gauge

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View of Experimental Model

Apparatus


Column fixed control

Fixed Support

Dead Weight

Dead weight in vertical direction in structural model

Dead weight in horizontal direction in structural model

Dead weight in both directions at the same time in structural model

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http://www.houserepairtalk.com/f37/tape-measures-6083/


Hand WeightsThere is a range of weights as follows:200 g, 0.5 kg, 1.0 kg, 2.0 kg, 4.0 kg, 5.0 kg, 8.0 kg

Steel measuring tape(Measure the length and width of the beam)

50 mm dial gauge(Reading the value of the dial shown, it is the central deflection depth y (mm); the data is record i.e. smaller pointer reading + bigger pointer reading × 0.01)

Micrometre(Measure the depth of the beam)

http://www.starrett.co.uk/shop/precision/micrometers/standard_outside_micrometer/



Structures III

3.5 Procedure for Experiment

1. Take photographs of the Initial Set Up and the Steel Portal Frame Test, before

experiments in three different conditions are started, and during the experiments;

2. Take the dial gauge measure data in

vertical direction dead load in the first

condition;

3. Zero the dial gauge on the outer dial by

making sure that the pointer is at 0 (if

could not take zero, we can record from an

initial value, and then modified values,

which is reading value – initial value);

4. This test is very sensitive to any disturbance;

5. Add weight to the load hanger (weight value is depended by ourselves, but we

need control a certain limit) and read the dial gauge (deflection of steel frame);

6. At each increment of load, read the dial gauge

and observe the shape changing in each

increment load.

7. When the structural model is collapse, stop

increase the dead load, which is collapse load

in the vertical direction load and record the

deflection value of dial gauge;

8. Change the dial gauge location to measure deflection in horizontal direction;


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9. Doing the same way which measured deflection

in vertical direction is to do horizontal direction

condition and record the horizontal collapse

load and a range of deflection;

10. Add

another dial

gauge put in vertical direction to measure

combine vertical direction load with

horizontal direction loads, which to do the

same way and then get the combined collapse

load and deflection reading.

11. Take off all the loads; record all observation in the experiment process.

Steel Test:


14.1kN 14.0kN

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Graph 3.1 The first steel frame test Graph 3.2

The second steel frame test

3.6 Results & Findings

3.6.1 Analysis Experimental Results

For Beam Collapse:

Table 3.1:

0.00 100.00 200.00 300.00 400.00 500.00 600.000.005.00

10.0015.0020.0025.0030.0035.0040.0045.0050.00

Beam Collapse Deflection

Beam Collapse Deflection

Load (N)

Def

lect

ion

(mm

)

Graph 3.3 The relation between Load and Deflection in Vertical load

In graph 3.1, it is shown that the relation between load and deflection during

experiment process. With load increasing, the deflection is increasing and we are able

to see elastic variation with load values from 0 to 441.45 N and then plastic deflection


14.1kN

Beam CollapseDead Weight

(kg)Load (N)

Deflection (mm)

0 0.00 0.009 88.29 1.2618 176.58 2.5327 264.87 3.8436 353.16 5.3745 441.45 8.0950 490.50 10.5555 539.55 45.00

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in between 441.45 N to 490.50 N, finally, there is quite increasing effectively to

failure which collapse load of 539.55 N. It should be having some errors in the

experiment process system. From experimental observation, the angle of the two

joints, which connect between column and beam, is still 90 degree during the

increment of load. Plastic variation is happened due to elastic deflection. The beam

collapse load is 539.55 N.

For Sway Collapse:

Graph 3.2:

Sway CollapseDead Weight (kg) Load (N) Deflection (mm) Modified Deflection (mm)

0 0 8.39 05 49.05 11.87 3.4810 98.1 15.74 7.3515 147.15 20.83 12.4420 196.2 25.46 17.0725 245.25 38 29.6126 255.06 41.5 33.1127 264.87 52 43.6128 274.68 56.81 48.42


Angle¿900 Angle¿900

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0 50 100 150 200 250 3000

10

20

30

40

50

60Sway Collapse Deflection

Sway Collapse De-flection

Load (N)

Def

lect

ion

(mm

)

Graph 3.4 The relation between Load and Deflection in Horizontal load

From experimental observation, the angle of the two joints, which connect between

column and beam, is still 90 degree during the increment of load when the load values

increasing from 0 to196.2 N. And then angle of the left joint, which close to load

point, is less than 90 degree as the same time as the right joint angle is more than 90

degree. When release the loading at the end processing, the frame go back the

structure with elastic deflection. The sway collapse load is 274.68 N.

For Combined Collapse:

H V

7d 4d

P p× 4 d = H × 11d

→ H = 411

p

P×7d = V × 11d

→ V = 411

p

So, VH

= 74


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Table 3.3:Combined Collapse

Dead Weight (kg)

Load (N) Deflection (mm) Modified Deflection (mm)

Direction Horizontal Vertical Horizontal Vertical0 0 9.15 24.06 0 09 88.29 11.45 24.81 2.3 0.7518 176.58 13.85 25.63 4.7 1.5727 264.87 16.46 26.5 7.31 2.4432 313.92 18.03 27.06 8.88 342 412.02 22.21 29.2 13.06 5.1447 461.07 26.5 35.62 17.35 11.5652 510.12 32.4 38.82 23.25 14.7654 529.74 35.1 40 25.95 15.9456 549.36 36.31 41 27.16 16.9458 568.98 36.64 41.86 27.49 17.860 588.6 36.86 42.72 29.71 18.66

0 100 200 300 400 500 600 7000

5

10

15

20

25

30

35

Combined Deflection

Horizontal DeflectionVectical Deflection

Load (N)

Defle

ction

(mm

)

Graph 3.5 The relation between Load and Deflection in Combined load

In Graph 3.5, we can observe deflection is increasing with a constant ratio by a nearly

straight line at load values, which is from 0 to 412.02, after that point there is a quite

increasing effectively jump. So we need to use 412.02 N as combined collapse load.

The combined collapse load, which is

412.02×(4/11) = 149.8 N in Horizontal load;

412.02×(7/11)= 262.2 N in Vertical load.


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3.6.2 Manual Theoretical Calculation

VH

6.0 mm

6.0 mm

Cross-section 301 mm

312 mm

156 mm 156 mm

Calculation of the plastic, M p b

Beam with cross section b = 6.0 mm d

d = 6.0 mm

The plastic modulus Zp = b d2

4 = 6 ×62

4 = 54 mm3

Therefore,

M p = Zp × σ y

The values of tensile testing of the steel from Graph shown as following:

q1 = 14.1 kN, q2 = 14.0 kN

σ y = qA

= 14+14.1

2×103

6 × 6 = 390.28 N/mm2

So, M p = Zp × σ y = 54 ×390.28 = 21075 N.mm = 0.021kNm


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V

H

Number of Redundancy R = 3

Number of Critical Section N = 5

Number of independent Mechanism M = N – R = 5 – 3 = 2

Number of Plastic hinges R +1 = 3 +1 = 4

M1:

Beam Collapse Mechanism

B C D

θ θθ θ

h = 301mm

L/2 = 156 mm L/2 = 156 mm A E

L = 312 mm


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The external work W E = V× ∆ = V×L2

θ

The internal virtual work observed by rotation of plastic hinges at B, C and D is:

W I =2M pθ + 2M pθ = 4M pθ

From Theorem of Virtual Work Collapse Load V c

W E = W I

V×L2

θ= 4M pθ → V = V c = 8 M p

L =

8 ×0.0210.312

= 0.5385 kN = 538.5 N

M2:Sway Collapse Mechanism

H B C D

θ θ h = 301 mmθ θ

L/2 = 156 mm L/2 = 156 mm

A E

L = 312 mm

The External work W E = H × = H × h× θ

The Internal work observed by rotation of plastic hinges at A, B, and D is:

W I = M pθ + M pθ +M pθ + M pθ

From Theorem of Virtual Work Collapse Load H c

W E = W I


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H × h× θ = 4M pθ → H = 4 M p

h =

4 × 0.0210.301

= 0.2791 kN = 279.1 N

M1 + M2:Combined Collapse Mechanism

B C D ∆H

θ θ2θ

∆V Right angle

h =301mm θ θ

A E

L/2 = 156 mm L/2 = 156 mm

L = 312 mm

The External work W E = V× ∆V + H× ∆H

= V×L2

θ + H×h×θ

= (VL2

+ Hh)θ

The Internal work observed by rotation of plastic hinges at A, C, D and E is:

W I = M pθ + M pθ + M pθ + 2 M pθ +M pθ = 6M pθ

Theorem of Virtual Work

W E = W I

(VL2

+ Hh)θ = 6M pθ

VL2

+ Hh = 6 M p →VL

2 M p +

HhM p

= 6

→VL2

+ Hh = 6 M p


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3.6.3 Interaction Diagram (ID)

VL/M p

Sway Collapse Hh/M p=4

12 10 Loading Path (s =4)

Beam Collapse VL/M p= 8

8 Loading Path (s¿1)

6

Loading Path (s =1)

4 Loading Path (s¿1)

2

Combine Collapse VL/2M p + Hh/M p = 6

0 Hh/M p

1 2 3 4 5 6

For a given load ratio ρ = VH

an incremental collapse analysis with load factor will

follow a loading path, which is a line with origin at ‘0’ and slope‘s’.

s =

VLM p

HhM p

= VLM p

×M p

Hh → s =

VH ×

Lh =ρ

Lh

Hence, s depends on load ratio ρ and ratio Lh

of frame dimensions.

For s ¿1 → Sway Collapse

For 1< s < 4 → Combined Collapse

For s > 4 → Beam Collapse

Using the ID, obtain the collapse loads

VH

= 74

= 1.75 (ρ = 1.75)

Lh

= 312301

= 1.04 → s =ρ ×1.04 = 1.82 ¿ 1


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Hence, combined is the collapse mechanism

VL2 M p

+ HhM p

= 6

→VL2

+ Hh = 6 M p

V = 1.35H →1.35 HL

2 + Hh = 6 M p → H =

6 M p

1.35 L+h =

6 × 0.0211.35× 0.312+0.301

= 0.1745

kN = 174.5 N

→ V = 1.35×174.5 = 235.6 N


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3.7 Discussion &Conclusions

Table 3.4:Experiment Result(N) Theoretical Result(N)

Beam Collapse Load (N) 539.55 538.5Sway Collapse Load (N) 274.68 279.1

Combined Collapse Load (N) Horizontal 149.8 174.5Vertical 262.2 235.6

These results from two methods of analysis are almost same or the difference between

them is relatively small from one answer to the other.

The one which is considerably bigger is probably due to the system that was set

during the combined case study and because that the proportion between vertical load

and horizontal load for the hand calculation was V/H = 1.75. The system used for the

experiment give us a proportion of V =7/11 times total load and H= 4/11 times total

load which result as V/H = 7/4 = 1.75.

For combined collapse load %err=Theototal−exp total

Theototal

×100=412−410.1412

×100=0.46 %

It can be seen from the table3.4 that the other results are relatively same which allowed us to conclude that both methods are converging.

The reasons of error happened in the experiment:

1. When we put the hand weights on the hanger, it is easy to make the hanger shake

slightly, therefore influence the reading from dial gauge by recorder;

2. The measure machine exists systematic error i.e. dial gauge;

3. There have some factors influence the material property of the steel beam leading

to the impact of the deflection i.e. temperature, air humidity and so on;

4. The weight of the hand weight is not quite exact;


Structures III

5. When making the hanger shake slightly, it exists reactive force in vertical

direction;

6. It has friction force between the parts of the model in experimental system.

4 General Conclusion

Structural analysis is the study of the behaviour of a structure considered as whole

under a loading condition and also, the behaviour of each element to determine how

structure act as whole to the loading and transmit forces down to the ground.

Before the size of structural element can be determined, the structural Mechanics

should come first to determine forces (shear, bending moment) acting on the element

and also their deformation which can result of the collapse of the structure.

Laboratorial experiments are conducted to simulate practical experiences and also

physical knowledge of materials used in construction. The laboratory

experimentations are almost the first practical experience for students specially,

student who did not have the relevant practice knowledge. Therefore, its mythology

and objectives defined to it are very important and should be object of more attention

from the students, tutors and laboratory technicians.


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5 References

1. http://www.scribd.com/doc/54422430/Chapter-35

2. http://elibrary.steel.org.au/shadomx/apps/fms/fmsdownload.cfm?file_uuid=7D299E8E-

1E4F-17FA-CD94-

A54DE88DB1C2&siteName=asi&CFID=1170405&CFTOKEN=53187847

3. Ouahid, Structure III Lecture notes, School of Engineering, University of Greenwich,

2011

4. P. Bhatt, Structures, University of Glasgow, pp567 – 571.


http://elibrary.steel.org.au/shadomx/apps/fms/fmsdownload.cfm?file_uuid=7D299E8E-1E4F-17FA-CD94-A54DE88DB1C2&siteName=asi&CFID=1170405&CFTOKEN=53187847



http://www.scribd.com/doc/54422430/Chapter-35

Structures III


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