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GLOBAL ACADEMY OF TECHNOLOGY Ideal Homes Township, Rajarajeshwarinagar,

Bangalore –560098

NAME :

USN :

SUBJECT :

SUBJECT CODE :

ClASS :

GLOBAL ACADEMY OF TECHNOLOGY Ideal Homes Township, Rajarajeshwarinagar,

Bangalore –560098

Laboratory Certificate

This is to certify that Mr/Ms …………………………………………..

bearing USN………………………of the Department of Civil Engineering–

M-tech Structural Engineering has satisfactorily completed the course of experiments in STRUCTURAL ENGINEERING LAB–II (16CSEL26)

prescribed by Visvesvaraya Technological University, Belgaum in the

laboratory of this college in the year 2016–17.

Signature of the Teacher In-Charge

Signature of Head of Department

Marks

Max. Obtained

Date: ..

CONTENTS SLNO PARTICULARS PAGE

NUMBERS

1 INTRODUCTION 1

2 ANALYSIS USING ETABS 2

3 PROCEDURE ETABS 3

4 STATIC ANALYSIS 11

5 DYNAMIC ANALYSIS 25

6 WIND ANALYSIS 29

7 EXPERIMENT -1 IN ETABS 34

8 EXPERIMENT-2 IN ETABS 36

9 ANALYSIS USING STADD PRO V8I 39

10 EXPERIMENT-1 TRUSS ANALYSIS 41

11 EXPERIMENT-2 TRUSS ANALYSIS 50

12 EXPERIMENT-3 PLATES ANALYSIS 77

13 EXPERIMENT-3 SHELLS ANALYSIS 89

14 EXCEL SHEET DESIGNS – DESIGN OF COLUMNS 95

15 EXCEL SHEET DESIGNS – DESIGN OF DOUBLY REINFORCEMENT BEAM 97

16 EXCEL SHEET DESIGNS – DESIGN OF ONE WAY CONTINUOUS SLAB 99

17 EXCEL SHEET DESIGNS – DESIGN OF ONE WAY SLAB 102

18 EXCEL SHEET DESIGNS – DESIGN OF SINGLY REINFORCEMENT BEAM 104

19 EXCEL SHEET DESIGNS – DESIGN OF TWO WAY SLAB 106

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DEPARMENT OF CIVIL ENGINEERING GAT-2016-17 Page 1

STRUCTURAL ANAYSIS AND DESIGN

In Structural Engineering Lab-2, the various structures are being analysed and

designed using the computer programmd software‟s like ETABS and STAAD Pro.

ETABS

Extended Three Dimensional Analysis of Building Systems, is a sophisticated, yet

easy to use, special purpose analysis and design program developed specifically for

building design.

ETABS is a powerful program that can greatly enhance an engineer's analysis and

design capabilities for structures. Part of that power lies in an array of options and

features. The other part lies in how simple it is to use. The basic approach for using

the program is very straightforward. The user establishes grid lines, places

structural objects relative to the grid lines using points, lines and areas, and assigns

loads and structural properties to those structural objects (for example, a line object

can be assigned section properties; a point object can be assigned spring properties;

an area object can be assigned slab or deck properties). Analysis and design are

then performed based on the structural objects and their assignments. Results are

generated in graphical or tabular form that can be printed to a printer or to a file for

use in other programs.

The Structural analysis is the ETABS software can be performed in 3 stages:

Modelling of structure.

Analysis of structure.

Design of the structural elements.

The following provides a broad overview of the basic

modelling, analysis, and design processes:

1. Set the units.

2. Open a file.

3. Set up grid lines.

4. Define story levels.

5. Draw structural objects.

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6. Define frame properties.

7. Define loads.

8. Edit the model geometry.

9. Assign properties.

10. View the model.

11. Analyse the model.

12. Display results for checking.

13. Design the model.

14. Generate output.

15. Save the model.

Modelling of Structure in ETABS

Modelling of a building in ETABS can be achieved in two ways:

1. Importing .DXF file.

2. Creating model with Grid lines.

Importing the .DXF file:

Run the ETABS software, select new model.

In Model Initialization tab, Input the use built in setting.

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In “New model quick

templates” tab, Go to File

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Slab sections (for slab, plates etc.)

Wall sections (for masonry wall, shear wall etc.)

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• Defining frame sections

Define, select section properties, select

frame sections In frame properties

Add new property, select section shape as per

requirement. Both concrete or steel can be

defined, even composite.

Select the

section, input

Property

name:

Example: B 230 X 450 mm

for beams C 230 X

600 mm for

columns

Material: Material defined in

previous step. Select colour for

display

Notes: Any special note if required

Input dimensions of member, i.e., depth and width

In reinforcement, select modify/show rebar, select design type (beam or

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column), select Assign rebar material, select input rebar and shear cover details.

• Defining slab sections:

Go to define, select section properties, select

slab sections “Slab properties” tab is displayed

Add new property, select input

Property name: 2-way slab (or) 1-way slab (as per

requirement) Slab material: Required concrete grade

Modelling type: Shell thin (or) shell thick (or) membrane

(or) layered Display colour: Choose colour

In property data: type = slab

thickness = 125mm (minimum)

• Defining wall sections:

Go to define, select section properties, select

wall sections “Wall properties” tab is displayed

Add new property,

select input

Property

name:

Property

type:

specified

Wall material: As per

requirement Modelling

type: As per requirement

Specify colour, notes if required

Property data: thickness= mm

In define, section properties other sections like deck, reinforcement bars,

link/support can also be defined.

After defining material and section properties, use left hand side tool bar to draw

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beams, slab, shear walls, columns etc.

After completing 3D model as per architectural plan, provide releases for

beams. Go to assign, select frame, select release fixity.

After completing the modelling and providing release check the model.

Go to analyse, select check model

In check model tab, enter the length

tolerance as 15mm Tick all conditions.

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No warnings or errors should be generated.

If warnings are generated, eliminate the errors

Go to options, select show model explorer and go to structural objects,

check the error of beam, slab (or) column in respective floors, rectify it by

re-drawing.

Zero errors and warning should be generated.

• Assigning loads to structure:

Go to, select define, select load patterns.

Dead load (only self-weight) is generated

automatically Self-weight multiplier

= 1

Add live load

Self-weight multiplier = 0

Add super dead load, self-weight multiplier = 0

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• To assign loads on members:

Go to assign, select frame loads, select type of load i.e., point or UDL or

temperature load, select load patterns as super dead or live and enter the value of

load, and click on apply.

Similarly, joint loads and shell loads can be applied.‟

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SEISMIC ANALYSIS IN ETABS

There are two types of analysis in ETABS

(i) Linear analysis

(ii) Non-

linear analysis

Linear analysis

consists of:

• Static analysis

• Dynamic analysis

o Response spectrum

o Time history analysis

Non-linear analysis consists of:

• Geometric non-linearity

o P-delta analysis

• Material non-linearity

o Creep and shrinkage analysis

STATIC ANALYSIS:

Go to define, select load patterns

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Add new load, name it as EQ-X, select seismic type, self-material multiplier = 0,

select auto lateral load and select IS 1893:20002, and click on add new load.

Similarly add new load as EQ-Y

Now select EQ-X, select modified lateral load, i.e., select direction only in X-

direction i.e., X direction, X-direction + eccentricity, X-direction – eccentricity.

Eccentric ratio = 5% = 0.05

Storey range

o Top is OHT and terrace

o Bottom is base

Factors

o Response reduction, R (Table 7 IS 1893:2002 part -1)

o Seismic zone factor, Z (Table 2 IS 183:2002 part -1)

o Site type (fig. 2 and clause 6.4.5 IS 1893:2002 part – 1)

o Importance factor (Table 6 IS 1893:2002 part – 1)

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Time period

o Select user defined, T (clause 7.6 IS 1893:2002 part – 1)

Modify in similar steps for EQ-Y

Add load combinations

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Go to define, select load combinations, select add default design combinations.

Run Analysis. After analysis

Go to display, select response

plots Select display

type: storey shears

Case/combo: EQ-X, get vb i.e.,

base shear EQ-Y, get vb

vb(base shear) for EQ-X

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vb(base shear) for EQ-Y

DYNAMIC RESPONSE SPECTRUM ANALYSIS IN ETABS:

To start up response spectrum complete the static analyse for structure and get Vb

values.

Steps in ETABS: Response spectrum

Go to define, select functions, select

response spectrum Define response spectrum

functions tab is displayed

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Choose Indian code IS 1893:2002

Add new function, select function name:

Earthquake Damping ratio = 0.05

Zone factor, Z Soil type

Plot option should be linear X – Linear Y always

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Go to define, select mass source, select add new mass source and name it as:

spectrum mass source

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Select spectrum mass source: specified load

patterns only Provide mass multipliers

Dead = 1

Live = 0.25 (or) 0.5 (table 8, IS

1893:2002 part – 1) Mass options: Tick

Include lateral mass and

Lump lateral mass @ storey levels

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Go to define, select load cases, select add new case

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Provide load case name: Spectrum-X

Load case type:

Response spectrum In

loads applied, select add

Load type – acceleration

Load name – U1 (for X and lie

for Y) Function: Earthquake (as

defined in step 1) Scale factor:

1000 (initially)

In other

parameters:

Modal load

case: Modal

Modal combination method: CQC

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Similarly create spectrum Y

Run analysis After analysis

Go to display, select response

plots Select display

type: storey shears

Case/combo: spectrum X, get Vb i.e., base shear

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Spectrum Y, get Vb

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The base shear obtained in static analysis (Vb) should be higher than base shear

in dynamic analysis (Vb) (refer clause 7.8.2 IS 1893:2002 part – 1)

Get ratios of Vb/vb for both X

and Y directions. For X-direction

= 761.28/745.33 = 1.021

For Y-direction = 568.86/562.97

= 1.014

Go to define, select load cases, select spectrum X, select

modify/show case Change scale factor to ratio value i.e.,

if Vb/vb = 1.021, enter scale factor Similarly change in

spectrum Y

Run analysis

After analysis, the maximum short-term displacement in a structure

shouldn‟t exceed 7mm. If exceeded 7mm, recheck the whole model.

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WIND ANALYSIS:

To apply wind load for

a structure Create a

diaphragm:

Go to define, select diaphragms, select add new diaphragm

o Name the diaphragm

o Select rigidity to be „rigid‟

Go to 3D view, select entire structure

Go to assign, select shell, select

diaphragms Select created

diaphragm

Apply

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Diaphragm of the structure:

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Go to define, select load patterns

Create „Wind X‟ in load, with

type „wind‟ Self-weight multiplier

„0‟, code IS 875:1987 Similarly,

for „Y‟

Modify lateral load by selecting wind X

Exposure and pressure co-efficients = exposure from extents of

diaphragms Wind co-efficients refer (IS 875:1987 part 2)

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Exposure height

Top storey = terrace/OHT

Bottom storey = GF/above

ground level Compulsorily include

parapet wall height Windward

exposure parameters

Windward co-efficient Cp (refer table 4 IS 875:1987

part 3) Leeward co-efficient Cp (refer table 4 IS

875:1987 part 3)

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EXPERIMENT:1

1. Run the program/software, select new model.

2. Provide country and code provisions as in previous method.

3. In “new model quick templates” tab,

4. Select grid dimensions, provide no. of grid lines in X and Y

directions based on architectural plan.

5. Specify spacing and labelling if necessary (or) Use custom grid spacing

6. Edit grid data, provide spacing between X and Y grids as per

structural layout of structure.

7. Specify no. of stories, storey height.

8. After importing or creating grid lines, the accurate 3D-model can

be created in ETABS.

9. To create the model, Material properties and section properties for

beam, column and slabs should be defined as shown in procedure in the

beginning.

10. Check the model, giving the tolerance as 15mm, no warnings should be

generated.

11. Assign the loads to the structure as shown in the procedure above and

carry on with the analysis.

12. The results are as shown below.

The plan of the structure in ETABS:

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Wall loads on the structure:

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Slab loads on structures with wall loads:

Deflections of the floor plan:

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Bending moments for the floor plan:

Steel distribution for the floor plan:

3D view of the structure:

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EXPERIMENT:2

1. Run the program/software, select new model.

2. Provide country and code provisions as in previous method.

3. In “new model quick templates” tab,

4. Select grid dimensions, provide no. of grid lines in X and Y

directions based on architectural plan.

5. Specify spacing and labelling if necessary (or) Use custom grid spacing

6. Edit grid data, provide spacing between X and Y grids as per

structural layout of structure.

7. Specify no. of stories, storey height.

8. After importing or creating grid lines, the accurate 3D-model can be

created in ETABS.

9. To create the model, Material properties and section properties for beam,

column and slabs should be defined as shown in procedure in the

beginning.

10. Check the model, giving the tolerance as 15mm, no warnings should be

generated.

11. Assign the loads to the structure as shown in the procedure above and

carry on with the analysis.

12. The results are as

shown below. The plan of

the structure in ETABS:

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Wall loads on the structure:

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Slab loads on structures with wall loads:

Deflections of the floor plan:

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Bending moments for the floor plan:

Steel distribution for the floor plan:

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3D view of the structure:

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ANALYSIS IN STAAD PRO V8i

The STAAD.Pro V8i Graphical User Interface (GUI) is normally used to create all input

specifications and all output reports and displays (See the Graphical Environment manual).

These structural modeling and analysis input specifications are stored in STAAD input file – a

text file with extension, .STD. When the GUI opens an existing model file, it reads all of the

information necessary from the STAAD input file. You may edit or create this STAAD input

file and then the GUI and the analysis engine will both reflect the changes. The STAAD input

file is processed by the STAAD analysis “engine” to produce results that are stored in several

files (with file extensions such as ANL, BMD, TMH, etc.). The STAAD analysis text file (file

extension .ANL) contains the printable output as created by the specifications in this manual.

The other files contain the results (displacements, member/element forces, mode shapes,

section forces/moments/displacements, etc.) that are used by the GUI in the post processing

mode.

In Structural Engineering Lab-2, Following structures are being analysed and designed using

STAAD Pro.

1. PLATE

A plate is a structural element which is characterized by two key properties. Firstly, its geometric

configuration is a three-dimensional solid whose thickness is very small when compared with other

dimensions. Secondly, the effects of the loads that are expected to be applied on it only generate

stresses whose resultants are, in practical terms, exclusively normal to the element's thickness. Thin

plates are initially flat structural members bounded by two parallel planes, called faces, and a

cylindrical surface, called an edge or boundary. The generators of the cylindrical surface are

perpendicular to the plane faces. The distance between the plane faces is called the thickness (h) of

the plate. It will be assumed that the plate thickness is small compared with other characteristic

dimensions of the faces (length, width, diameter, etc.). Geometrically, plates are bounded either by

straight or curved boundaries. The static or dynamic loads carried by plates are predominantly

perpendicular to the plate faces

2. SHELLS

A shell is a curved surface, which by virtue of their shape can withstand both membrane and

bending forces. A shell structure can take higher loads if, membrane stresses are predominant,

which is primarily caused due to in-plane forces (plane stress condition). However, localized

bending stresses will appear near load concentrations or geometric discontinuities. The shells are

analogous to cable or arch structure depending on whether the shell resists tensile or,

compressive stresses respectively. Few advantages using shell elements are given below.

https://en.wikipedia.org/wiki/Structural_element

https://en.wikipedia.org/wiki/Net_force

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A. Higher load carrying capacity

B. Lesser thickness and hence lesser dead load

C. Lesser support requirement

D. Larger useful space

E. Higher aesthetic value.

3. TRUSS

In engineering, a truss is a structure that "consists of two-force members only, where the

members are organized so that the assemblage as a whole behaves as a single object".A "two-

force member" is a structural component where force is applied to only two points. Although

this rigorous definition allows the members to have any shape connected in any stable

configuration, trusses typically comprise five or more triangular units constructed with straight

members whose ends are connected at joints referred to as nodes.

In this typical context, external forces and reactions to those forces are considered to act only at

the nodes and result in forces in the members that are either tensile or compressive. For straight

members, moments (torques) are explicitly excluded because, and only because, all the joints in

a truss are treated as revolutes, as is necessary for the links to be two-force members.

A planar truss is one where all members and nodes lie within a two dimensional plane, while a

space truss has members and nodes that extend into three dimensions. The top beams in a truss

are called top chords and are typically in compression, the bottom beams are called bottom

chords, and are typically in tension. The interior beams are called webs, and the areas inside the

webs are called panels.

4. IN PLANE AND OUT PLANE LOADING:

In plane bending moment means the plate bends in its own plane such as a shear wall with

horizontal and vertical forces which are applied to its plane and thus produce in plane bending

moments. Out of plane bending moments are those which are caused by out of plane forces

such a building slab.

https://en.wikipedia.org/wiki/Architectural_structure

https://en.wikipedia.org/wiki/Vertex_(geometry)

https://en.wikipedia.org/wiki/Tension_(physics)

https://en.wikipedia.org/wiki/Compression_(physics)

https://en.wikipedia.org/wiki/Torque

https://en.wikipedia.org/wiki/Revolute_joint

https://en.wikipedia.org/wiki/Compression_(physics)

https://en.wikipedia.org/wiki/Tension_(physics)

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EXPERMENT NO: 1 TRUSSES WITH WIND LOAD

Analyse the truss each of 3m height and 10m width for wind load. the structure has to be

designed in zone-ii with basic wind speed 33 m/sec. terrain category-2, class of structure

category-b, topography flat(θ<3°). design the frame for wind load and various

combinations by using relevant software.

AIM:-

To analyse the given structure for various combinations of load analyzing for wind load

resistant design of the structure.

DATA:-

According to is875 (part-

3):1987 Zone ii

Basic wind speed=33 m/sec

Assuming the life of structure as 50

years K1= 1 (according to table 1)

K3= 1 (topography flat (θ<3°))

K2=0.98upto 10m height (according to table 2), k2=0.996 for 12m height (according to

table 2)

HEIGHT (m)

Vb

(m/sec

)

K1 K2 K3 Vz=Vb*K1*K2*K

3 (N/m2) Pz=0.6*Vz2(N/m2) Pz(KN/m2)

3 33 1 0.98 1 32.34 627.53 0.627

6 33 1 0.98 1 32.34 627.53 0.627

9 33 1 0.98 1 32.34 627.53 0.627

12 33 1 0.996 1 32.868 648.183 0.648

`SOFTWARE PACKAGE USED

STAAD–PRO V8i

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PROCEDURE

1. Draw height of the truss as 6m and width as 10m

2. Go to geometry and select howe roof apply parameters

3. Select all and generate translation repeat along global z direction

4. Go to support and create fixed support at bottom and pinned at joints

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5. Go to loads and definitions and define wind load 1st in definitions.

6. Then in loads and definitions define types of loads in load case details.

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7. Then assign the material as concrete for columns and steel for truss

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8. Analyses the structure and go to post processing mode to take out the maximum

bending and shear force.

`

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RESULTS OF SELECTED FRAME:-

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Axial Shear Torsion Bending

Beam Node L/C Fx

(kN)

Fy

(kN)

Fz

(kN)

Mx

(kNm)

My

(kNm)

Mz

(kNm)

1 3 8:COMBINATION LOAD CASE 1.077 2.806 - -0.936 1.267 0.864 9 8:COMBINATION LOAD CASE -1.077 2.072 1.255 0.936 0.824 -0.253 2 1 8:COMBINATION LOAD CASE 8.779 - 0.831 0.206 -1.659 -0.427 3 8:COMBINATION LOAD CASE 8.779 0.215 - -0.206 -3.327 -0.864 3 2 8:COMBINATION LOAD CASE 8.779 0.215 0.831 -0.206 -1.659 0.427 4 8:COMBINATION LOAD CASE 8.779 - - 0.206 -3.327 0.864 4 7 8:COMBINATION LOAD CASE 1.916 2.804 - 0.936 1.266 0.864 19 8:COMBINATION LOAD CASE -1.916 2.073 1.253 -0.936 0.823 -0.255 5 5 8:COMBINATION LOAD CASE 8.779 - - 0.206 1.659 -0.427 7 8:COMBINATION LOAD CASE 8.779 0.215 0.831 -0.206 3.327 -0.864 6 6 8:COMBINATION LOAD CASE 8.779 0.215 - -0.206 1.659 0.427 8 8:COMBINATION LOAD CASE 8.779 - 0.831 0.206 3.327 0.864 7 3 8:COMBINATION LOAD CASE 0.000 8.779 0.353 -0.000 -1.061 2.391 7 8:COMBINATION LOAD CASE 0.000 8.779 - 0.000 -1.060 -2.391 8 4 8:COMBINATION LOAD CASE 0.000 8.779 - 0.000 1.061 2.391 8 8:COMBINATION LOAD CASE 0.000 8.779 0.353 -0.000 1.060 -2.391 9 9 8:COMBINATION LOAD CASE 1.553 2.595 - -0.259 0.603 0.253 10 8:COMBINATION LOAD CASE -1.553 2.283 1.044 0.259 1.136 0.007 10 10 8:COMBINATION LOAD CASE -2.630 1.753 - -0.053 -0.153 -0.007

11 8:COMBINATION LOAD CASE 2.630 3.124 0.530 0.053 1.036 -1.135 11 11 8:COMBINATION LOAD CASE -2.630 3.124 0.530 0.053 -1.036 1.135

12 8:COMBINATION LOAD CASE 2.630 1.753 - -0.053 0.153 0.007 12 12 8:COMBINATION LOAD CASE 1.553 2.283 1.044 0.259 -1.136 -0.007

13 8:COMBINATION LOAD CASE -1.553 2.595 - -0.259 -0.603 -0.253 13 13 8:COMBINATION LOAD CASE 1.077 2.072 1.255 0.936 -0.824 0.253

4 8:COMBINATION LOAD CASE -1.077 2.806 - -0.936 -1.267 -0.864 14 19 8:COMBINATION LOAD CASE 1.441 2.599 - 0.258 0.603 0.255

20 8:COMBINATION LOAD CASE -1.441 2.278 1.046 -0.258 1.141 0.012 15 20 8:COMBINATION LOAD CASE -3.357 1.748 - 0.054 -0.161 -0.012

21 8:COMBINATION LOAD CASE 3.357 3.129 0.515 -0.054 1.019 -1.139 16 21 8:COMBINATION LOAD CASE -3.357 3.129 0.515 -0.054 -1.019 1.139

22 8:COMBINATION LOAD CASE 3.357 1.748 - 0.054 0.161 0.012 17 22 8:COMBINATION LOAD CASE 1.441 2.278 1.046 -0.258 -1.141 -0.012

23 8:COMBINATION LOAD CASE -1.441 2.599 - 0.258 -0.603 -0.255 18 23 8:COMBINATION LOAD CASE 1.916 2.073 1.253 -0.936 -0.823 0.255

8 8:COMBINATION LOAD CASE -1.916 2.804 - 0.936 -1.266 -0.864 19 3 8:COMBINATION LOAD CASE 0.053 0.088 0.000 0.000 0.000 0.000

14 8:COMBINATION LOAD CASE 0.053 0.088 0.000 0.000 0.000 0.000 20 14 8:COMBINATION LOAD CASE 0.053 0.088 0.000 0.000 0.000 0.000


16 8:COMBINATION LOAD CASE 0.053 0.088 0.000 0.000 0.000 0.000 22 16 8:COMBINATION LOAD CASE -0.053 0.088 0.000 0.000 0.000 0.000

17 8:COMBINATION LOAD CASE -0.053 0.088 0.000 0.000 0.000 0.000 23 17 8:COMBINATION LOAD CASE -0.053 0.088 0.000 0.000 0.000 0.000


4 8:COMBINATION LOAD CASE -0.053 0.088 0.000 0.000 0.000 0.000 25 9 8:COMBINATION LOAD CASE - 0.000 0.000 0.000 0.000 0.000

14 8:COMBINATION LOAD CASE 13.550 0.000 0.000 0.000 0.000 0.000 26 10 8:COMBINATION LOAD CASE - 0.000 0.000 0.000 0.000 0.000








`

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8 8:COMBINATION LOAD CASE -0.053 0.088 0.000 0.000 0.000 0.000 40 19 8:COMBINATION LOAD CASE - 0.000 0.000 0.000 0.000 0.000









28 8:COMBINATION LOAD CASE 5.265 0.088 0.000 0.000 0.000 0.000 49 9 8:COMBINATION LOAD CASE 0.207 8.779 0.476 0.000 -1.427 0.678

19 8:COMBINATION LOAD CASE -0.207 8.779 - -0.000 -1.426 -0.678 50 10 8:COMBINATION LOAD CASE 0.532 8.779 0.327 0.000 -0.983 0.205

20 8:COMBINATION LOAD CASE -0.532 8.779 - -0.000 -0.981 -0.205 51 11 8:COMBINATION LOAD CASE 1.030 8.779 0.000 -0.000 -0.000 0.107

21 8:COMBINATION LOAD CASE -1.030 8.779 - 0.000 -0.000 -0.107 52 12 8:COMBINATION LOAD CASE 0.532 8.779 - -0.000 0.983 0.205

22 8:COMBINATION LOAD CASE -0.532 8.779 0.327 0.000 0.981 -0.205 53 13 8:COMBINATION LOAD CASE 0.207 8.779 - -0.000 1.427 0.678

23 8:COMBINATION LOAD CASE -0.207 8.779 0.476 0.000 1.426 -0.678 54 14 8:COMBINATION LOAD CASE 0.000 0.315 0.000 0.000 0.000 0.000





28 8:COMBINATION LOAD CASE 0.000 0.315 0.000 0.000 0.000 0.000

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`

EXPERMENT NO: 2

TRUSSES WITH WIND LOAD AND STEEL DESIGN

PROBLEM:-

Analyse the 2 storey truss each of 3m height and 10m width for wind load. the structure has to

be designed in zone-ii with basic wind speed 33 m/sec. terrain category-2, class of structure

category-b, topography flat(θ<3°). design the frame for wind load and various combinations

by using relevant software.

AIM:-

To analyse the given structure for various combinations of load analyzing for wind load

resistant design of the structure.

DATA:-

According to is875 (part-3):1987

Zone (ii)

Basic wind speed=33 m/sec

Assuming the life of structure as 50 years

K1= 1 (according to table 1)

K3= 1 (topography flat (θ<3°))

K2=0.98upto 10m height (according to table 2), k2=0.996 for 12m height (according to table

2)

HEIGHT

(m) Vb

(m/sec)

K1 K2 K3 Vz=Vb*K1*K2*K3

(N/m2)

Pz=0.6*Vz2(N/m2) Pz(KN/m2)

3 33 1 0.98 1 32.34 627.53 0.627

6 33 1 0.98 1 32.34 627.53 0.627

9 33 1 0.98 1 32.34 627.53 0.627

12 33 1 0.996 1 32.868 648.183 0.648

STRUCTURAL LAB II


`

SOFTWARE PACKAGE USED

STAAD–PRO V8i

PROCEDURE:-

1. Select the space specification and select the units (kn & m).

2. Click on next and select the open structure wizard to create geometry.

STRUCTURAL LAB II


`

3. Go to general and assign the steel section using section data base for beams and

columns and angles for trusses in property to the member and trusses.

4. In specifications assign the members as trusses so that its designed as truss member

STRUCTURAL LAB II


`

5. Go to supports and create fixed, pinned and roller supports for different end

conditions.

6. Go to loads and definitions and define wind load 1st in definitions.

STRUCTURAL LAB II


`

7. Then in loads and definitions define types of loads in load case details.

8. Define the combinations according to is 456-2000

A. 1.2dl+1.2wlx+ve

B. 1.2dl+1.2wlx-ve

C. 1.2dl+1.2wlz+ve

D. 1.2dl+1.2wlz-ve

9. Then assign the material as steel.

STRUCTURAL LAB II


STRUCTURAL LAB II


`

10. Analyse the structure and go to design of steel design.

11. In the steel design

A. check code

B. member take off

C. select

D. take off

assign all these to design the member and take out the results

STRUCTURAL LAB II


`

12. Results from output file sheet.

STRUCTURAL LAB II


STRUCTURAL LAB II


`

STRUCTURAL LAB II


PROBLEM:-

EXPERIMENT No: 3

PLATE ANALYSIS

Analyse a square plate of size 10m * 10m with two edges are fixed and subjected to uniform

distributed load over entire surface and to find out membrane forces and by using relevant

software.

AIM:-

To analyse the given plate for in-plane loading and out-plane loading.


STAAD–PRO V8i

PROCEDURE:-

INPLANE LOADING:


STRUCTURAL LAB II


`


3. Go to general and assign the rectangular property to the member.

STRUCTURAL LAB II


`

4. Go to supports and create a fixed support at the bottom.

5. Then in loads and definitions define types of loads in load case details out plane load applied

along local co-ordinates

STRUCTURAL LAB II


6. Then assign the material as concrete.

7. Analyse the structure and go to post processing mode to take out the maximum force.

`

STRUCTURAL LAB II


STRUCTURAL LAB II


RESULTS OF OUT-PLANE LOADING:-

Shear Membrane Bending

Plate L/C Qx

(N/mm2)

Qy

(N/mm2)

Sx

(N/mm2)

Sy

(N/mm2)

Sxy

(N/mm2) Mx

(kNm/m)

My

(kNm/m)

Mxy

(kNm/m)

2 1:DL 0.158 -0.009 0.000 0.000 0.000 -41.516 -5.569 -0.181

3 1:DL 0.127 -0.006 0.000 0.000 0.000 -14.071 -1.110 1.789

4 1:DL 0.096 0.002 0.000 0.000 0.000 6.626 -0.439 1.942

5 1:DL 0.060 0.004 0.000 0.000 0.000 20.441 0.022 1.407

6 1:DL 0.020 0.006 0.000 0.000 0.000 27.350 0.211 0.506

7 1:DL -0.020 0.006 0.000 0.000 0.000 27.350 0.211 -0.506

8 1:DL -0.060 0.004 0.000 0.000 0.000 20.440 0.022 -1.407

9 1:DL -0.096 0.002 0.000 0.000 0.000 6.626 -0.439 -1.942

10 1:DL -0.127 -0.006 0.000 0.000 0.000 -14.071 -1.110 -1.789

11 1:DL -0.158 -0.009 0.000 0.000 0.000 -41.516 -5.569 0.181

12 1:DL 0.154 0.005 0.000 0.000 0.000 -40.933 -7.112 0.418

13 1:DL 0.114 -0.001 0.000 0.000 0.000 -13.659 -2.802 0.893

14 1:DL 0.076 0.001 0.000 0.000 0.000 6.585 0.020 1.053

15 1:DL 0.044 0.003 0.000 0.000 0.000 20.098 1.523 0.820

16 1:DL 0.015 0.004 0.000 0.000 0.000 26.859 2.198 0.303

17 1:DL -0.015 0.004 0.000 0.000 0.000 26.859 2.198 -0.303

18 1:DL -0.044 0.003 0.000 0.000 0.000 20.098 1.523 -0.820

19 1:DL -0.076 0.001 0.000 0.000 0.000 6.585 0.020 -1.053

20 1:DL -0.114 -0.001 0.000 0.000 0.000 -13.659 -2.802 -0.893

21 1:DL -0.154 0.005 0.000 0.000 0.000 -40.933 -7.112 -0.418

22 1:DL 0.148 0.004 0.000 0.000 0.000 -40.187 -6.930 0.226

23 1:DL 0.116 0.001 0.000 0.000 0.000 -13.449 -2.687 0.500

24 1:DL 0.083 0.001 0.000 0.000 0.000 6.572 0.325 0.464

25 1:DL 0.050 0.002 0.000 0.000 0.000 19.874 2.243 0.335

26 1:DL 0.017 0.002 0.000 0.000 0.000 26.529 3.159 0.123

27 1:DL -0.017 0.002 0.000 0.000 0.000 26.529 3.159 -0.123

28 1:DL -0.050 0.002 0.000 0.000 0.000 19.874 2.243 -0.335

29 1:DL -0.083 0.001 0.000 0.000 0.000 6.572 0.325 -0.464

30 1:DL -0.116 0.001 0.000 0.000 0.000 -13.449 -2.687 -0.500

31 1:DL -0.148 0.004 0.000 0.000 0.000 -40.187 -6.930 -0.226

32 1:DL 0.148 0.002 0.000 0.000 0.000 -39.796 -6.821 0.099

33 1:DL 0.115 0.001 0.000 0.000 0.000 -13.360 -2.512 0.241

34 1:DL 0.082 0.001 0.000 0.000 0.000 6.539 0.629 0.241

35 1:DL 0.049 0.001 0.000 0.000 0.000 19.801 2.690 0.167

36 1:DL 0.016 0.001 0.000 0.000 0.000 26.424 3.706 0.059

37 1:DL -0.016 0.001 0.000 0.000 0.000 26.424 3.706 -0.059

38 1:DL -0.049 0.001 0.000 0.000 0.000 19.801 2.690 -0.167

39 1:DL -0.082 0.001 0.000 0.000 0.000 6.539 0.629 -0.241

40 1:DL -0.115 0.001 0.000 0.000 0.000 -13.360 -2.512 -0.241

41 1:DL -0.148 0.002 0.000 0.000 0.000 -39.796 -6.821 -0.099

42 1:DL 0.148 0.001 0.000 0.000 0.000 -39.644 -6.770 0.026

43 1:DL 0.115 0.000 0.000 0.000 0.000 -13.292 -2.405 0.067

44 1:DL 0.082 0.000 0.000 0.000 0.000 6.533 0.795 0.069

45 1:DL 0.049 0.000 0.000 0.000 0.000 19.764 2.899 0.047

46 1:DL 0.016 0.000 0.000 0.000 0.000 26.378 3.945 0.016

47 1:DL -0.016 0.000 0.000 0.000 0.000 26.378 3.945 -0.016

48 1:DL -0.049 0.000 0.000 0.000 0.000 19.764 2.899 -0.047

49 1:DL -0.082 0.000 0.000 0.000 0.000 6.533 0.795 -0.069

50 1:DL -0.115 0.000 0.000 0.000 0.000 -13.292 -2.405 -0.067

51 1:DL -0.148 0.001 0.000 0.000 0.000 -39.644 -6.770 -0.026

52 1:DL 0.148 -0.001 0.000 0.000 0.000 -39.644 -6.770 -0.026

53 1:DL 0.115 -0.000 0.000 0.000 0.000 -13.292 -2.405 -0.067

54 1:DL 0.082 -0.000 0.000 0.000 0.000 6.533 0.795 -0.069

55 1:DL 0.049 -0.000 0.000 0.000 0.000 19.764 2.899 -0.047

56 1:DL 0.016 -0.000 0.000 0.000 0.000 26.378 3.945 -0.016

57 1:DL -0.016 -0.000 0.000 0.000 0.000 26.378 3.945 0.016

58 1:DL -0.049 -0.000 0.000 0.000 0.000 19.764 2.899 0.047

59 1:DL -0.082 -0.000 0.000 0.000 0.000 6.533 0.795 0.069

60 1:DL -0.115 -0.000 0.000 0.000 0.000 -13.292 -2.405 0.067

61 1:DL -0.148 -0.001 0.000 0.000 0.000 -39.644 -6.770 0.026

62 1:DL 0.148 -0.002 0.000 0.000 0.000 -39.796 -6.821 -0.099

STRUCTURAL LAB II


`

63 1:DL 0.115 -0.001 0.000 0.000 0.000 -13.360 -2.512 -0.241

64 1:DL 0.082 -0.001 0.000 0.000 0.000 6.539 0.629 -0.241

65 1:DL 0.049 -0.001 0.000 0.000 0.000 19.801 2.690 -0.167

66 1:DL 0.016 -0.001 0.000 0.000 0.000 26.424 3.706 -0.059

67 1:DL -0.016 -0.001 0.000 0.000 0.000 26.424 3.706 0.059

68 1:DL -0.049 -0.001 0.000 0.000 0.000 19.801 2.690 0.167

69 1:DL -0.082 -0.001 0.000 0.000 0.000 6.539 0.629 0.241

70 1:DL -0.115 -0.001 0.000 0.000 0.000 -13.360 -2.512 0.241

71 1:DL -0.148 -0.002 0.000 0.000 0.000 -39.796 -6.821 0.099

72 1:DL 0.148 -0.004 0.000 0.000 0.000 -40.187 -6.930 -0.226

73 1:DL 0.116 -0.001 0.000 0.000 0.000 -13.449 -2.687 -0.500

74 1:DL 0.083 -0.001 0.000 0.000 0.000 6.572 0.325 -0.464

75 1:DL 0.050 -0.002 0.000 0.000 0.000 19.874 2.243 -0.335

76 1:DL 0.017 -0.002 0.000 0.000 0.000 26.529 3.159 -0.123

77 1:DL -0.017 -0.002 0.000 0.000 0.000 26.529 3.159 0.123

78 1:DL -0.050 -0.002 0.000 0.000 0.000 19.874 2.243 0.335

79 1:DL -0.083 -0.001 0.000 0.000 0.000 6.572 0.325 0.464

80 1:DL -0.116 -0.001 0.000 0.000 0.000 -13.449 -2.687 0.500

81 1:DL -0.148 -0.004 0.000 0.000 0.000 -40.187 -6.930 0.226

82 1:DL 0.154 -0.005 0.000 0.000 0.000 -40.933 -7.112 -0.418

83 1:DL 0.114 0.001 0.000 0.000 0.000 -13.659 -2.802 -0.893

84 1:DL 0.076 -0.001 0.000 0.000 0.000 6.585 0.020 -1.053

85 1:DL 0.044 -0.003 0.000 0.000 0.000 20.098 1.523 -0.820

86 1:DL 0.015 -0.004 0.000 0.000 0.000 26.859 2.198 -0.303

87 1:DL -0.015 -0.004 0.000 0.000 0.000 26.859 2.198 0.303

88 1:DL -0.044 -0.003 0.000 0.000 0.000 20.098 1.523 0.820

89 1:DL -0.076 -0.001 0.000 0.000 0.000 6.585 0.020 1.053

90 1:DL -0.114 0.001 0.000 0.000 0.000 -13.659 -2.802 0.893

91 1:DL -0.154 -0.005 0.000 0.000 0.000 -40.933 -7.112 0.418

92 1:DL 0.158 0.009 0.000 0.000 0.000 -41.516 -5.569 0.181

93 1:DL 0.127 0.006 0.000 0.000 0.000 -14.071 -1.110 -1.789

94 1:DL 0.096 -0.002 0.000 0.000 0.000 6.626 -0.439 -1.942

95 1:DL 0.060 -0.004 0.000 0.000 0.000 20.441 0.022 -1.407

96 1:DL 0.020 -0.006 0.000 0.000 0.000 27.350 0.211 -0.506

97 1:DL -0.020 -0.006 0.000 0.000 0.000 27.350 0.211 0.506

98 1:DL -0.060 -0.004 0.000 0.000 0.000 20.440 0.022 1.407

99 1:DL -0.096 -0.002 0.000 0.000 0.000 6.626 -0.439 1.942

100 1:DL -0.127 0.006 0.000 0.000 0.000 -14.071 -1.110 1.789

101 1:DL -0.158 0.009 0.000 0.000 0.000 -41.516 -5.569 -0.181

STRUCTURAL LAB II


PROCEDURE:-

INPLANE LOADING



STRUCTURAL LAB II


3. Go to general and assign the rectangular property to the member.


5. Then in loads and definitions define types of loads in load case details in plane load

applied along local co-ordinates

STRUCTURAL LAB II


6. Then assign the material as concrete.


STRUCTURAL LAB II


RESULTS OF IN-PLANE LOADING:-

Shear Membrane Bending

Plate L/C Qx

(N/mm2)

Qy

(N/mm2)

Sx

(N/mm2)

Sy

(N/mm2)

Sxy

(N/mm2) Mx

(kNm/m) My

(kNm/m) Mxy

(kNm/m)

2 1:DL 0.000 0.000 -0.062 -0.008 -0.005 0.000 0.000 0.000 3 1:DL 0.000 0.000 -0.045 -0.000 0.000 0.000 0.000 0.000 4 1:DL 0.000 0.000 -0.031 -0.000 0.001 0.000 0.000 0.000 5 1:DL 0.000 0.000 -0.018 -0.000 0.001 0.000 0.000 0.000 6 1:DL 0.000 0.000 -0.006 -0.000 0.001 0.000 0.000 0.000 7 1:DL 0.000 0.000 0.006 0.000 0.001 0.000 0.000 0.000 8 1:DL 0.000 0.000 0.018 0.000 0.001 0.000 0.000 0.000 9 1:DL 0.000 0.000 0.031 0.000 0.001 0.000 0.000 0.000 10 1:DL 0.000 0.000 0.045 0.000 0.000 0.000 0.000 0.000 11 1:DL 0.000 0.000 0.062 0.008 -0.005 0.000 0.000 0.000 12 1:DL 0.000 0.000 -0.060 -0.009 -0.003 0.000 0.000 0.000 13 1:DL 0.000 0.000 -0.048 -0.004 -0.002 0.000 0.000 0.000 14 1:DL 0.000 0.000 -0.034 -0.001 0.001 0.000 0.000 0.000 15 1:DL 0.000 0.000 -0.020 -0.000 0.001 0.000 0.000 0.000 16 1:DL 0.000 0.000 -0.007 -0.000 0.002 0.000 0.000 0.000 17 1:DL 0.000 0.000 0.007 0.000 0.002 0.000 0.000 0.000 18 1:DL 0.000 0.000 0.020 0.000 0.001 0.000 0.000 0.000 19 1:DL 0.000 0.000 0.034 0.001 0.001 0.000 0.000 0.000 20 1:DL 0.000 0.000 0.048 0.004 -0.002 0.000 0.000 0.000 21 1:DL 0.000 0.000 0.060 0.009 -0.003 0.000 0.000 0.000 22 1:DL 0.000 0.000 -0.061 -0.010 -0.002 0.000 0.000 0.000 23 1:DL 0.000 0.000 -0.048 -0.005 -0.001 0.000 0.000 0.000 24 1:DL 0.000 0.000 -0.035 -0.002 0.000 0.000 0.000 0.000 25 1:DL 0.000 0.000 -0.021 -0.001 0.001 0.000 0.000 0.000 26 1:DL 0.000 0.000 -0.007 -0.000 0.002 0.000 0.000 0.000 27 1:DL 0.000 0.000 0.007 0.000 0.002 0.000 0.000 0.000 28 1:DL 0.000 0.000 0.021 0.001 0.001 0.000 0.000 0.000 29 1:DL 0.000 0.000 0.035 0.002 0.000 0.000 0.000 0.000 30 1:DL 0.000 0.000 0.048 0.005 -0.001 0.000 0.000 0.000 31 1:DL 0.000 0.000 0.061 0.010 -0.002 0.000 0.000 0.000 32 1:DL 0.000 0.000 -0.061 -0.010 -0.001 0.000 0.000 0.000 33 1:DL 0.000 0.000 -0.048 -0.006 -0.001 0.000 0.000 0.000 34 1:DL 0.000 0.000 -0.035 -0.003 0.000 0.000 0.000 0.000 35 1:DL 0.000 0.000 -0.021 -0.002 0.001 0.000 0.000 0.000 36 1:DL 0.000 0.000 -0.007 -0.000 0.001 0.000 0.000 0.000 37 1:DL 0.000 0.000 0.007 0.000 0.001 0.000 0.000 0.000 38 1:DL 0.000 0.000 0.021 0.002 0.001 0.000 0.000 0.000 39 1:DL 0.000 0.000 0.035 0.003 0.000 0.000 0.000 0.000 40 1:DL 0.000 0.000 0.048 0.006 -0.001 0.000 0.000 0.000 41 1:DL 0.000 0.000 0.061 0.010 -0.001 0.000 0.000 0.000 42 1:DL 0.000 0.000 -0.061 -0.010 -0.000 0.000 0.000 0.000 43 1:DL 0.000 0.000 -0.048 -0.007 -0.000 0.000 0.000 0.000

STRUCTURAL LAB II


`

44 1:DL 0.000 0.000 -0.035 -0.004 0.000 0.000 0.000 0.000 45 1:DL 0.000 0.000 -0.021 -0.002 0.000 0.000 0.000 0.000 46 1:DL 0.000 0.000 -0.007 -0.001 0.000 0.000 0.000 0.000 47 1:DL 0.000 0.000 0.007 0.001 0.000 0.000 0.000 0.000 48 1:DL 0.000 0.000 0.021 0.002 0.000 0.000 0.000 0.000 49 1:DL 0.000 0.000 0.035 0.004 0.000 0.000 0.000 0.000 50 1:DL 0.000 0.000 0.048 0.007 -0.000 0.000 0.000 0.000 51 1:DL 0.000 0.000 0.061 0.010 -0.000 0.000 0.000 0.000 52 1:DL 0.000 0.000 -0.061 -0.010 0.000 0.000 0.000 0.000 53 1:DL 0.000 0.000 -0.048 -0.007 0.000 0.000 0.000 0.000 54 1:DL 0.000 0.000 -0.035 -0.004 -0.000 0.000 0.000 0.000 55 1:DL 0.000 0.000 -0.021 -0.002 -0.000 0.000 0.000 0.000 56 1:DL 0.000 0.000 -0.007 -0.001 -0.000 0.000 0.000 0.000 57 1:DL 0.000 0.000 0.007 0.001 -0.000 0.000 0.000 0.000 58 1:DL 0.000 0.000 0.021 0.002 -0.000 0.000 0.000 0.000 59 1:DL 0.000 0.000 0.035 0.004 -0.000 0.000 0.000 0.000 60 1:DL 0.000 0.000 0.048 0.007 0.000 0.000 0.000 0.000 61 1:DL 0.000 0.000 0.061 0.010 0.000 0.000 0.000 0.000 62 1:DL 0.000 0.000 -0.061 -0.010 0.001 0.000 0.000 0.000 63 1:DL 0.000 0.000 -0.048 -0.006 0.001 0.000 0.000 0.000 64 1:DL 0.000 0.000 -0.035 -0.003 -0.000 0.000 0.000 0.000 65 1:DL 0.000 0.000 -0.021 -0.002 -0.001 0.000 0.000 0.000 66 1:DL 0.000 0.000 -0.007 -0.000 -0.001 0.000 0.000 0.000 67 1:DL 0.000 0.000 0.007 0.000 -0.001 0.000 0.000 0.000 68 1:DL 0.000 0.000 0.021 0.002 -0.001 0.000 0.000 0.000 69 1:DL 0.000 0.000 0.035 0.003 -0.000 0.000 0.000 0.000 70 1:DL 0.000 0.000 0.048 0.006 0.001 0.000 0.000 0.000 71 1:DL 0.000 0.000 0.061 0.010 0.001 0.000 0.000 0.000 72 1:DL 0.000 0.000 -0.061 -0.010 0.002 0.000 0.000 0.000 73 1:DL 0.000 0.000 -0.048 -0.005 0.001 0.000 0.000 0.000 74 1:DL 0.000 0.000 -0.035 -0.002 -0.000 0.000 0.000 0.000 75 1:DL 0.000 0.000 -0.021 -0.001 -0.001 0.000 0.000 0.000 76 1:DL 0.000 0.000 -0.007 -0.000 -0.002 0.000 0.000 0.000 77 1:DL 0.000 0.000 0.007 0.000 -0.002 0.000 0.000 0.000 78 1:DL 0.000 0.000 0.021 0.001 -0.001 0.000 0.000 0.000 79 1:DL 0.000 0.000 0.035 0.002 -0.000 0.000 0.000 0.000 80 1:DL 0.000 0.000 0.048 0.005 0.001 0.000 0.000 0.000 81 1:DL 0.000 0.000 0.061 0.010 0.002 0.000 0.000 0.000 82 1:DL 0.000 0.000 -0.060 -0.009 0.003 0.000 0.000 0.000 83 1:DL 0.000 0.000 -0.048 -0.004 0.002 0.000 0.000 0.000 84 1:DL 0.000 0.000 -0.034 -0.001 -0.001 0.000 0.000 0.000 85 1:DL 0.000 0.000 -0.020 -0.000 -0.001 0.000 0.000 0.000 86 1:DL 0.000 0.000 -0.007 -0.000 -0.002 0.000 0.000 0.000 87 1:DL 0.000 0.000 0.007 0.000 -0.002 0.000 0.000 0.000 88 1:DL 0.000 0.000 0.020 0.000 -0.001 0.000 0.000 0.000 89 1:DL 0.000 0.000 0.034 0.001 -0.001 0.000 0.000 0.000 90 1:DL 0.000 0.000 0.048 0.004 0.002 0.000 0.000 0.000 91 1:DL 0.000 0.000 0.060 0.009 0.003 0.000 0.000 0.000 92 1:DL 0.000 0.000 -0.062 -0.008 0.005 0.000 0.000 0.000 93 1:DL 0.000 0.000 -0.045 -0.000 -0.000 0.000 0.000 0.000 94 1:DL 0.000 0.000 -0.031 -0.000 -0.001 0.000 0.000 0.000 95 1:DL 0.000 0.000 -0.018 -0.000 -0.001 0.000 0.000 0.000 96 1:DL 0.000 0.000 -0.006 -0.000 -0.001 0.000 0.000 0.000 97 1:DL 0.000 0.000 0.006 0.000 -0.001 0.000 0.000 0.000 98 1:DL 0.000 0.000 0.018 0.000 -0.001 0.000 0.000 0.000 99 1:DL 0.000 0.000 0.031 0.000 -0.001 0.000 0.000 0.000 100 1:DL 0.000 0.000 0.045 0.000 -0.000 0.000 0.000 0.000 101 1:DL 0.000 0.000 0.062 0.008 0.005 0.000 0.000 0.000

STRUCTURAL LAB II


EXPERIMENT No: 4 SHELL ANALYSIS

Analyse a shell where edges are fixed and subjected to uniform distributed load over entire

surface and to find out membrane forces and by using relevant software.

AIM:-

To analyse the given shell structure.


STAAD–PRO V8i

PROCEDURE:-


2. Click on next and select the open structure wizard to create geometry

STRUCTURAL LAB II


3. Go to general and assign the shell property to the member.


STRUCTURAL LAB II


5. Then in Assign material property

6. Then in loads and definitions define types of loads in load case

STRUCTURAL LAB II



STRUCTURAL LAB-II


`

STRUCTURAL LAB-II


E

`

EXCEL DESIGN SHEETS

1. DESIGN OF COLUMN:

20

N/mm2

fy 415 N/mm2

L 3000 mm

Le=0.65L 1950 mm

980 KN

Pu = 1.5P 1470 KN

Dia of bars 16 mm

Ast = (π/4)*d2 200.96 mm2

Ag = Asc + Ac

Asc = 1% Ag 0.01 Ag

Ac = Ag - Asc 0.99 Ag

Ultimate load carried by the coulmn,

Pu = 0.4 fck Ac + 0.67 fy Asc

1470 7.92 Ag 2.7805 Ag

10.7005 Ag

Ag 137376.76 mm2

Let us design a square column

B =D =√Ag 370.64 mm

B = D 375 mm

min eccentricity 20 mm

emin/D 0.053 >0.05 0.05

Therefore, B = D 400 mm

Area of steel required in Asc 1373.77 mm2

Number of bars 6.84

Provide 8 bars of #16mm 8 16mm2

Steel provided 1607.68 mm2

Actual percentage of steel = 100Asc/BD 0.976901391

( more than 0.8% and less than 6%,hence ok.)

Design of Tranverse reinforcement

Diameter of tie = 1/4 dia of main steel 4

provide 6mm lateral ties

spacing 300 mm

16*# 256 mm

least lateral dimension

400

mm

STRUCTURAL LAB-II


spacing 256 mm

provide 2L-6# @ 300mm c/c

STRUCTURAL LAB-II


`

2. DOUBLY REINFORCED BEAM:

fck

fy

N/mm2

N/mm2

L mm 15 m

bearing mm

Clear cover mm

width,b mm 0 m

dia,d mm

density KN/mm3

FOS

span depth ratio

step 1

dimensions

effective depth=length/span

depth ratio # mm

provide effective depth,d ` # mm 0.75

effective cover 30 mm

overall depth # mm 0.78 m

provide overall depth,D # mm 0.8 m

step2 effective Span

effective span=length+2(bearing/2) # mm

effective span=length+effective depth # mm 15.75 m

step3 loads

dead load=width*depth*density of concrete 4 kn/m2

assume live load 2 kn/m2

floor finish 1 kn/m2

total load 7 kn/m2

factored load 10 kn/m2

step 4 bending moment

mu=Factored load* length2 /8

#

N-mm

300.1 KN- m

mulim=0.138fckbd2

# n-mm

# KN-m

Mu lim > Mu, design as under reinfored section

step 5 check for depth , d=√Mu/(0.38*fck*b)

# mm

depth required < depth provided hence safe

STRUCTURAL LAB-II


`

step 6 shear force

shear force,v=factored load*length/2

#

N

77.63

KN

step 7 area of steel

mu/bd2 3 2.7

pt 1

pc 0

Ast=pt*b*d/100 # mm2

ast=(pi/4)d2

no of bars

provide 6-20# bars

#

5

mm2

Asc=Pc*b*d/100 3 mm2

asc=(pi/4)d2

no of bars

provide 2-16# bars

#

0

mm2

step 8 check for shear

nominal shear stress Շv= Vu/bd 1 N/mm2

permisible stress Շ c

Շ c < Շv provide minimum shear

reinforcement Assume 2l-8mm dia stirrups

0

8

area of stirrups, Asv=2((∏/4)d2)

spacing of bars , Sv = 0.87fyAsv/0.4b

shear reinforcement least of 3 values

#

#

#

mm2

mm

0.75d

minimum spacing

provide 2l-8#@300mm c/c

#

#

mm

mm

STRUCTURAL LAB-II


`

3. ONE WAY CONTINOUS SLAB:

GIVEN DATA

Fck 20 N/mm2

Fy 415 N/mm2

L 3000 mm 3 m

Bearing 230 mm

CLEAR COVER 20 mm

Width 1 m 1000 mm

Dia 10 mm

Density of concrete 25 KN/m3

Factor of safety 1.5

Assume live load 2 KN/m2

Floor finish 1 KN/m2

Span/Depth ratio (Ref.is 456-2000 CL23-2.1)

Simply Continuous Cantilever

20 26 7

STEP-1 Dimensions

Effective Depth=Length/(Span/Depth ratio)

0 m

Say effective depth=

0.125 m

125 mm

Effective cover 25 mm

Overall Depth 150 mm

0.15 m

STEP-2 Effective span (Ref: IS 456-2000 CL 22.2)

Effective span=length+2(bearing/2) 0 mm

0 m

Effective span=length+effective depth 125 mm

0.125 m

STEP-3 LOADS

Dead load=width*depth*density of concrete 0 KN/m2


STRUCTURAL LAB-II


`

Floor Finish 1 KN/m2

Total load 3 KN/m2

Factored load ( Wu) 4.5 KN/m2

STEP 4 BENDING MOMENT CO-EFFICIENT ( IS456-2000 )

Type of load

Bending Moment (DL+LL)

LL

Total Moments

Max of both values of middle span and support , Mu

Span moments and support moments for

Bending moment (DL + LL) = 0

LL = 0

Total moments = 0

Max of both values of middle span and support, Mu = 0

STEP 5 AREA OF STEEL FOR MIDDLE SPAN

Mu/bd2 #DIV/0!

pt 0.172

Ast=pt*b*d/100 0 mm2

ast=(∏/4)d2 78.5 mm2 10 mm

Spacing of bars (s= 1000*ast/Ast) #DIV/0! mm c/c

minimum spacing 300 mm c/c

3d 0 mm c/c

( least of 3 values )

hence spacing of bars 300 mm c/c

Mu/bd2 #DIV/0!

STRUCTURAL LAB-II


`

pt 0.172

Ast=pt*b*d/100 0 mm2

ast=(∏/4)d2 78.5 mm2

Spacing of bars (s= 1000*ast/Ast) #DIV/0! mm c/c

minimum spacing 300 mm c/c

3d 0 mm c/c

( least of 3 values )

hence spacing of bars 300 mm c/c

STRUCTURAL LAB-II


`

4. ONE WAY SLAB:

GIVEN DATA

Fck 20 N/mm2

Fy 415 N/mm2

Bearing 230 mm

Clear cover 20 mm

Width (b) 1 m 1000 mm

Dia , d 10 mm

Density of

concrete 25 KN/mm3


Ly 6000 mm 6 m

Lx 3000 mm 3 m

Ly/Lx 2 >= 2 hence one way slab



20 26 7

STEP-1 Dimensions

Effective Depth=Length/(Span/Depth ratio) 115.4 m

Say effective depth ( d) = 125 mm

0.125 m



0.15 m



3.23 m


3.125 m

STEP-3 LOADS

Dead load=width * depth*density of

concrete 3.75

KN/m2

STRUCTURAL LAB-II


`



Total load 6.75 KN/m2


STEP 4 BENDING MOMENTS AND SHEAR FORCE

Bending moment , Mu=( Factored load*(length2/8)) 12.36 KN-m

1E+07 N-mm

Mu, lim = 0.138fckbd2 4E+07 N-mm

Mulim > Mu , design as under reinforced section

STEP 5 CHECK FOR DEPTH , d = √Mu/(0.138*fck*b) 66.919

depth required < depth provided , under reinforced section

STEP 6 SHEAR FORCE

Shear force , Vu = Factored load * length / 2 15.188 KN

STEP 7 AREA OF STEEL

Mux/bd2

0.791016

pt 0.233

Ast=pt*b*d/100 291.25 mm2

ast=(pi/4)d2 78.5 mm2

Spacing = ( 1000*ast/Ast) 269.5279 mm c/c

SAY S 250 mm c/c

3d 375 mm c/c

least of 3 values

provide 10 mm dia @ 250 mm c/c

STEP 8

check for shear

nominal shear stress Շ v= Vu/bd 0.1215

permisible Shear strength Շ c 0.36 N/mm3

Շ c> Շ v Hence its k

STRUCTURAL LAB-II


`

5. SINGLY REINFORCED BEAM:

fck 20 N/mm2

fy 415 N/mm2

L 7000 mm

7 m

Bearing 230 mm

clear cover 25 mm

width , b 230 mm

0.23 m

Dia, d 16 mm

Density 25 KN/m3

FOS 1.5

SPAN DEPTH RATIO

Simply supported Continous Cantilever

20 26 7

STEP 1

DIMENSIONS

Effective depth = length/span depth ratio 350 mm

provide Effective depth , d 400 mm

0.4 m


Over all depth 430 mm

0.43 m

STEP 2 EFFECTIVE SPAN


effective span=length + effective depth 7400 mm

STEP 3 LOADS

Dead load = width * depth * density of concrete 2.4725 KN/m2

Assume Live load 2 KN/m2

Floor finish 1 KN/m2


Factored load 8.20875 KN/m2


Bending moment , Mu=( Factored load*(length2/8)) 53636896 Nmm

53.636896 KNm

STRUCTURAL LAB-II


`

Mu, lim = 0.138fckbd2 101568000 Nmm

101.568 KNm

Mulim > Mu , design as under reinforced section

STEP 5 Check for depth , d=√Mu/(0.138fckb) 290.6788 mm


STEP 6 SHEAR FORCE

Shear force , Vu = Factored load * length / 2 28730.625 N

28.730625 KN


Mu/bd2

1.4575243

Pt (sp-16) 0.46

Ast = Pt*b*d/100 423.2 mm2

ast = (∏/4)d2 200.96 mm2

No of bars , n = ( Ast / ast) 2.1058917

provide 3-16 # bars

STEP 8 CHECK FOR SHEAR

Nominal shear stress τv = Vu/bd 0.312289 N/mm2

Permissible shear stress τc 0.46

τc > τv provide minimum shear reinforcement

assume 2L-8mm dia stirrups 8

Area of stirupps , Asv=2((∏/4)d2) 100.48 mm2

spacing of bars , Sv = 0.87fyAsv/0.4b 394.3294

shear reinforcement least of 3 values

s 394.32939 mm

0.75d 300 mm

minimum spacing 300 mm

provide minimum spacing 300mm

provide 2L-8#@300mmc/c

STRUCTURAL LAB-II


`

6. TWO WAY SLAB:

GIVEN DATA

Fck 20 N/mm2

Fy 415 N/mm2

Bearing 230 mm

Clear cover 20 mm

Width (b) 1 m 1000 mm

Dia. d 10 mm

Density of

concrete

25

KN/m3


Ly 6000 mm 6 m

Lx 5500 mm 5.5 m

Ly/Lx 1.090909091 < 2 hence two-way slab



20 26 7

STEP-1 Dimensions

Effective Depth=Length/(Span/Depth ratio) 211.54 m

Say effective depth ( d) = 175 mm

0.175 m



0.2 m



5.73 m


5.675 m

STEP-3 LOADS

Dead load=width *depth*density of

concrete 4.375

KN/m2


STRUCTURAL LAB-II


STRUCTURAL LAB-II


`





αx -ve 0.043

αy -ve 0.037

αx +ve 0.033

αy +ve 0.028

Negative Moment Mux=αx*Wu*Lx2

14389547

KN-m

Muy=αy*Wu*Lx2 12381703 KN-m

Positive Moment Mux=αx*Wu*Lx2 11043141 KN-m

Muy=αy*Wu*Lx2 9369938 KN-m

STEP 5 CHECK FOR DEPTH , d =

√Mu/(0.138*fck*b)

72.20529

mm



Mux/bd2

0.469863

pt 0.143

Ast=pt*b*d/100 250.25 mm2

ast=(pi/4)d2 78.5 mm2


provide 10mm dia @ 300mm c/c

Muy/bd2

0.404301

pt 0.114

Ast=pt*b*d/100 199.5 mm2

ast=(pi/4)d2 78.5 mm2


provide 10mm dia @ 300mm c/c

STEP7 check for shear

Vu =(Wx*L/2) 30421.88 N

nominal shear stress Շ v= Vu/bd 0.173839 N/mm2

STRUCTURAL LAB-II


`

permisible Shear strength Շ c 0.28 N/mm2

Շ c> Շ v Hence its k

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