INDUSTRIAL BUILDINGS

INDUSTRIAL BUILDINGS

1.INTRODUCTION:

Industrial Buildings are low rise steel structures, housing workshops or industries and characterized by their low height and absence of interior walls and partitions.

Usually, the roofs system of industrial buildings consists of trusses with light roof coverings. Now-a-days, new materials for both cladding and roofing are available. These are lighter compared to conventional Asbestos  Cement Sheets and Galvanized Iron Sheets.

However, few industrial buildings may be multistoreyed (not more than four storeys) to house these industries which involve some processes predominantly in vertical plane. More commonly, industrial buildings have single storey or single storey with mezzanine floors. To provide more working space, interior columns are either all together avoided, or else are very widely spaced in big industrial buildings.

The industrial buildings are classified into two categories :

1. Normal or Simple Industrial Buildings
2. Special type of Industrial Buildings

Normal or simple industrial buildings :

• These are common for short span layouts. These consist of simple single storeyed industrial sheds, with or without gantry girders, to house workshops, warehouses or factories, and do not contain intermediate columns.
• These structures provide more protected areas. These buildings require large and clear areas unobstructed by the columns. The industrial buildings are provided with adequate head room for the use of an over head travelling cranes.

Special type of industrial buildings :

• These special type of industrial buildings are usually called steel mill buildings. These are used to house big industries in which some manufacturing processes need spaces with specific and controlled environmental conditions. These may contain mezzanine floors, and may be multi-storeyed also, in some portion of the total floor area.
• This type consists of trusses framed in the long dimensions of the bay with purlins framing between the trusses in the short dimensions. This system is used where long span required in one direction is more than 21 m, while much shorter span is acceptable in the other direction. Trusses are not likely to be economical for spans less than about 21 m.
• When the roof truss is attached to, and is supported on steel columns at both the ends, the assembly is known as a  bent or a mill bent or transverse bent. Bents are braced together at intervals depending on the character of the building, Its length, Width, covering, exposure etc. Two Bents braced together form a braced bay. The space between two column lines is called an aisle. 

2. PLANNING AND STRUCTURAL FRAMING :

The planning and design of industrial buildings requires the knowledge of the following areas :

1. Site information
2. Soil Conditions
3. Future expansion plans
4. Plant layout and work flow
5. Preferred bay sizes
6. Crane types and capacity.
7. Availability of raw materials
8. Roofing, side cladding and wall material preferences.
9. Heating, ventilation, air conditioning equipment loads.
10. Parking facilities
11. Preferred fabricators/contractors
12. Availability of waste disposal and sanitary facilities
13. Budget and project schedule, etc.

The planning of an industrial building is based on functional requirements, i.e. on the operations to be performed inside the building. In the planning of a particular type of industrial building, the following should be given due consideration :

• wide area free of columns
• large height
• large doors and windows
• large span of trusses
• minimum weight of trusses, purlins, beams, columns, etc
• lighting
• sanitary arrangements
• safety during fire

Factors Considered While Selecting Site For Industrial Building:

1. The Site should be located on an aerial road.
2. Local availability of raw materials.
3. Facilities like watersupply, electricity.
4. Topography of an area.
5. Soil conditions with respect to foundation design.
6. Waste disposal facilities.
7. Transport facilities.
8. Sufficient space for storage of raw materials.
9. Space for future expansion.

  

Response Spectrum Analysis of a Multi-storied Structure for Elcentro ground Motion & IS:1893-2002 method

clc
clear all
fprintf('-----------------STRUCTURAL DYNAMIC-------------------------------\n')
fprintf('-----------------Response Spectrum Analysis------------------------\n')
N=6;                                         %number of lumped Masses/storey's
H=3.2;                                   %height of each storey
M=[373710 0 0 0 0 0;
        0 373710 0 0 0 0;
        0 0 373710 0 0 0;
        0 0 0 373710 0 0;
        0 0 0 0 269040 0;
        0 0 0 0 0 26510];                   %Mass Matrix
   
K=[1152600000 -576300000 0 0 0 0;
    -576300000 1152600000 -576300000 0 0 0;
    0 -576300000 1152600000 -576300000 0 0;
    0 0 -576300000 1152600000 -576300000 0;
    0 0 0 -113200000 689500000 -113200000;
    0 0 0 0 -113200000 113200000];          %stiffness matrix
%Finding Eigen Vector and Eigen Value

fprintf('\nEigen Value and Eigen Vector :');

[Modeshape,omega]=eig(K,M);
fprintf('\nEigen Vector\n');
disp(Modeshape);
fprintf('\nEigen Value\n');
disp(omega);
m=diag(M);
%Modal Expansion of effective EQ forces (Chopra pg.no.551)

for j=1:N
    for i=1
        num(i,j)=m(i)*(Modeshape(i,j));
        den(i,j)=m(i)*Modeshape(i,j)*Modeshape(i,j);
    end
    for i=2:N
        num(i,j)= num(i-1,j)+m(i)*(Modeshape(i,j));
        den(i,j)=den(i-1,j)+m(i)*Modeshape(i,j)*Modeshape(i,j);
    end
    mod_exp(j)=num(N,j)/den(N,j);
end

%Modal Mass calculation from IS1893
for i=1:1:N
    modal_mass(i)=((num(N,i)*9.81)^2/(9.81*9.81*den(N,i)))
end

%Calculation of Time Peritod of each Mode
for i=1:1:N
    Timemode(i)=(2*pi)/sqrt(omega(i,i))
end

importance_fact=input('\nImportance factor Of Building:')
response_red_fact=input('\nEnter Response Reduction Factor R:')
zonefact=input('\nEnter Zone Factor Z:')
selection=input('\nSelect any one method for calculation of base shear\n 1)IS Code Method \n2)Elecentro Ground motion')
if selection==1
%Obtaining Sa/g Values for Diffrent terrains using IS1893
fprintf('\nSelect Type of Terrain:\n');
fprintf('\n1)Rocky\n2)Medium Soils\n3)Soft Soils')
sel_terrain=input('\nEnter Type of Terrain 1,2 or 3')
%According to IS1893
if sel_terrain==1
        for i=1:N
            if Timemode(i)>=0 && Timemode(i)<=0.10
                 Sabyg(i)=1+15*Timemode(i);
            elseif Timemode(i)>=0.10 && Timemode(i)<=0.40
                 Sabyg(i)=2.5;
            elseif Timemode(i)>=0.40 && Timemode(i)<=4.0
                 Sabyg(i)=1/Timemode(i);
            end
        end
elseif sel_terrain==2
    for i=1:N
            if Timemode(i)>=0 && Timemode(i)<=0.10
                 Sabyg(i)=1+15*Timemode(i);
            elseif Timemode(i)>=0.10 && Timemode(i)<=0.40
                 Sabyg(i)=2.5;
            elseif Timemode(i)>=0.40 && Timemode(i)<=4.0
                 Sabyg(i)=1.36/Timemode(i);
            end
    end
else
    for i=1:N
            if Timemode(i)>=0 && Timemode(i)<=0.10
                 Sabyg(i)=1+15*Timemode(i);
            elseif Timemode(i)>=0.10 && Timemode(i)<=0.40
                 Sabyg(i)=2.5;
            elseif T(i)>=0.40 && Timemode(i)<=4.0
                 Sabyg(i)=1.67/Timemode(i);
            end
    end
end
else
    %Calculation of Sabyg Starts - using elcentro motion

    K = 45*10^4;
%T = 31.18;
 Z = 0.05;                  % Case 1 ( 5% Damping)
 dt = 0.02;
 matno=1;
 maxpga=0.515;
for lkn=1:1:1
            load elecentro.txt;
            a1=elecentro;
            ug1_pga=max(a1);
            a1=a1.*(maxpga/ug1_pga);
            ts=0.02;
            fprintf('1')
       

T=0.01:0.01:4;                % building time period
for i=1:1:length(T)
     U(2) = 0;                 %  SD
     V(2) = 0;                 %  SV
     V1(2) = 0;                %  PSV
     A1(2) = 0;                %  PSA
   
     W(i) = ((2*pi)./T(i));
     m(i) = ((K)./(W(i).^2));
     P = m(i).*(a1);
     c(i) = (2*Z.*m(i).*(W(i)));
     K1(i) = (m(i)./(dt^2))+(c(i)./(2*dt));
     a(i) = (m(i)./(dt^2))-(c(i)./(2*dt));
     b(i) = K-(2*(m(i))./(dt^2));    
     A(2) = ((P(1)-(c(i).*V(2))-(K.*U(2)))./m(i));                                  %  SA
     U(1) = U(2)-(dt.*V(2))+(A(2).*(dt^2)/2);

     t=0:0.02:31.18;
     for j=2:1:length(t)-1
         P7(j) = P(j)-(a(i).*U(j-1))-(b(i).*U(j));
       
         U(j+1) = P7(j)./K1(i);                            %  SD
                 
         V(j+1) = ((U(j+1)-U(j-1))./(2*dt));               %  SV
         V1(j+1)= U(j+1).*W(i);                            %  PSV
       
         A(j+1) = (((U(j+1)-(2*U(j))+U(j-1)))./(dt^2));    %  SA
         A1(j+1)= V(j+1).*(W(i));                          %  PSA
       
     end
     %display(U);
   % Maximum Values:
     Umax(i)=max(abs(U));
   
     Vmax(i)=max(abs(V));
     V1max(i)=Umax(i).*W(i);                        
         
     Amax(i)=max(abs(A));
     A1max(i)=V1max(i).*W(i);

     Amaxall(:,matno)=mean(Amax(i));
     matno=matno+1;
end
     
end
meanamaxall=Amaxall;
stddevumaxall=std(Amaxall,0,1);
plotamaxall=meanamaxall+stddevumaxall;
T=0.01:0.01:4;    
          plot(T,plotamaxall);
          title('Design Spectrum Diagram');
          xlabel('Time period')
          ylabel('Spectral Acceleration in m')
hold off



for j=1:1:N
               
 Sabyg(j)=plotamaxall((round(Timemode(j).*100)));
               
end



%Calculation of Sa/g Ends
end
%Horizontal Seismic Coefficient Ah
for i=1:1:N
    Ah(i)=zonefact*importance_fact*Sabyg(i)/(2*response_red_fact);
end

%Design Lateral Force at each floor in each mode
for k=1:N
 
    for i=1:N
        Qik(i,k)=(Ah(k)*Modeshape(i,k)*mod_exp(k)*modal_mass(k)*9.81)*1*10^-4
    end
end

%Storey shear forces in each mode

for j=1:N
    for i=1:N
        if i==1
            storey_shear(i,j)=Qik(1,j)+Qik(2,j)+Qik(3,j)+Qik(4,j)+Qik(5,j)+Qik(6,j)
        end
        if i==2
            storey_shear(i,j)=Qik(2,j)+Qik(3,j)+Qik(4,j)+Qik(5,j)+Qik(6,j)
        end
        if i==3
            storey_shear(i,j)=Qik(3,j)+Qik(4,j)+Qik(5,j)+Qik(6,j)
        end
        if i==4
            storey_shear(i,j)=Qik(4,j)+Qik(5,j)+Qik(6,j)
        end
        if i==5
            storey_shear(i,j)=Qik(5,j)+Qik(6,j)
        end
        if i==6
            storey_shear(i,j)=Qik(6,j)
        end
       
    end
end

%Using SRSS Method to find Peak Modal Responses of each storey
for i=1:N
    peak_storey_shear(i)=sqrt(storey_shear(i,1)^2+storey_shear(i,2)^2+storey_shear(i,3)^2+storey_shear(i,4)^2+storey_shear(i,5)^2)
end

%Lateral Forces at each storey due to all modes considered
for i=1:N
    if i~=N%For non terrace floors
    force(i)=peak_storey_shear(i)-peak_storey_shear(i+1)
    else%For terrace/roof floor
    force(i)=peak_storey_shear(i)
    end
end

%calculation for base shear
for i=1:N
    if i==1%For ground floor
        base_shear=force(1)
    else
        base_shear=base_shear+force(i)
    end
end

for i=1:N
    if i==1
        base_moment=force(1)*H
    elseif i>1
        base_moment=base_moment+force(i)*H*i
    end
end

tot_mass=sum(diag(M));
base_reaction=tot_mass/950

display(base_shear)
display(base_moment)
display(base_reaction)

% % % output:
%base_shear =
%
%   116.1745
%
%
% base_moment =
%
%   1.0834e+003
%
%
% base_reaction =
%
%   1.8846e+003

MATLAB coding for Plotting SFD's and BMD's for different cases in continuous beams

a)

% CONTINUOUS SPAN BEAMS

%- SFD

%- BMD

clear all

clc

fprintf('Title:continuous beam with half UDL and Moment\n');

%diamensions of beam

L=2000; %Total length of the beam

L1=1000;%Length of Span AB

L2=L1;%Length of span BC

B=200; %breath of the beam

D=200;  %depth of the beam

M20=20;%grade of concrete

w=2000; %load

E=5000*sqrt(M20);

a=1000;

b=1000;

L=L1+L2;

Ra=(7*w*L1)/16;

Rb=(5*w*L1)/8;

Rc=(w*L1)/16;

I= (B*D^3)/12;

j=0;

for x=0:1:L

    j=j+1;

    if x<L1

    sf(j)=Ra-w*x;

    sf1(j)=0;

    bm(j)=Ra*x-(w*x^2)/2;

    bm1(j)=0;

    else

    sf(j)=Rc;

    sf1(j)=0;

    bm(j)=Rc*(L-x);

    bm1(j)=0;

    end

end

x=0:1:L

subplot(3,1,1)

plot(x,sf,x,sf1,'linewidth',2);

title('shear force');

xlabel('length of the beam in mm');

ylabel('shear force in n');

subplot(3,1,2)

plot(x,bm,x,bm1,'linewidth',3);

title('bending moment');

xlabel('length of the beam in mm');

ylabel('bending moment in n-mm');

---------------------------------------------------------------------------

b)

clear all   

clc

fprintf('Title:continuous beam with half UDL and Moment\n');

%Dimensions

L=input('Length of the beam in m: ');

%Loads

w=input('Enter the Load in KN/m : ');

M=input('Enter the Moment in KN-m : ');

%Moments

Mb=((-w*L^2)/(24))-((24*M)/(150));

Ma=((w*L^2)/(8))-((Mb)/(2));

Mc=((-2*M)/(50))-((Mb)/(2));

%Reactions

Ra=((Ma-Mb)/(L))+((w*L)/(2));

Rc=((Mc-Mb)/(L))+(M/L);

Rb=(w*L)-(Ra)-(Rc);

i=0;

for X=0:0.02:(2*L);

    i=i+1;

    if X<=L

        SF(i)=Ra-(w*X);  

        SF1(i)=0;

        BM(i)=(Ra*X)-((w*X^2)*0.5)-Ma;  

        BM1(i)=0;   

    elseif X>=L&&X<=(1.6*L)

        SF(i)=Ra+Rb-(w*L); 

        SF1(i)=0;

        BM(i)=(Rc*((2*L)-X))-Mc-M; 

        BM1(i)=0;

    else

        SF(i)=Ra+Rb-(w*L); 

        SF1(i)=0;

        BM(i)=-Mc+(Rc*((2*L)-X)); 

        BM1(i)=0;

    end

end

X=0:0.02:(2*L);

% ploting the SF,&BM

subplot(2,1,1);

plot(X,SF,X,SF1);

title('Shear Force Diagram');

xlabel('Length of the Beam in m');

ylabel('Shear Force in KN');

% ploting the SF,&BM

subplot(2,1,2);

plot(X,BM,X,BM1);

title('Bending Moment Diagram');

xlabel('Length of the Beam in m');

ylabel('Bending Moment in KN-m');

---------------------------------------------------------------------------

c)

clear all

clc

fprintf('Title:continuous beam with point load\n');

% Dimensions of the Beam

L=input('length of the Beam in m: ');

%Load

w=input('Enter the Load in KN/m: ');

%Moments

Ma=w*L/6;

Mb=w*L/24;

Mc=-w*L/48;

%Reactions

Ra=((Ma-Mb)/L)+(w/2);

Rc=((Mc-Mb)/(2*L));

Rb=w-Ra-Rc;

i=0;

for X=0:0.002:(3*L)

    i=i+1;

   if X<L/2

       SF(i)=Ra;

       SF1(i)=0;

       BM(i)=(Ra*X)-Ma;

       BM1(i)=0;

   elseif X>=L/2&&X<=L

       SF(i)=Ra-w;

       SF1(i)=0;

       BM(i)=(Ra*X)-w*(X-(L/2))-Ma;

       BM1(i)=0;

   else

       SF(i)=Ra+Rb-w;

       SF1(i)=0;

       BM(i)=(Ra*X)+(Rb*(X-L))-(w/2*(2*X-L))-Ma;

       BM1(i)=0;

   end

end

X=0:0.002:(3*L);

% ploting the SF,&BM

subplot(2,1,1)

plot(X,SF,X,SF1)

xlabel('Length of the beam in m')

ylabel('Shear Force in KN')

title('Shear force diagram')

subplot(2,1,2)

plot(X,BM,X,BM1)

xlabel('Length of the beam in m')

ylabel('Bending Moment in KN-m')

title('Bending Moment diagram')

---------------------------------------------------------------------------

d)

clear all

clc

fprintf('Title:continuous beam with two point loads and UDL \n');

% Dimensions of the Beam

L=input('length of the Beam in m: ');

%Load

w1=input('Enter the UDL in KN/m: ');

w2=input('Enter the point load in KN/m: ');

%Moments

Ma=5*(w1*L^2)/48-1/32*w2*L;

Mb=(w1*(L^2)/24)+(w2*L/16);

Mc=5*(w2*L)/32-1/48*w1*(L^2);

%Reactions

Ra=((Ma-Mb)/L)+(w1*L/2);

Rc=((Mc-Mb)/L)+w2/2;

Rb=(w1*L)+w2-Ra-Rc;

i=0;

for X=0:0.002:(3*L)

    i=i+1;

   if X<L

       SF(i)=Ra-(w1*X);

       SF1(i)=0;

       BM(i)=(Ra*X)-Ma-(w1*(X^2)*0.5);

       BM1(i)=0;

   elseif X>=L&&X<1.5*L

       SF(i)=Ra-(w1*L)+Rb;

       SF1(i)=0;

       BM(i)=(Ra*X)+(Rb*(X-L))-w1*L*((0.5*L)+(X-L))-Ma;

       BM1(i)=0;

   else

       SF(i)=Ra+Rb-(w1*L)-w2;

       SF1(i)=0;

       BM(i)=(Ra*X)+(Rb*(X-L))-w1*L*((0.5*L)+(X-L))-Ma-(w2*(X-1.5*L));

       BM1(i)=0; 

   end

  

end

X=0:0.002:(3*L);

% ploting the SF,&BM

subplot(2,1,1)

plot(X,SF,X,SF1)

xlabel('Length of the beam in m)')

ylabel('Shear Force in KN')

title('Shear force diagram')

subplot(2,1,2)

plot(X,BM,X,BM1)

xlabel('Length of the beam in m)')

ylabel('Bending Moment in KN-m')

title('Bending Moment diagram')

---------------------------------------------------------------------------

e)

clear all

clc

fprintf('Title:continuous beam with UDL at centre and two points loads at some distance from free end\n');

%Dimensions

L=input('Length of the beam in m: ');

%Loads

w=input('Enter the Load in KN : ');

w1=input('Enter the Load in KN/m : ');

%Moments

Mc=((58*w*L)/(1215))+((11*w1*L^2)/(180));

Md=((10*w*L)/(54))-((Mc)/(2));

Mb=((w1*L^2)/(4))+((10*w*L)/(54))-((7*Mc)/(2));

Ma=((10*w)/(54))-((Mb)/(2));

%Reactions

Ra=((Ma-Mb)/(L))+((2*w)/(3));

Rd=((Md-Mc)/(L))+(w/3);

Rb=((Ma-Mc)/(L))-(2*Ra)+((5*w)/(3))+((w1*L)/(2));

Rc=((Md-Mb)/(L))-(2*Rd*L)-((4*w)/(3))+((w1*L)/(2));

i=0;

for X=0:0.01:(3*L)

    i=i+1;

    if X<(L/3)

        SF(i)=Ra;

        SF1(i)=0;

        BM(i)=(Ra*X)-(Ma);  

        BM1(i)=0;   

    elseif X>=(L/3)&&X<=L

        SF(i)=Ra-(w); 

        SF1(i)=0;

        BM(i)=(Ra*X)-(w*(X-(L/3)))-(Ma); 

        BM1(i)=0;

    elseif X>=L&&X<=(2*L) 

        SF(i)=Ra-(w)+Rb-(w1*(X-L)); 

        SF1(i)=0;

        BM(i)=(Ra*X)-(w*(X-(L/3)))-(Ma)+(Rb*(X-L))-(w1*((X-L)^2)*0.5);

        BM1(i)=0;

    elseif X>=(2*L)&&X<=((7/3)*L) 

        SF(i)=-Rd+w; 

        SF1(i)=0;

        BM(i)=(Rd*((3*L)-X))-Md-(w*((2.33*L)-X)); 

        BM1(i)=0;

    else

       

        SF(i)=-Rd; 

        SF1(i)=0;

        BM(i)=(Rd*((3*L)-X))-Md;

        BM1(i)=0;

   

    end

    end

X=0:0.01:(3*L)

%Plot Shear Force

subplot(2,1,1);

plot(X,SF,X,SF1);

title('Shear Force Diagram');

xlabel('Length of the Beam in m');

ylabel('Shear Force in KN');

%Plot Bending Moment

subplot(2,1,2);

plot(X,BM,X,BM1);

title('Bending Moment Diagram');

xlabel('Length of the Beam in m');

ylabel('Bending Moment in KN-m');

---------------------------------------------------------------------------

f)

clear all

clc

fprintf('Title: continuous beam with UDL at centre and point loads at some distance from left end\n');

% Dimensions of the Beam

l=input('length of the Beam in m: ');

%Load

w=input('Enter the UDL in KN/m: ');

w1=input('Enter the PointLoad in KN/m: ');

%Moments

Ma=((-3*w1*l^2)/104)+((3*w*l)/208);

Mb=(-2*Ma);

Mc=((5*w1*(l^2))/104)+((21*w*l)/208);

%Reactions

Ra=(Ma-Mb)/l;

Rb=((Ma-Mc)/l)-(2*Ra)+((w1*l)/2);

Rd=(-Mc/l)+(w/2);

Rc=(-Mb/l)-(2*Rd)+(3*w/2)+(w1*l/2);

i=0;

for X=0:0.02:(3*l);

    i=i+1;

    if(X<l)

        SF(i)=Ra;

        SF1(i)=0;

        BM(i)=(Ra*X)-Ma;

        BM1(i)=0;

    elseif(X>=l&&X<2*l)

        SF(i)=Ra+Rb-(w1*(X-l));

        SF1(i)=0;

        BM(i)=(Ra*X)+Rb*(X-l)-(w1*((X-l)^2)*0.5)-Ma

        BM1(i)=0;

    elseif(X>=2*l&&X<2.5*l)

        SF(i)=-Rd+w;

        SF1(i)=0;

        BM(i)=(Ra*X)+Rb*(X-l)-Ma-w1*l*((l*0.5)+(X-(2*l)))+(Rc*(X-(2*l)));

        BM1(i)=0;

       

    else

        SF(i)=-Rd;

        SF1(i)=0;

        BM(i)=Rd*((3*l)-X);

        BM1(i)=0;

    end

end

X=0:0.02:(3*l);

subplot(2,1,1);

plot(X,SF,X,SF1)

xlabel('Length of the beam in m')

ylabel('Shear Force in KN')

title('Shear force diagram')

subplot(2,1,2)

plot(X,BM,X,BM1)

xlabel('Length of the beam in m')

ylabel('Bending Moment in KN-m')

title('Bending Moment diagram')

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