diff options
Diffstat (limited to '698/CH2')
36 files changed, 1381 insertions, 0 deletions
diff --git a/698/CH2/EX2.001/nature.sci b/698/CH2/EX2.001/nature.sci new file mode 100644 index 000000000..0bcef8584 --- /dev/null +++ b/698/CH2/EX2.001/nature.sci @@ -0,0 +1,10 @@ +//Function to determine nature of normal stresses, i.e, compressive or tensile
+
+funcprot(0)
+function nature(Sn)
+ if Sn<0
+ disp(" compressive")
+else
+ disp(" tensile")
+end
+endfunction
\ No newline at end of file diff --git a/698/CH2/EX2.002/standard_dimensions.sci b/698/CH2/EX2.002/standard_dimensions.sci new file mode 100644 index 000000000..2ea28a316 --- /dev/null +++ b/698/CH2/EX2.002/standard_dimensions.sci @@ -0,0 +1,17 @@ +//Function to standardize the dimensions
+
+funcprot(0)
+function stddim(x)
+ x=x*(10^3)
+ std=[1 2 3 4 5 6 8 10 12 15 18 20 24 26 30:5:400]
+ n=length(std)
+ for i=1:n
+ if (x<std(i)) then
+ y=std(i)
+ break
+ else
+ continue
+ end
+ end
+ y=y*(10^-3)
+endfunction
\ No newline at end of file diff --git a/698/CH2/EX2.1/1_factor_of_safety.txt b/698/CH2/EX2.1/1_factor_of_safety.txt new file mode 100644 index 000000000..0183cecaa --- /dev/null +++ b/698/CH2/EX2.1/1_factor_of_safety.txt @@ -0,0 +1,7 @@ +The stress in the bar is given by:
+
+ Sx=P/A
+
+ Factor of safety=(Yield Strength)/(Induced Strength)
+
+The factor of safety is 5
\ No newline at end of file diff --git a/698/CH2/EX2.1/P1_factor_of_safety.sce b/698/CH2/EX2.1/P1_factor_of_safety.sce new file mode 100644 index 000000000..682f423a8 --- /dev/null +++ b/698/CH2/EX2.1/P1_factor_of_safety.sce @@ -0,0 +1,35 @@ +clc
+//Example 2.1
+//Factor of safety
+
+//------------------------------------------------------------------------------
+//Given Data:
+//Dimensions
+b=0.04//m
+h=0.05//m
+// the value of h is given 20mm in the problem statement but taken 50mm while solving, therefore we will take 50mm as the value of h
+//Force
+P=200000//N
+
+//------------------------------------------------------------------------------
+//Stress induced
+S=P/(b*h)
+S=S*(10^-6)//To convert units into Mpa
+
+//Yield strength
+Sy=500//Mpa
+
+//Factor of safety
+fos=Sy/S
+
+//------------------------------------------------------------------------------
+//Printing the result file to .txt
+res1=mopen(TMPDIR+'1_factor_of_safety.txt','wt')
+mfprintf(res1,"The stress in the bar is given by:\n")
+mfprintf(res1,"\n\tSx=P/A\n")
+mfprintf(res1,"\n\tFactor of safety=(Yield Strength)/(Induced Strength)\n\n")
+mfprintf(res1,"The factor of safety is %d",fos)
+mclose(res1)
+editor(TMPDIR+'1_factor_of_safety.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.10/10_thickness_of_bracket_at_section.txt b/698/CH2/EX2.10/10_thickness_of_bracket_at_section.txt new file mode 100644 index 000000000..747b7fb5e --- /dev/null +++ b/698/CH2/EX2.10/10_thickness_of_bracket_at_section.txt @@ -0,0 +1,9 @@ +Assuming the thickness of the bracket as b,
+The area of cross section will be A=0.05*b
+Moment of inertia will be I=b*(0.05)^3/12
+The tensile stress is given by equation:
+ Sn_max=(P/A)+((M*c)/I)
+
+b=0.00900 m
+
+9 mm is required to limit the stress at 70 MN/m^2
\ No newline at end of file diff --git a/698/CH2/EX2.10/P10_thickness_of_bracket_at_section.sce b/698/CH2/EX2.10/P10_thickness_of_bracket_at_section.sce new file mode 100644 index 000000000..bf250b7bd --- /dev/null +++ b/698/CH2/EX2.10/P10_thickness_of_bracket_at_section.sce @@ -0,0 +1,40 @@ +clc
+//Example 2.10
+//Thickness of bracket at section
+//------------------------------------------------------------------------------
+
+//Given Data:
+//Loads
+P=4500 // N
+
+//Distance of line of action from horizontal axis of bracket
+c=0.025 // m
+
+//Distance of vertical axis of bracket from bottom
+
+//Height of section
+h=0.05 // m
+
+//Permissible tensile stress
+S_perm=70*(10^6)
+
+//------------------------------------------------------------------------------
+//Printing result file to .txt
+//Moment of force about vertical axis
+M=P*h
+
+res10=mopen(TMPDIR+'10_thickness_of_bracket_at_section.txt','wt')
+//Assuming the thickness of the bracket is b
+mfprintf(res10,"Assuming the thickness of the bracket as b,\n")
+mfprintf(res10,"The area of cross section will be A=0.05*b\n")
+mfprintf(res10,"Moment of inertia will be I=b*(0.05)^3/12\n")
+
+mfprintf(res10,"The tensile stress is given by equation:\n\tSn_max=(P/A)+((M*c)/I)\n\n")
+b=(1/S_perm)*((P/h)+((M*c*12)/(h^3)))
+
+mfprintf(res10,"b=%0.5f m",b)
+mfprintf(res10,"\n\n%d mm is required to limit the stress at %d MN/m^2",(b*1000),(S_perm*(10^-6)))
+mclose(res10)
+editor(TMPDIR+'10_thickness_of_bracket_at_section.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.11/11_thickness_of_bracket.txt b/698/CH2/EX2.11/11_thickness_of_bracket.txt new file mode 100644 index 000000000..3274d4018 --- /dev/null +++ b/698/CH2/EX2.11/11_thickness_of_bracket.txt @@ -0,0 +1,20 @@ +Assuming the thickness of the bracket as t
+Area of cross section A= b*t = 0.04*t
+Direct stress Sd= P/A
+Bending stress Sb=(M*c)/I
+ where c=w+ t/2 i.e., 0.05+ t/2
+ M=P*t/2
+ I=(b*(t^3))/12
+
+Maximum tensile stress:
+ S=Sb+Sd
+ S=(P/A)+((M*c)/I)
+
+On simplifying above equation,
+ we get
+ (S*b)t^2 - (4P)t - 6Pw = 0
+ 30=(500/t)+(155*(100+t) / t^2
+
+Standardizing the dimension, we get t=111.5 mm
+
+The corresponding stress induced is 30.00 MN/m^2
\ No newline at end of file diff --git a/698/CH2/EX2.11/P11_thickness_of_bracket.sce b/698/CH2/EX2.11/P11_thickness_of_bracket.sce new file mode 100644 index 000000000..595476e39 --- /dev/null +++ b/698/CH2/EX2.11/P11_thickness_of_bracket.sce @@ -0,0 +1,103 @@ +clc
+//Example 2.11
+//Thickness of bracket
+
+//Given Data:
+
+funcprot(0)
+function [y]=stddim(x)
+ x=x*(10^3)
+ standard=[1 2 3 4 5 6 8 9 10 30:2:60 65:5120]
+ n=length(standard)
+ for i=1:n
+ if (x<standard(i)) then
+ y=standard(i)
+ break
+ else
+ continue
+ end
+ end
+ y=y*(10^-3)
+endfunction
+
+
+//Maximum tensile stress
+S=30*(10^6) // MN/m^2
+
+//Load
+P=20000 // N
+
+//Dimensions
+//width of section
+b=0.04 // m
+//Distance of line of action from bottom section
+w=0.05 // m
+
+res11=mopen(TMPDIR+'11_thickness_of_bracket.txt','wt')
+
+mfprintf(res11,"Assuming the thickness of the bracket as t\n")
+mfprintf(res11,"Area of cross section A= b*t = 0.04*t\n")
+
+//Direct Stress
+//Sd=P/A
+mfprintf(res11,"Direct stress Sd= P/A \n")
+//Bending Stress
+//Sb=(M*c)/I
+mfprintf(res11,"Bending stress Sb=(M*c)/I\n\twhere c=w+ t/2 i.e., 0.05+ t/2\n\t")
+mfprintf(res11,"M=P*t/2\n\tI=(b*(t^3))/12\n\n")
+
+//Maximum tensile stress
+//S=Sb+Sd
+mfprintf(res11,"Maximum tensile stress:\n\tS=Sb+Sd\n")
+mfprintf(res11,"\tS=(P/A)+((M*c)/I)\n\n")
+mfprintf(res11,"On simplifying above equation,\n\t")
+
+mfprintf(res11,"we get\n\t (S*b)t^2 - (4P)t - 6Pw = 0\n\t")
+mfprintf(res11,"30=(500/t)+(155*(100+t) / t^2\n\n")
+
+//Solving the equation
+
+p=poly([(-6*P*w) (-4*P) (S*b)],'thk','c')
+thk=roots(p)
+disp (thk)
+n1=length(thk)
+//mfprintf(res11,"Equation:\t %s",p)
+
+//selecting only positive roots
+
+
+// if (thk(1)<=0)
+// t=thk(2)
+ // else
+ // t=thk(1)
+ //end
+
+t=thk(2)
+
+disp(t)
+
+//mfprintf (res11,"\n\nThickness t=%f mm\n",t*(1000))
+
+//Standardizing the dimension
+//Function to standardize the dimensions
+
+//ts=t
+
+//disp(t)
+
+//[ts]=stddim(t)
+
+//disp (ts)
+
+mfprintf(res11,"Standardizing the dimension, we get t=%0.1f mm\n\n",ts*(1000))
+
+//Calculation the corresponding stress induced,
+A=b*ts
+M=P*(ts/2)
+c=w+(ts/2)
+I=(b*(ts^3))/12
+Sa=(P/A)+((M*c)/I)
+mfprintf(res11,"The corresponding stress induced is %0.2f MN/m^2",Sa*(10^-6))
+mclose(res11)
+editor(TMPDIR+'11_thickness_of_bracket.txt')
+disp (Sa)
\ No newline at end of file diff --git a/698/CH2/EX2.12/12_stresses_in_locomotive.txt b/698/CH2/EX2.12/12_stresses_in_locomotive.txt new file mode 100644 index 000000000..21d9fdffa --- /dev/null +++ b/698/CH2/EX2.12/12_stresses_in_locomotive.txt @@ -0,0 +1,27 @@ +The angular speed of wheels is 29.3 rad/s
+All points on the side rod have a downward acceleration ap
+The locomotive is moving with uniform velocity, therefore acceleration ao=0
+
+ ap=ao+apo
+ =apo=322.51m/s^2
+
+Inertia force acting upward on the rod, Fi=58051.15 N
+
+Net force acting upward on the rod, Fup=56285.35 N
+
+Axial force F can be determined by assuming rear wheel
+and rod as free bodies and taking moments about centre of wheel.
+ Cl*F= Et*R
+The axial force F is 109800.00 N
+
+Maximum bending moment for simple beam carrying unifornmly distributed load is
+ (W*L)/8
+
+Normal Stress
+ Sx= (P/A) + (M*c)/I
+
+The maximum normal stress is
+ Sn_max = Sx = 59.8 MN/m^2
+
+The maximum shear stress is
+ Tmax = Sx/2 = 29.9 MN/m^2
\ No newline at end of file diff --git a/698/CH2/EX2.12/P12_stresses_in_locomotive.sce b/698/CH2/EX2.12/P12_stresses_in_locomotive.sce new file mode 100644 index 000000000..99cab88cd --- /dev/null +++ b/698/CH2/EX2.12/P12_stresses_in_locomotive.sce @@ -0,0 +1,82 @@ +clc
+//Example 2.12
+//Stresses in a locomotive
+//------------------------------------------------------------------------------
+
+//Given Data:
+
+//Speed
+
+V=96.6 //kmph
+
+//Crank length
+Cl=0.375 // m
+//Length of side rod
+L=2 // m
+//Radius of driver
+R=0.915 // m
+
+//Tractive effort per wheel
+Et=45000 // N
+//Weight of the rod
+Wl=90 // kg/m
+//Total weight
+Wt=Wl*L // kg
+
+//Cross section of rod
+b= 0.075 // m
+t= 0.15 // m
+
+//------------------------------------------------------------------------------
+res12=mopen(TMPDIR+'12_stresses_in_locomotive.txt','wt')
+
+//angular speed of wheels
+omega=((V*5)/18)/R
+
+mfprintf (res12,"The angular speed of wheels is %0.1f rad/s\n",omega)
+mfprintf (res12,"All points on the side rod have a downward acceleration ap\n")
+mfprintf(res12,"The locomotive is moving with uniform velocity, therefore acceleration ao=0\n\n")
+
+//Acceleration
+apo=Cl* omega^2
+mfprintf(res12,"\tap=ao+apo\n\t =apo=%0.2fm/s^2\n\n",apo)
+
+//Inertai Force
+Fi=Wt*apo
+mfprintf(res12,"Inertia force acting upward on the rod, Fi=%0.2f N\n\n",Fi)
+
+//Net upward force
+W=Fi-(Wt*9.81)
+mfprintf(res12,"Net force acting upward on the rod, Fup=%0.2f N\n\n",W)
+
+mfprintf(res12,"Axial force F can be determined by assuming rear wheel\nand rod as free bodies and taking moments about centre of wheel.\n")
+
+mfprintf(res12,"\tCl*F= Et*R\n")
+
+F=(Et*R)/Cl
+
+mfprintf(res12,"The axial force F is %0.2f N\n\n",F)
+
+mfprintf(res12,"Maximum bending moment for simple beam carrying unifornmly distributed load is\n\t")
+mfprintf(res12,"(W*L)/8\n\n")
+
+M=(W*L)/8
+
+mfprintf(res12,"Normal Stress \n\tSx= (P/A) + (M*c)/I\n\n")
+P=F
+A=b*t
+c=b
+I=(b* (t^3))/12
+Sx= (P/A) + (M*c)/I
+
+Sn_max = Sx
+Tmax = Sx/2
+
+mfprintf(res12,"The maximum normal stress is \n\tSn_max = Sx = %0.1f MN/m^2\n\n",Sn_max* 10^-6)
+
+mfprintf(res12,"The maximum shear stress is \n\tTmax = Sx/2 = %0.1f MN/m^2",Tmax* 10^-6)
+
+mclose(res12)
+editor(TMPDIR+'12_stresses_in_locomotive.txt')
+//------------------------------------------------------------------------------
+//---------------------------End of Program-------------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.13/13_shear_stress_in_Z_bracket.txt b/698/CH2/EX2.13/13_shear_stress_in_Z_bracket.txt new file mode 100644 index 000000000..5a701a5d1 --- /dev/null +++ b/698/CH2/EX2.13/13_shear_stress_in_Z_bracket.txt @@ -0,0 +1,14 @@ +At section A-A, Sy=0 and Txy=0
+
+ Sx=(P/A) + ((M1*c)/I)
+ Tmax = Sx/2
+
+Normal stress Sx = 158.4 MPa
+Shear stress Tmax = 79.2 MPa
+
+At section B-B, Sy=0 and Txy=0
+
+ Sx=((M1*c)/I)
+
+Normal stress Sx = 194.4 MPa
+Shear stress Tmax = 97.2 MPa
diff --git a/698/CH2/EX2.13/P13_shear_stress_in_Z_bracket.sce b/698/CH2/EX2.13/P13_shear_stress_in_Z_bracket.sce new file mode 100644 index 000000000..66c694777 --- /dev/null +++ b/698/CH2/EX2.13/P13_shear_stress_in_Z_bracket.sce @@ -0,0 +1,61 @@ +clc
+//Example 2.13
+//Shear Stress in Z-bracket
+//------------------------------------------------------------------------------
+
+//Given dta:
+//Load
+P=45000 // N
+
+//Dimensions
+//Cross Section
+d= 0.05 // m
+b = 0.125 // m
+A=b*d
+I=(b*(d^3))/12
+c = d/2
+
+//Total Height of bracket
+h= 0.3// m
+//Distance of section A-A from bottom
+a = 0.2 // m
+//Distance of mid point of section A-A from line of action of P
+l1 = 0.175 // m
+//Distance of section B-B from line of action of P
+l2 = 0.225 // m
+
+
+//Bending Moment
+M1=P*l1 // Bending moment at section A-A
+M2=P*l2 // Bending moment at section B-B
+
+//------------------------------------------------------------------------------
+//Printing the result file to .txt
+res13=mopen(TMPDIR+'13_shear_stress_in_Z_bracket.txt','wt')
+
+mfprintf(res13,'At section A-A, Sy=0 and Txy=0\n')
+
+Sx1=(P/A) + ((M1*c)/I)
+mfprintf(res13,'\n\tSx=(P/A) + ((M1*c)/I)')
+
+Tmax1 = Sx1/2
+mfprintf(res13,'\n\tTmax = Sx/2')
+
+mfprintf(res13,'\n\nNormal stress Sx = %0.1f MPa\n',Sx1* (10^-6))
+mfprintf(res13,'Shear stress Tmax = %0.1f MPa\n',Tmax1* (10^-6))
+
+
+mfprintf(res13,'\nAt section B-B, Sy=0 and Txy=0\n')
+
+Sx2=((M2*c)/I)
+mfprintf(res13,'\n\tSx=((M1*c)/I)')
+
+Tmax2 = Sx2/2
+
+mfprintf(res13,'\n\nNormal stress Sx = %0.1f MPa\n',Sx2* (10^-6))
+mfprintf(res13,'Shear stress Tmax = %0.1f MPa\n',Tmax2* (10^-6))
+
+mclose(res13)
+editor(TMPDIR+'13_shear_stress_in_Z_bracket.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.14/14_stress_in_eccentric_loading.txt b/698/CH2/EX2.14/14_stress_in_eccentric_loading.txt new file mode 100644 index 000000000..c97ee0f80 --- /dev/null +++ b/698/CH2/EX2.14/14_stress_in_eccentric_loading.txt @@ -0,0 +1,8 @@ +The member is subjected to direct stress and
+also bending stresses due to eccentricity of the load
+ Total stress
+ S=Sd + Sb --- This will be compressive, since Sd is compressive, and Sb is acting in the direction of Sd
+ S=Sd - Sb --- This will be compressive, since Sb is opposing Sd
+
+Stresses induced are -20.833 MPa compressive and 4.167 MPa tensile
+Maximum Tensile Stress is 4.167 MPa
\ No newline at end of file diff --git a/698/CH2/EX2.14/P14_stress_in_eccentric_loading.sce b/698/CH2/EX2.14/P14_stress_in_eccentric_loading.sce new file mode 100644 index 000000000..98d449179 --- /dev/null +++ b/698/CH2/EX2.14/P14_stress_in_eccentric_loading.sce @@ -0,0 +1,50 @@ +clc
+//Example 2.14
+//Stress in Eccentric Loading
+//------------------------------------------------------------------------------
+
+//Given dta:
+//Load
+P=500000 // N
+
+//Dimensions
+//Eccentricity
+e=0.05 // m
+
+//Cross section
+b=0.3 // m
+d=0.2 // m
+A = b*d
+I=(b*(d^3))/12
+c = d/2
+
+//Bending moment
+M=P*e
+
+//------------------------------------------------------------------------------
+//Printing the result file to .txt
+res14=mopen(TMPDIR+'14_stress_in_eccentric_loading.txt','wt')
+
+mfprintf(res14,'The member is subjected to direct stress and \nalso bending stresses due to eccentricity of the load')
+
+mfprintf(res14,'\n Total stress \n\tS=Sd + Sb --- This will be compressive, since Sd is compressive, and Sb is acting in the direction of Sd')
+mfprintf(res14,'\n\tS=Sd - Sb --- This will be compressive, since Sb is opposing Sd')
+
+Sd=(P/A)
+
+Sb=(M*c)/I
+
+S1=-(Sd+Sb) // This will be compressive
+
+S2=-(Sd-Sb) // This will be tensile
+
+mfprintf(res14,'\n\nStresses induced are %0.3f MPa',S1* (10^-6))
+nature(S1)
+mfprintf(res14,' and %0.3f MPa',S2* (10^-6))
+nature(S2)
+mfprintf(res14,'\nMaximum Tensile Stress is %0.3f MPa',S2* (10^-6))
+
+mclose(res14)
+editor(TMPDIR+'14_stress_in_eccentric_loading.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.15/15_stresses_in_steel_latch.txt b/698/CH2/EX2.15/15_stresses_in_steel_latch.txt new file mode 100644 index 000000000..0d5bf200e --- /dev/null +++ b/698/CH2/EX2.15/15_stresses_in_steel_latch.txt @@ -0,0 +1,27 @@ + AT SECTION A-A
+Top fibres are critical.
+
+ Sn(max) = ((M*c)/I) + (P/A)
+Maximum normal stress Sn(max)=49.5 MN/m^2 tensile at top fibres of section A-A
+
+ Sn(min) = -((M*c)/I) + (P/A)
+Minimum normal stress Sn(min)=-31.5 MN/m^2 compressive at bottom fibres of section A-A
+
+Maximum shear stress T(max)=24.7 MN/m^2 at top fibres of section A-A
+
+ AT POINT B
+(neglecting stress concentration)
+
+ Sx = ((M*c)/I) + (P/A)
+Sx=49.5 MN/m^2 tensile
+ Sy = ((M*c)/I)
+Sy=54.0 MN/m^2 tensile
+ Sz=0 and Txy=0
+
+ Sn(max) =((Sx+Sy)/2) + sqrt( ((Sx-Sy)/2)^2 - (Txy^2) )
+Maximum normal stress Sn(max)=54.0 MN/m^2 tensile
+ Sn(min) =((Sx+Sy)/2) - sqrt( ((Sx-Sy)/2)^2 - (Txy^2) )
+Minimum normal stress Sn(min)=49.5 MN/m^2 tensile
+ T(max)=(Sn(max)-0)/2
+Maximum shear stress T(max)=27.0 MN/m^2
+
diff --git a/698/CH2/EX2.15/P15_stresses_in_steel_latch.sce b/698/CH2/EX2.15/P15_stresses_in_steel_latch.sce new file mode 100644 index 000000000..beb64a02b --- /dev/null +++ b/698/CH2/EX2.15/P15_stresses_in_steel_latch.sce @@ -0,0 +1,82 @@ +clc
+//Example 2.15
+//Stresses in steel latch
+//------------------------------------------------------------------------------
+
+//Given data:
+//Loads
+P=2700 // N
+
+//Dimensions
+b=0.006 // m
+d=0.05 // m
+c=d/2
+A = b*d
+I1=(b*(d^3))/12
+
+R=0.025 // m
+I2=(b*(R^3))/12
+
+//distance of line of action from horizontal axis
+l1=0.0375 // m
+//distance of line of action from vertical axis
+l2=R/2 // m
+
+//
+M1=P*l1
+M2=P*l2
+
+
+res15=mopen(TMPDIR+'15_stresses_in_steel_latch.txt','wt')
+
+mfprintf(res15,'\tAT SECTION A-A\nTop fibres are critical.\n\n')
+
+Sn_maxa=((M1*c)/I1)+(P/A)
+
+Sn_mina=-((M1*c)/I1) + (P/A)
+
+Tmaxa=Sn_maxa/2
+
+mfprintf(res15,'\tSn(max) = ((M*c)/I) + (P/A)\n')
+mfprintf(res15,'Maximum normal stress Sn(max)=%0.1f MN/m^2',Sn_maxa* (10^-6))
+nature(Sn_maxa)
+mfprintf(res15,' at top fibres of section A-A\n')
+mfprintf(res15,'\n\tSn(min) = -((M*c)/I) + (P/A)\n')
+mfprintf(res15,'Minimum normal stress Sn(min)=%0.1f MN/m^2',Sn_mina* (10^-6))
+nature(Sn_mina)
+mfprintf(res15,' at bottom fibres of section A-A\n')
+mfprintf(res15,'\nMaximum shear stress T(max)=%0.1f MN/m^2',Tmaxa* (10^-6))
+mfprintf(res15,' at top fibres of section A-A\n\n')
+
+
+mfprintf(res15,'\tAT POINT B\n(neglecting stress concentration)\n\n')
+
+Sx=((M1*c)/I1)+(P/A)
+mfprintf(res15,'\tSx = ((M*c)/I) + (P/A)\n')
+mfprintf(res15,'Sx=%0.1f MN/m^2',Sx* (10^-6))
+nature(Sx)
+
+Sy=((M2*l2)/I2)
+mfprintf(res15,'\n\tSy = ((M*c)/I)\n')
+mfprintf(res15,'Sy=%0.1f MN/m^2',Sy* (10^-6))
+nature(Sy)
+
+mfprintf(res15,'\n Sz=0 and Txy=0\n\n')
+
+Sn_maxb=((Sx+Sy)/2) + sqrt((((Sx-Sy)/2)^2)-0)
+mfprintf(res15,'\tSn(max) =((Sx+Sy)/2) + sqrt( ((Sx-Sy)/2)^2 - (Txy^2) )\n')
+mfprintf(res15,'Maximum normal stress Sn(max)=%0.1f MN/m^2',Sn_maxb* (10^-6))
+nature(Sn_maxb)
+
+Sn_minb=((Sx+Sy)/2) - sqrt((((Sx-Sy)/2)^2)-0)
+mfprintf(res15,'\n\tSn(min) =((Sx+Sy)/2) - sqrt( ((Sx-Sy)/2)^2 - (Txy^2) )\n')
+mfprintf(res15,'Minimum normal stress Sn(min)=%0.1f MN/m^2',Sn_minb* (10^-6))
+nature(Sn_minb)
+
+Tmaxb=(Sn_maxb-0)/2
+mfprintf(res15,'\n\tT(max)=(Sn(max)-0)/2')
+mfprintf(res15,'\nMaximum shear stress T(max)=%0.1f MN/m^2\n\n',Tmaxb* (10^-6))
+mclose(res15)
+editor(TMPDIR+'15_stresses_in_steel_latch.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.16/16_stresses_in_crank.txt b/698/CH2/EX2.16/16_stresses_in_crank.txt new file mode 100644 index 000000000..c0952850b --- /dev/null +++ b/698/CH2/EX2.16/16_stresses_in_crank.txt @@ -0,0 +1,12 @@ +The critical points are at front and back fibres of the section.
+
+ Sx= (M*c) /I
+Sx=21.7 MN/m^2
+
+ Txy=(T*r)/J
+Txy=15.1 MN/m^2
+
+Maximum normal stress=29.5 MN/m^2
+
+Maximum normal stress=18.6 MN/m^2
+
diff --git a/698/CH2/EX2.16/P16_stresses_in_crank.sce b/698/CH2/EX2.16/P16_stresses_in_crank.sce new file mode 100644 index 000000000..b8173a756 --- /dev/null +++ b/698/CH2/EX2.16/P16_stresses_in_crank.sce @@ -0,0 +1,48 @@ +clc
+//Example 2.15
+//Stresses in crank
+//------------------------------------------------------------------------------
+
+//Given dta:
+//Loads
+P=10000 // N
+
+//Dimensions
+//distance of section A-A from plane of action of load P
+l1=0.025 + 0.025 + 0.04 // m
+//distance of centre of section A-A from line of action of P
+l2=0.125 // m
+//Cross section A-A
+d=0.075 // m
+c=d/2
+r=d/2
+I=(%pi/64)*(d^4)
+J=2*I
+
+//Bending moment
+M=P*l1
+//Torque
+T=P*l2
+
+res16=mopen(TMPDIR+'16_stresses_in_crank.txt','wt')
+mfprintf(res16,'The critical points are at front and back fibres of the section.\n')
+
+Sx=(M*c)/I
+Sy=0
+mfprintf(res16,'\n\tSx= (M*c) /I\n')
+mfprintf(res16,'Sx=%0.1f MN/m^2\n',Sx*(10^-6))
+
+Txy=(T*r)/J
+mfprintf(res16,'\n\tTxy=(T*r)/J\n')
+mfprintf(res16,'Txy=%0.1f MN/m^2\n\n',Txy*(10^-6))
+
+Sn_max=((Sx+Sy)/2) + sqrt( (((Sx-Sy)/2)^2) + (Txy^2) )
+mfprintf(res16,'Maximum normal stress=%0.1f MN/m^2\n\n',Sn_max*(10^-6))
+
+Tmax=sqrt( (((Sx-Sy)/2)^2) + (Txy^2) )
+mfprintf(res16,'Maximum normal stress=%0.1f MN/m^2\n\n',Tmax*(10^-6))
+
+mclose(res16)
+editor(TMPDIR+'16_stresses_in_crank.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.17/17_design_of_overhanging_crank.txt b/698/CH2/EX2.17/17_design_of_overhanging_crank.txt new file mode 100644 index 000000000..af945197c --- /dev/null +++ b/698/CH2/EX2.17/17_design_of_overhanging_crank.txt @@ -0,0 +1,20 @@ +Let the diameter of shaft be d
+
+Bending Stress Sx= M*c/I
+ where c=d/2
+ I=(pi/64)*d^4
+
+Shear stress Txy=T*r/J
+ where r=d/2
+ J=(pi/32)*d^4
+
+Maximum shear stress
+ Tmax=sqrt( ((Sx-Sy)/2)^2 + (Txy^2) )
+where Tmax is 75MN/m^2 and Sy=0
+diameter
+ d=( (16* sqrt( M^2 + T^2 )) / (Ss*pi) )^1/3
+
+The calculated diameter is 0.0357 m
+Standardizing the above value, we get
+
+ d=0.0360 m = 36.0 mm
\ No newline at end of file diff --git a/698/CH2/EX2.17/P17_design_of_overhanging_crank.sce b/698/CH2/EX2.17/P17_design_of_overhanging_crank.sce new file mode 100644 index 000000000..dd547b195 --- /dev/null +++ b/698/CH2/EX2.17/P17_design_of_overhanging_crank.sce @@ -0,0 +1,65 @@ +clc
+//Example 2.17
+//Design of shaft of overhanging crank
+//------------------------------------------------------------------------------
+
+//Given data:
+//Loads
+P=2000 // N
+//Stresses
+Ss=75*(10^6) // MN/m^2
+
+//Distance of shaft from plane of action
+l1=0.3 // m
+//Distance of shaft from line of action
+l2=0.15 // m
+
+//Bending moment
+M=P*l2
+//Torsion
+T=P*l1
+
+//------------------------------------------------------------------------------
+//Function to standardize dimensions by choosing from available range
+funcprot(0)
+function [y]=stddim(x)
+ x=x*(10^3)
+ std=[1 2 3 4 5 6 8 9 10 12 15 18 20 24:2:60]
+ n=length(std)
+ for i=1:n
+ if (x<std(i)) then
+ y=std(i)
+ break
+ else
+ continue
+ end
+ end
+ y=y*(10^-3)
+endfunction
+//------------------------------------------------------------------------------
+
+res17=mopen(TMPDIR+'17_design_of_overhanging_crank.txt','wt')
+mfprintf(res17,"Let the diameter of shaft be d\n\n")
+//Bending stress
+mfprintf(res17,"Bending Stress Sx= M*c/I\n\twhere c=d/2\n\t\t I=(pi/64)*d^4\n\n")
+//Torsional shear Stress
+mfprintf(res17,'Shear stress Txy=T*r/J\n\twhere r=d/2\n\t\t J=(pi/32)*d^4\n\n')
+
+mfprintf(res17,'Maximum shear stress\n\tTmax=sqrt( ((Sx-Sy)/2)^2 + (Txy^2) )\n')
+mfprintf(res17,'where Tmax is 75MN/m^2 and Sy=0\n')
+
+mfprintf(res17,'diameter\n\td=( (16* sqrt( M^2 + T^2 )) / (Ss*pi) )^1/3\n\n')
+
+d=( (16* sqrt( M^2 + T^2 )) / (%pi*Ss) ) ^ (1/3)
+disp (d)
+mfprintf(res17,'The calculated diameter is %0.4f m',d)
+
+mfprintf(res17,'\nStandardizing the above value, we get\n')
+
+[ds]=stddim(d)
+
+mfprintf(res17,'\n\td=%0.4f m = %0.1f mm',ds,ds*10^3)
+mclose(res17)
+editor(TMPDIR+'17_design_of_overhanging_crank.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.18/18_stress_in_gas_turbine.txt b/698/CH2/EX2.18/18_stress_in_gas_turbine.txt new file mode 100644 index 000000000..00bef6fd9 --- /dev/null +++ b/698/CH2/EX2.18/18_stress_in_gas_turbine.txt @@ -0,0 +1,10 @@ +Given
+ Sx= 20 MPa and
+ Sy= 50 MPa
+
+Maximum normal stress
+ Sn(max)=Sy
+
+Maximumshear stress
+ Tmax=(Sn_max-0)/2
+ =25.0 MPa (shear)
\ No newline at end of file diff --git a/698/CH2/EX2.18/P18_stress_in_gas_turbine.sce b/698/CH2/EX2.18/P18_stress_in_gas_turbine.sce new file mode 100644 index 000000000..07206de7b --- /dev/null +++ b/698/CH2/EX2.18/P18_stress_in_gas_turbine.sce @@ -0,0 +1,28 @@ +clc
+//Example 2.18
+//Stress in gas turbine
+//------------------------------------------------------------------------------
+
+//Given data:
+//Stresses
+//Radial Stress
+Sx=20* (10^6) // MPa
+//Tangential Stress
+Sy=50* (10^6) // MPa
+
+res18=mopen(TMPDIR+'18_stress_in_gas_turbine.txt','wt')
+
+mfprintf(res18,'Given \n\tSx= 20 MPa and \n\tSy= 50 MPa\n\n')
+
+Sn_max=Sy
+
+mfprintf(res18,'Maximum normal stress\n\tSn(max)=Sy\n\n')
+
+Tmax=(Sn_max-0)/2
+
+mfprintf(res18,'Maximumshear stress\n\tTmax=(Sn_max-0)/2\n\t=%0.1f MPa (shear)',Tmax*(10^-6))
+
+mclose(res18)
+editor(TMPDIR+'18_stress_in_gas_turbine.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.2/2_response_to_various_loadings.txt b/698/CH2/EX2.2/2_response_to_various_loadings.txt new file mode 100644 index 000000000..f9f8994e3 --- /dev/null +++ b/698/CH2/EX2.2/2_response_to_various_loadings.txt @@ -0,0 +1,108 @@ + NOTE: Sy=0 in all cases
+
+ Sn_max=(Sx/2)+sqrt(((Sx/2)^2)+(Txy^2))
+ Sn_min=(Sx/2)-sqrt(((Sx/2)^2)+(Txy^2))
+ T_max=[Sn_max+Sn_min]/2
+
+
+(a) AXIAL LOAD ONLY
+ Sx=7.64 MN/m^2
+ Txy=0.00 MN/m^2
+ Maximum normal stress=7.64 MN/m^2
+ Minimum normal stress=0.00 MN/m^2
+ Maximum shear stress=3.82 MN/m^2
+
+
+(b) BENDING ONLY
+ Section at which cantilever is fixed is critical
+ {Sx=(M*c)/I at topmost point of section at which cantilever is fixed}
+ At the topmost point of section at which cantilever is fixed,
+ Sx=61.12 MN/m^2
+ Txy=0.00 MN/m^2
+ Maximum normal stress=61.12 MN/m^2
+ Minimum normal stress=0.00 MN/m^2
+ Maximum shear stress=30.56 MN/m^2
+
+ {Sx=-(M*c)/I at topmost point of section at which cantilever is fixed}
+ At the bottom point of section at which cantilever is fixed,
+ Sx=-61.12 MN/m^2
+ Txy=0.00 MN/m^2
+ Maximum normal stress=0.00 MN/m^2
+ Minimum normal stress=-61.12 MN/m^2
+ Maximum shear stress=30.56 MN/m^2
+
+
+(c) TORSION ONLY
+ Here, the critical points occur along the outer surface of the member
+ Sx=0.00 MN/m^2
+ Txy=40.74 MN/m^2
+ Maximum normal stress=40.74 MN/m^2
+ Minimum normal stress=-40.74 MN/m^2
+ Maximum shear stress=40.74 MN/m^2
+
+
+(d) BENDING AND TORSION
+ Section at which cantilever is fixed is critical
+ At the topmost point of section at which cantilever is fixed,
+ Sx=61.12 MN/m^2
+ Txy=40.74 MN/m^2
+ Maximum normal stress=81.49 MN/m^2
+ Minimum normal stress=-20.37 MN/m^2
+ Maximum shear stress=50.93 MN/m^2
+
+ At the bottom point of section at which cantilever is fixed,
+ Sx=-61.12 MN/m^2
+ Txy=40.74 MN/m^2
+ Maximum normal stress=20.37 MN/m^2
+ Minimum normal stress=-81.49 MN/m^2
+ Maximum shear stress=50.93 MN/m^2
+
+ The magnitudes of stresses at the two points are same.
+ The different signs of normal stresses indicate that while topmost fibres are in
+ tension,the bottom fibres are in compression.
+ The different signs of shear stress are of no consequence.
+
+
+(e) BENDING AND AXIAL LOAD
+ Section at which cantilever is fixed is critical
+ At the topmost point of section at which cantilever is fixed,
+ Sx=68.75 MN/m^2
+ Txy=0.00 MN/m^2
+ Maximum normal stress=68.75 MN/m^2
+ Minimum normal stress=0.00 MN/m^2
+ Maximum shear stress=34.38 MN/m^2
+
+ At the bottom point of section at which cantilever is fixed,
+ Sx=-53.48 MN/m^2
+ Txy=0.00 MN/m^2
+ Maximum normal stress=0.00 MN/m^2
+ Minimum normal stress=-53.48 MN/m^2
+ Maximum shear stress=26.74 MN/m^2
+
+
+(f) TORSION AND AXIAL LOAD
+ Critical points are those on the outer surface of the cantilever
+ At the topmost point of section at which cantilever is fixed,
+ Sx=7.64 MN/m^2
+ Txy=40.74 MN/m^2
+ Maximum normal stress=44.74 MN/m^2
+ Minimum normal stress=-37.10 MN/m^2
+ Maximum shear stress=40.92 MN/m^2
+
+
+(g) BENDING, TORSION AND AXIAL LOAD
+ Section at which cantilever is fixed is critical
+ At the topmost point of section at which cantilever is fixed,
+ Sx=68.75 MN/m^2
+ Txy=40.74 MN/m^2
+ Maximum normal stress=87.69 MN/m^2
+ Minimum normal stress=-18.93 MN/m^2
+ Maximum shear stress=53.31 MN/m^2
+
+ At the bottom point of section at which cantilever is fixed,
+ Sx=-53.48 MN/m^2
+ Txy=40.74 MN/m^2
+ Maximum normal stress=22.00 MN/m^2
+ Minimum normal stress=-75.47 MN/m^2
+ Maximum shear stress=48.73 MN/m^2
+
diff --git a/698/CH2/EX2.2/P2_response_to_various_loadings.sce b/698/CH2/EX2.2/P2_response_to_various_loadings.sce new file mode 100644 index 000000000..83902bb10 --- /dev/null +++ b/698/CH2/EX2.2/P2_response_to_various_loadings.sce @@ -0,0 +1,130 @@ +clc
+//Example 2.2
+//Response of cantilever to various uniaxial loadings
+//------------------------------------------------------------------------------
+
+//Given Data:
+//Dimensions
+d=0.05//m
+L=0.25//m
+A=(%pi/4)* d^2
+I=(%pi/64)* d^4
+J=(%pi/32)* d^4
+P=15000//N
+F=3000//N
+M=F*L
+T=1000//Nm
+r=d/2
+
+//------------------------------------------------------------------------------
+//Printing result file to .txt
+res2=mopen(TMPDIR+'2_response_to_various_loadings.txt','wt')
+mfprintf(res2,"\t\tNOTE: Sy=0 in all cases\n")
+mfprintf(res2,"\n\tSn_max=(Sx/2)+sqrt(((Sx/2)^2)+(Txy^2))")
+mfprintf(res2,"\n\tSn_min=(Sx/2)-sqrt(((Sx/2)^2)+(Txy^2))")
+mfprintf(res2,"\n\tT_max=[Sn_max+Sn_min]/2\n\n")
+
+//------------------------------------------------------------------------------
+//Function to find maximun and minimum normal stres and shear stress
+
+funcprot (0)
+
+function [Sn_max,Sn_min,T_max]=stresses(Sx,Txy)
+ Sn_max=(Sx/2)+(sqrt(((Sx/2)^2)+(Txy^2)))
+ Sn_min=(Sx/2)-(sqrt(((Sx/2)^2)+(Txy^2)))
+ T_max=(Sn_max-Sn_min)/2
+
+ mfprintf(res2," Sx=%0.2f MN/m^2\n",Sx*(10^-6))
+ mfprintf(res2," Txy=%0.2f MN/m^2\n",Txy*(10^-6))
+
+ mfprintf(res2," Maximum normal stress=%0.2f MN/m^2\n",Sn_max*(10^-6))
+ mfprintf(res2," Minimum normal stress=%0.2f MN/m^2\n",Sn_min*(10^-6))
+ mfprintf(res2," Maximum shear stress=%0.2f MN/m^2\n\n",T_max*(10^-6))
+
+endfunction
+//------------------------------------------------------------------------------
+
+//(a) AXIAL LOAD ONLY
+mfprintf(res2,"\n(a) AXIAL LOAD ONLY\n")
+Sx=P/A
+Txy=0
+[Sn_max,Sn_min,T_max]=stresses(Sx,Txy)
+//------------------------------------------------------------------------------
+
+//(b) BENDING ONLY
+mfprintf(res2,"\n(b) BENDING ONLY\n")
+mfprintf(res2," Section at which cantilever is fixed is critical\n")
+SxA=(M*r)/I
+Txy=0
+mfprintf(res2," \t{Sx=(M*c)/I at topmost point of section at which cantilever is fixed}\n")
+mfprintf(res2," At the topmost point of section at which cantilever is fixed,\n")
+[Sn_maxA,Sn_minA,T_maxA]=stresses(SxA,Txy)
+
+//------------------------------------------------------------------------------
+SxB=-(M*r)/I
+mfprintf(res2," \t{Sx=-(M*c)/I at topmost point of section at which cantilever is fixed}\n")
+mfprintf(res2," At the bottom point of section at which cantilever is fixed,\n")
+[Sn_maxB,Sn_minB,T_maxB]=stresses(SxB,Txy)
+//------------------------------------------------------------------------------
+
+//(c) TORSION ONLY
+mfprintf(res2,"\n(c) TORSION ONLY\n")
+mfprintf(res2," Here, the critical points occur along the outer surface of the member\n")
+Sx=0
+Txy=(T*r)/J
+[Sn_max,Sn_min,T_max]=stresses(Sx,Txy)
+//------------------------------------------------------------------------------
+
+//(d) BENDING AND TORSION
+mfprintf(res2,"\n(d) BENDING AND TORSION\n")
+mfprintf(res2," Section at which cantilever is fixed is critical\n")
+SxA=(M*r)/I // at topmost point of fixed section.
+//Sx will be negative of above value at bottom point
+Txy=(T*r)/J
+mfprintf(res2," At the topmost point of section at which cantilever is fixed,\n")
+[Sn_maxA,Sn_minA,T_maxA]=stresses(SxA,Txy)
+
+mfprintf(res2," At the bottom point of section at which cantilever is fixed,\n")
+SxB=-(M*r)/I
+[Sn_maxB,Sn_minB,T_maxB]=stresses(SxB,Txy)
+mfprintf(res2," The magnitudes of stresses at the two points are same.\n The different signs of normal stresses indicate that while topmost fibres are in\n tension,the bottom fibres are in compression.\n The different signs of shear stress are of no consequence.\n\n")
+//------------------------------------------------------------------------------
+
+//(e) BENDING AND AXIAL LOAD
+mfprintf(res2,"\n(e) BENDING AND AXIAL LOAD\n")
+mfprintf(res2," Section at which cantilever is fixed is critical\n")
+SxA=(P/A)+((M*r)/I) // at topmost point of fixed section.
+Txy=0
+mfprintf(res2," At the topmost point of section at which cantilever is fixed,\n")
+[Sn_maxA,Sn_minA,T_maxA]=stresses(SxA,Txy)
+
+mfprintf(res2," At the bottom point of section at which cantilever is fixed,\n")
+SxB=(P/A)-(M*r)/I
+[Sn_maxB,Sn_minB,T_maxB]=stresses(SxB,Txy)
+//------------------------------------------------------------------------------
+
+//(f) TORSION AND AXIAL LOAD
+mfprintf(res2,"\n(f) TORSION AND AXIAL LOAD\n")
+mfprintf(res2," Critical points are those on the outer surface of the cantilever\n")
+Sx=(P/A)
+Txy=(T*r)/J
+mfprintf(res2," At the topmost point of section at which cantilever is fixed,\n")
+[Sn_max,Sn_min,T_max]=stresses(Sx,Txy)
+//------------------------------------------------------------------------------
+
+//(g) BENDING, TORSION AND AXIAL LOAD
+mfprintf(res2,"\n(g) BENDING, TORSION AND AXIAL LOAD\n")
+mfprintf(res2," Section at which cantilever is fixed is critical\n")
+SxA=(P/A)+((M*r)/I)
+Txy=(T*r)/J
+mfprintf(res2," At the topmost point of section at which cantilever is fixed,\n")
+[Sn_maxA,Sn_minA,T_maxA]=stresses(SxA,Txy)
+
+mfprintf(res2," At the bottom point of section at which cantilever is fixed,\n")
+SxB=(P/A)-(M*r)/I
+[Sn_maxB,Sn_minB,T_maxB]=stresses(SxB,Txy)
+
+mclose(res2)
+editor(TMPDIR+'2_response_to_various_loadings.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.3/3_max_shear_stress.txt b/698/CH2/EX2.3/3_max_shear_stress.txt new file mode 100644 index 000000000..f0c5823ef --- /dev/null +++ b/698/CH2/EX2.3/3_max_shear_stress.txt @@ -0,0 +1,11 @@ +The maximum shear stress may occur at two locations:
+ a) Surface of the cantilever (due to bending moment)
+ OR
+ b) Midsection of the cantilever (due to transverse shear load)
+
+
+Shear stress on the surface is 3.6 MN/m^2
+Shear stress at the midsection is 4.5 MN/m^2
+
+
+Maximum shear stress occurs at the midsection and is equal to 4.5 MN/m^2
\ No newline at end of file diff --git a/698/CH2/EX2.3/P3_max_shear_stress.sce b/698/CH2/EX2.3/P3_max_shear_stress.sce new file mode 100644 index 000000000..74c46e240 --- /dev/null +++ b/698/CH2/EX2.3/P3_max_shear_stress.sce @@ -0,0 +1,46 @@ +clc
+//Example 2.3
+//Maximum shear stress
+//------------------------------------------------------------------------------
+
+//Given Data:
+//Dimensions
+L=0.04//m
+b=0.02//m
+h=0.1//m
+//Load
+W=6000//N
+
+//The maximum shear stress may occur at two locations:
+//a) Surface of the cantilever (due to bending moment) OR
+//b) Midsection of the cantilever (due to transverse shear load)
+
+//On the surface,
+//Tmax=(1/2)*(Mc/I); Tmax=maximum shear stress, Mc=bending moment; I=moment of area
+I=(b*(h^3))/12
+M=W*L
+c=h/2
+Tmax1=(1/2)*((M*c)/I)
+
+//At the midsection
+//Tmax=(3/2)*(V/A); V=Load=W; A=area of cross section
+Tmax2=(3/2)*(W/(b*h))
+
+//------------------------------------------------------------------------------
+//Printing result file to .txt
+res3=mopen(TMPDIR+'3_max_shear_stress.txt','wt')
+mfprintf(res3,"The maximum shear stress may occur at two locations:\n")
+mfprintf(res3,"\ta) Surface of the cantilever (due to bending moment)\n\t\t OR\n")
+mfprintf(res3,"\tb) Midsection of the cantilever (due to transverse shear load)\n")
+mfprintf(res3,"\n\nShear stress on the surface is %0.1f MN/m^2",Tmax1*(10^-6))
+mfprintf(res3,"\nShear stress at the midsection is %0.1f MN/m^2\n",Tmax2*(10^-6))
+Tmax=max(Tmax1,Tmax2)
+if Tmax==Tmax1
+ mfprintf(res3,"\n\nMaximum shear stress occurs on the surface and is equal to %0.1f MN/m^2",Tmax1*(10^-6))
+else
+ mfprintf(res3,"\n\nMaximum shear stress occurs at the midsection and is equal to %0.1f MN/m^2",Tmax2*(10^-6))
+end
+mclose(res3)
+editor(TMPDIR+'3_max_shear_stress.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.4/4_principle_stresses.txt b/698/CH2/EX2.4/4_principle_stresses.txt new file mode 100644 index 000000000..9bd0ce2a0 --- /dev/null +++ b/698/CH2/EX2.4/4_principle_stresses.txt @@ -0,0 +1,9 @@ + Sn_max=(Sx/2)+sqrt(((Sx/2)^2)+(Txy^2))
+ Sn_min=(Sx/2)-sqrt(((Sx/2)^2)+(Txy^2))
+ T_max=[Sn_max+Sn_min]/2
+
+The maximum normal stress is -300 N/m^2 and its nature is compressive
+
+The minimum normal stress is -1300 N/m^2 and its nature is compressive
+
+The maximum shear stress is -650 N/m^2
\ No newline at end of file diff --git a/698/CH2/EX2.4/P4_principle_stresses.sce b/698/CH2/EX2.4/P4_principle_stresses.sce new file mode 100644 index 000000000..03bb843c3 --- /dev/null +++ b/698/CH2/EX2.4/P4_principle_stresses.sce @@ -0,0 +1,38 @@ +clc
+//Example 2.4
+//Normal and shear stresses
+
+//------------------------------------------------------------------------------
+//Given Data:
+//Stresses
+Sx=-400//N/m^2(compressive)
+Sy=-1200//N/m^2(compressive)
+Txy=300//N/m^2
+
+res4=mopen(TMPDIR+'4_principle_stresses.txt','wt')
+mfprintf(res4,"\tSn_max=(Sx/2)+sqrt(((Sx/2)^2)+(Txy^2))")
+mfprintf(res4,"\n\tSn_min=(Sx/2)-sqrt(((Sx/2)^2)+(Txy^2))")
+mfprintf(res4,"\n\tT_max=[Sn_max+Sn_min]/2\n\n")
+
+//------------------------------------------------------------------------------
+
+//Maximum normal stress
+Sn_max=((Sx+Sy)/2)+sqrt((((Sx-Sy)/2)^2)+((Txy)^2))
+mfprintf(res4,"The maximum normal stress is %d N/m^2 and its nature is",Sn_max)
+nature(Sn_max)
+
+//Minimum normal stress
+Sn_min=((Sx+Sy)/2)-sqrt((((Sx-Sy)/2)^2)+((Txy)^2))
+mfprintf(res4,"\n\nThe minimum normal stress is %d N/m^2 and its nature is",Sn_min)
+nature(Sn_min)
+
+//Third principal stress=0
+Sz=0
+
+//Maximum shear stress
+Tmax=(Sn_min-Sz)/2
+mfprintf(res4,"\n\nThe maximum shear stress is %d N/m^2",Tmax)
+mclose(res4)
+editor(TMPDIR+'4_principle_stresses.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.6/6_max_tensile_and_shear_stress.txt b/698/CH2/EX2.6/6_max_tensile_and_shear_stress.txt new file mode 100644 index 000000000..cf9b0e498 --- /dev/null +++ b/698/CH2/EX2.6/6_max_tensile_and_shear_stress.txt @@ -0,0 +1,9 @@ +The critical stress is at the section at which beam is fixed
+ Sy=0
+
+ Sx=(P/A)+(Mc/I).....Tensile stress
+
+ Txy=(T*(d/2))/J.....Shear stress
+
+The maximum tensile stress is 25.8 MN/m^2 (tension)
+The maximum shear stress is 13.2 MN/m^2 (shear)
\ No newline at end of file diff --git a/698/CH2/EX2.6/P6_max_tensile_and_shear_stress.sce b/698/CH2/EX2.6/P6_max_tensile_and_shear_stress.sce new file mode 100644 index 000000000..f70d738cd --- /dev/null +++ b/698/CH2/EX2.6/P6_max_tensile_and_shear_stress.sce @@ -0,0 +1,52 @@ +clc
+//Example 2.6
+//Maximum tensile and shear stress
+//------------------------------------------------------------------------------
+
+//Given Data:
+//Dimensions
+d=0.05//m
+//Area
+A=(%pi/4)*(d^2)
+//Load
+P=9000//N
+//Torsional Moment
+T=100//Nm
+//Eccentricity[distance by which line of action of load is offset from axis of shaft]
+e=0.028//m
+
+//------------------------------------------------------------------------------
+//Moment of area
+I=(%pi/64)*(d^4)
+//Polar moment
+J=(%pi/32)*(d^4)
+
+//Bending moment
+Mc=P*e*(d/2)
+
+//The critical stress is at the section at which beam is fixed
+Sx=(P/A)+(Mc/I)
+//Shaft is not loaded along X-direction
+Sy=0
+
+//Shear stress
+Txy=(T*(d/2))/J
+//Maximum tensile stress
+Sn_max=(Sx/2)+sqrt(((Sx/2)^2)+(Txy^2))
+//Maximum shear stress
+T_max=sqrt(((Sx/2)^2)+(Txy^2))
+
+//------------------------------------------------------------------------------
+//Printing result file to .txt
+res6=mopen(TMPDIR+'6_max_tensile_and_shear_stress.txt','wt')
+mfprintf(res6,"The critical stress is at the section at which beam is fixed\n")
+mfprintf(res6,"\tSy=0\n")
+mfprintf(res6,"\n\tSx=(P/A)+(Mc/I).....Tensile stress\n")
+mfprintf(res6,"\n\tTxy=(T*(d/2))/J.....Shear stress\n\n")
+mfprintf(res6,"The maximum tensile stress is %0.1f MN/m^2",(Sn_max*(10^-6)))
+nature(Sn_max)
+mfprintf(res6,"\nThe maximum shear stress is %0.1f MN/m^2 (shear)",(T_max*(10^-6)))
+mclose(res6)
+editor(TMPDIR+'6_max_tensile_and_shear_stress.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
diff --git a/698/CH2/EX2.7/7_stress_in_stepped_bar.txt b/698/CH2/EX2.7/7_stress_in_stepped_bar.txt new file mode 100644 index 000000000..be4e8f6a3 --- /dev/null +++ b/698/CH2/EX2.7/7_stress_in_stepped_bar.txt @@ -0,0 +1,12 @@ +The force produces deflection in both the portions of steel.
+Total deflection will be sum of inividual deflections.
+The deflection is given by:
+ x=(F*L)/(A*E)
+ x=x1+x2
+
+The total deflection is 3.183 mm
+
+ Stress=Force/Area
+
+The tensile stress at larger cross section area is 25.46 MPa
+The tensile stress at smaller cross section area is 101.86 MPa
\ No newline at end of file diff --git a/698/CH2/EX2.7/P7_stress_in_stepped_bar.sce b/698/CH2/EX2.7/P7_stress_in_stepped_bar.sce new file mode 100644 index 000000000..59cd45163 --- /dev/null +++ b/698/CH2/EX2.7/P7_stress_in_stepped_bar.sce @@ -0,0 +1,49 @@ +clc
+//Example 2.7
+//Stresses in stepped bar
+//------------------------------------------------------------------------------
+
+//Given Data:
+//Load
+F=50000//N
+
+//Dimensions
+d1=0.05//m
+L1=5//m
+d2=0.025//m
+L2=L1
+A1=(%pi/4)*(d1^2)
+A2=(%pi/4)*(d2^2)
+
+//Modulus of rigidity
+E=200e9
+//------------------------------------------------------------------------------
+
+//Total extension of the bar will be the sum of individual extensions
+//Extension of larger part
+x1=(F*L1)/(A1*E)
+//Extension of the smaller part
+x2=(F*L2)/(A2*E)
+
+//Total extension of the bar
+x=x1+x2
+
+//Stress in part of larger diameter
+S1=F/A1
+//Stress in part of smaller diameter
+S2=F/A2
+
+//------------------------------------------------------------------------------
+//Printing result file to .txt
+res7=mopen(TMPDIR+'7_stress_in_stepped_bar.txt','wt')
+mfprintf(res7,"The force produces deflection in both the portions of steel.\nTotal deflection will be sum of inividual deflections.\n")
+mfprintf(res7,"The deflection is given by:\n\t\tx=(F*L)/(A*E)\n")
+mfprintf(res7,"\t\tx=x1+x2\n\n")
+mfprintf(res7,"The total deflection is %0.3f mm\n",x*(10^3))
+mfprintf(res7,"\n\tStress=Force/Area\n\n")
+mfprintf(res7,"The tensile stress at larger cross section area is %0.2f MPa\n",S1*(10^-6))
+mfprintf(res7,"The tensile stress at smaller cross section area is %0.2f MPa",S2*(10^-6))
+mclose(res7)
+editor(TMPDIR+'7_stress_in_stepped_bar.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.8/8_extreme_normal_stresses.txt b/698/CH2/EX2.8/8_extreme_normal_stresses.txt new file mode 100644 index 000000000..e4314abf6 --- /dev/null +++ b/698/CH2/EX2.8/8_extreme_normal_stresses.txt @@ -0,0 +1,10 @@ +Sx=-12.45 MN/m^2 compressive
+
+Txy=3.62 MN/m^2 shear
+
+ Sy=0
+
+The maximum normal stress is 0.98 MN/m^2 and its nature is tensile
+
+The minimum normal stress is -13.43 MN/m^2 and its nature is compressive
+
diff --git a/698/CH2/EX2.8/P8_extreme_normal_stresses.sce b/698/CH2/EX2.8/P8_extreme_normal_stresses.sce new file mode 100644 index 000000000..d4a75899f --- /dev/null +++ b/698/CH2/EX2.8/P8_extreme_normal_stresses.sce @@ -0,0 +1,43 @@ +clc
+//Example 2.8
+//Extreme normal stresses
+//------------------------------------------------------------------------------
+
+//Given Data:
+//Dimensions
+d=0.075//m
+L=0.25//m
+
+//Loads
+P=-55000//N (compression)
+T=300//Nm
+
+//Polar moment of inertia
+J=(%pi/32)*(d^4)
+
+//------------------------------------------------------------------------------
+//Printing result file to .txt
+res8=mopen(TMPDIR+'8_extreme_normal_stresses.txt','wt')
+//Normal stress along Y direction Sy=0
+//Normal stress along X direction
+Sx=P/((%pi/4)*(d^2))
+mfprintf(res8,"Sx=%0.2f MN/m^2 ",Sx*(10^-6))
+nature(Sx)
+//Shear stress
+Txy=(T*(d/2))/J
+mfprintf(res8,"\nTxy=%0.2f MN/m^2 shear",Txy*(10^-6))
+
+mfprintf(res8,"\n\n\tSy=0\n\n")
+//Maximum normal stress
+Sn_max=(Sx/2)+sqrt(((Sx/2)^2)+(Txy^2))
+mfprintf(res8,"The maximum normal stress is %0.2f MN/m^2 and its nature is",Sn_max*(10^-6))
+nature(Sn_max)
+
+//Minimum normal stress
+Sn_min=(Sx/2)-sqrt(((Sx/2)^2)+(Txy^2))
+mfprintf(res8,"\nThe minimum normal stress is %0.2f MN/m^2 and its nature is",Sn_min*(10^-6))
+nature(Sn_min)
+mclose(res8)
+editor(TMPDIR+'8_extreme_normal_stresses.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file diff --git a/698/CH2/EX2.9/9_stresses_at_section.txt b/698/CH2/EX2.9/9_stresses_at_section.txt new file mode 100644 index 000000000..4f52a6063 --- /dev/null +++ b/698/CH2/EX2.9/9_stresses_at_section.txt @@ -0,0 +1,19 @@ +Bending moment due to horizontal load about midpoint Mx= 250.00 Nm
+Bending moment due to vertical load about midpoint My= -400.00 Nm
+The total bending moment:
+ M=sqrt((Mx^2)+(My^2))
+The total bending moment M = 471.70 Nm
+
+The normal stress is given by
+ Sx=Py/A-((M*(d/2))/I)
+Normal stress acting at the section is -39.46 MN/m^2
+
+The shear stress is given by
+ Txy=(T*(d/2))/J
+Normal stress acting at the section is 8.15 MN/m^2
+
+Maximum normal stress is 1.62 MN/m^2
+Minimum normal stress is -41.07 MN/m^2
+
+The maximum numerical normal stress is 41.07 MN/m^2
+The maximum shear stress is 21.344732 MN/m^2
\ No newline at end of file diff --git a/698/CH2/EX2.9/P9_stresses_at_section.sce b/698/CH2/EX2.9/P9_stresses_at_section.sce new file mode 100644 index 000000000..776d56ded --- /dev/null +++ b/698/CH2/EX2.9/P9_stresses_at_section.sce @@ -0,0 +1,70 @@ +clc
+//Example 2.9
+//Stresses at a section
+//------------------------------------------------------------------------------
+
+//Given Data:
+//Loads
+//Load acting vertically downward
+Py=-2000//N (compressive)
+//Load acting horizontally
+Px=1000//N
+
+//Dimensions
+//Diameter
+d=0.05//m
+//length of neck region
+a=0.125//m
+//distance between midpoint of section and line of action of vertical load
+b=0.2//m
+//distance between midpoint of section and line of horizontal load
+c=0.25//m
+//Moment of inertia
+I=(%pi/64)*(d^4)
+//Polar moment of inertia
+J=(%pi/32)*(d^4)
+//Area of cross section
+A=(%pi/4)*(d^2)
+//------------------------------------------------------------------------------
+
+//Printing steps and result file to .txt
+res9=mopen(TMPDIR+'9_stresses_at_section.txt','wt')
+//Bending moment due to horizontal load about midpoint
+Mx=Px*c
+mfprintf(res9,"Bending moment due to horizontal load about midpoint Mx= %0.2f Nm\n",Mx)
+//Bending moment due to vertical load about midpoint
+My=Py*b
+mfprintf(res9,"Bending moment due to vertical load about midpoint My= %0.2f Nm\n",My)
+//Total bending moment
+M=sqrt((Mx^2)+(My^2))
+mfprintf(res9,"The total bending moment:\n\tM=sqrt((Mx^2)+(My^2))\n")
+mfprintf(res9,"The total bending moment M = %0.2f Nm\n\n",M)
+//Normal stress in X direction
+Sx=Py/A-((M*(d/2))/I)
+mfprintf(res9,"The normal stress is given by\n\tSx=Py/A-((M*(d/2))/I)\n")
+mfprintf(res9,"Normal stress acting at the section is %0.2f MN/m^2\n\n",Sx*(10^-6))
+
+//Couple moment due to horizontal load about midpoint
+T=Px*b
+//Shear stress
+Txy=(T*(d/2))/J
+mfprintf(res9,"The shear stress is given by\n\tTxy=(T*(d/2))/J\n")
+mfprintf(res9,"Normal stress acting at the section is %0.2f MN/m^2\n\n",Txy*(10^-6))
+//------------------------------------------------------------------------------
+
+//Extreme streeses:
+//Maximum normal stress
+Sn_max=(Sx/2)+sqrt(((Sx/2)^2)+(Txy^2))
+mfprintf(res9,"Maximum normal stress is %0.2f MN/m^2\n",Sn_max*(10^-6))
+//Minimum normal stress
+Sn_min=(Sx/2)-sqrt(((Sx/2)^2)+(Txy^2))
+mfprintf(res9,"Minimum normal stress is %0.2f MN/m^2\n\n",Sn_min*(10^-6))
+S=max((abs(Sn_max)),(abs(Sn_min)))
+mfprintf(res9,"The maximum numerical normal stress is %0.2f MN/m^2",S*(10^-6))
+//Maximum shear stress
+T_max=sqrt(((Sx/2)^2)+(Txy^2))
+mfprintf(res9,"\nThe maximum shear stress is %f MN/m^2",T_max*(10^-6))
+mclose(res9)
+editor(TMPDIR+'9_stresses_at_section.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program-----------------------------------
\ No newline at end of file |