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-rw-r--r--3772/CH5/EX5.1/Ex5_1.sce23
-rw-r--r--3772/CH5/EX5.11/Ex5_11.sce41
-rw-r--r--3772/CH5/EX5.12/Ex5_12.sce42
-rw-r--r--3772/CH5/EX5.13/Ex5_13.sce40
-rw-r--r--3772/CH5/EX5.14/Ex5_14.sce39
-rw-r--r--3772/CH5/EX5.15/Ex5_15.sce39
-rw-r--r--3772/CH5/EX5.2/Ex5_2.sce36
-rw-r--r--3772/CH5/EX5.3/Ex5_3.sce30
-rw-r--r--3772/CH5/EX5.4/Ex5_4.sce28
-rw-r--r--3772/CH5/EX5.8/Ex5_8.sce32
-rw-r--r--3772/CH5/EX5.9/Ex5_9.sce27
11 files changed, 377 insertions, 0 deletions
diff --git a/3772/CH5/EX5.1/Ex5_1.sce b/3772/CH5/EX5.1/Ex5_1.sce
new file mode 100644
index 000000000..943575825
--- /dev/null
+++ b/3772/CH5/EX5.1/Ex5_1.sce
@@ -0,0 +1,23 @@
+// Problem 5.1,Page no.121
+
+clc;clear;
+close;
+
+b=100 //mm //width of timber joist
+d=200 //mm //depth of joist
+L=3 //m //Length of beam
+sigma=7 //KN/mm**2 //bending stress
+w_1=5 //KN/mm**2 //unit weight of timber
+
+//Calculations
+w=0.1*0.2*1*5*100 //N/m //self weight of the joist
+I_xx=1*12**-1*100*200**3 //mm**4 //M.I of section about N.A
+
+//M=W*L+w*L**2*2**-1 //Max Bending moment
+//Therefore,M=(3*W+450)
+
+//using the relation M*I**-1=sigma*y**-1,we get
+W=(((7*2*10**8)*(100*10**3*3)**-1)-450)*3**-1 //N //Max Load applied
+
+//Result
+printf("The Max value of Load applied is %.2f N",W)
diff --git a/3772/CH5/EX5.11/Ex5_11.sce b/3772/CH5/EX5.11/Ex5_11.sce
new file mode 100644
index 000000000..c4eecfb1b
--- /dev/null
+++ b/3772/CH5/EX5.11/Ex5_11.sce
@@ -0,0 +1,41 @@
+// Problem 5.11,Page no.131
+
+clc;clear;
+close;
+
+D=4 //cm //Outside diameter
+d=3 //cm //inside diamter
+L=2 //m //span of beam
+W=1000 //N //Max safe Load
+
+//Calculations
+
+I=%pi*64**-1*(D**4-d**4) //cm**4 //M.I
+A=%pi*4**-1*(D**2-d**2) //cm**2 //Area
+y=2
+Z=I*y**-1 //cm**3 //Section modulus
+
+M=W*L*4**-1 //N*cm //Max bending moment
+
+//From Flexural Formula
+sigma=M*Z**-1 //N/cm**2
+
+//For Tubes
+//M.I about x-x axis
+I_1=4*(8.59+5.492*2**2) //cm**4
+
+Z_1=122.32*4**-1 //cm**3
+
+//M=W_1*200*4**-1 //N*cm
+//After substituting values we get
+//M=50*W_1 (equation 1)
+
+//Again from Flexural Formula
+M=sigma*Z_1
+
+//substitute value of M in equation 1
+
+W=11640*30.58*50**-1 //N
+
+//Result
+printf("Max central load is %.2f N",W)
diff --git a/3772/CH5/EX5.12/Ex5_12.sce b/3772/CH5/EX5.12/Ex5_12.sce
new file mode 100644
index 000000000..8267df0ea
--- /dev/null
+++ b/3772/CH5/EX5.12/Ex5_12.sce
@@ -0,0 +1,42 @@
+// Problem 5.12,Page no.133
+
+clc;clear;
+close;
+
+b=200 //mm //width of beam
+d=300 //mm //depth of beam
+t=12 //mm //thickness of beam
+E_s=220 //KN/m**2 //modulus of elasticity of steel
+E_w=11 //KN/m**2 //modulus of elasticity of timber
+sigma_s=115 //MN/m**2 //stress of steel
+sigma_w=9.2 //MN/m**2 //stress of timber
+L=2 //m //Span of beam
+
+//Calculations
+
+//E_w*E_s**-1=1*20**-1 //ratio of Modulus of elasticity of timber to steel
+
+
+//(Part-1)
+b_1=b*20**-1 //mm //web thickness of transformed section
+stress=20*sigma_w //MN/m**2 //Allowable stress in web of equivalen beam
+//But allowable stress in flanges is sigma_s is 115 KN/m**2 and therefore taken into consideration
+
+
+d_1=324 //mm //depth of beam with thickness in consideration
+I=1*12**-1*0.2*0.324**3-2*1*12**-1*0.095*0.3**3 //m**4 //M.I of transformed section
+
+//Using Relation, M*I**-1=sigma*y**-1 we get
+
+//Part-2
+M_max=I*(324*10**-3*2**-1)**-1*sigma_s*10**6 //N*m //Max allowable Bending moment for steel section
+
+//Part-3
+//As beam is simply supported at the ends and the load is applied at the centre of beam
+//M_max=W*L*4**-1 //Max Bending moment
+W=M_max*4*L**-1 //N //Allowable stress Load
+
+//Result
+printf("Web thickness of Equivalent steel section is %.2f mm",b_1)
+printf("\n Max Allowable bending moment for section is %.2f N-m",M_max)
+printf("\n Allowable safe Load is %.2f N",W)
diff --git a/3772/CH5/EX5.13/Ex5_13.sce b/3772/CH5/EX5.13/Ex5_13.sce
new file mode 100644
index 000000000..69d2eb537
--- /dev/null
+++ b/3772/CH5/EX5.13/Ex5_13.sce
@@ -0,0 +1,40 @@
+// Problem 5.13,Page no.135
+
+clc;clear;
+close;
+
+d=10 //cm //distance between joists
+t=2 //cm //thickness of steel plate
+d_2=20 //cm //depth of beam
+sigma_t=8.5 //N/mm**2 //stress in timber
+E_s=2*10**5 //N/mm**2 //Modulus of elasticity of steel
+E_t=10**4 //N/mm**2 ////Modulus of elasticity of timber
+L=5 //cm //span of beam
+
+//calculation
+sigma=10*15**-1*sigma_t //stress in timber at distance of 10 cm from XX (N/mm**2)
+
+dell=sigma*E_t**-1 //strain in timber at 10 cm from XX (N/mm**2)
+
+sigma_s=dell*E_s //N/mm**2 //Max stress
+
+//For Timber
+Z_w=1*6**-1*10*30**2*2 //cm**3 //section modulus of timber
+M_w=sigma_t*100*Z_w //moment of resistance of timber (N-cm)
+
+//For steel
+Z_s=1*6**-1*2*20**2 //cm**3 //section modulus of steel
+M_s=sigma_s*Z_s*100 //moment of resistance of steel (N-cm)
+
+M=(M_w+M_s)*10**-5 //total moment of resistance(N-cm)
+
+//M=w*L**2*8**-1 //N*cm //Max bending moment
+w=8*M*(L**2)**-1 //kN/m //Max uniform distributed Load
+
+//Result kN/m
+printf("Moment of resistance is %.3f N-cm",M)
+printf("\n Max uniform distributed Load = %.3f kN/m",w)
+// answer in the textbook is not accurate.
+
+
+
diff --git a/3772/CH5/EX5.14/Ex5_14.sce b/3772/CH5/EX5.14/Ex5_14.sce
new file mode 100644
index 000000000..ac0d42a62
--- /dev/null
+++ b/3772/CH5/EX5.14/Ex5_14.sce
@@ -0,0 +1,39 @@
+// Problem 5.14,Page no.136
+
+clc;clear;
+close;
+
+B=10 //cm //width of timber section
+D=15 //cm //depth of timber section
+b=10 //cm //width of steel plate
+t=12 //mm //thickness
+w=3 //KN/m //Uniformly distributed Load
+L=4 //m //Span of beam
+m=20 //Ratio of modulus of elasticity of steel to timber
+W=3 //KN/m //Load
+
+//Calculations
+
+y_1=15*2**-1 //C.G of timber
+y_2=1.2*2**-1 //C.G of steel plate
+b_s=10*m**-1 //cm //Equivalent width of steel
+Y_bar=(10*1.2*0.6+15*0.5*8.7)*(10*1.2+15*0.5)**-1 //cm //distance of C.G from bottom edge
+
+I=1*12**-1*10*(1.2)**3+10*1.2*(3.72-0.6)**2+1*12**-1*0.5*(15)**3+0.5*15*(7.5-3.72)**2
+M=W*10**5*L**2*8**-1 //N*m
+
+Y_bar_1=3.72 //cm //C.G from bottom edge
+Y_bar_2=16.2-Y_bar //cm //C.G from top edge
+
+sigma_1=(M*I**-1*Y_bar_1)*10**-2 //N/mm**2 //stress at bottom
+
+sigma_2=(M*I**-1*Y_bar_2)*10**-2 //N/mm**2 //stress at top
+
+sigma_max=sigma_2*m**-1
+
+//The Answers in book for Moment of Inertia about x-x axis onwards are incorrect
+
+//Result
+printf("Moment of Inertia = %.f N-m",M)
+printf("\n The Max Stress in steel is %.2f N/mm^2",sigma_1)
+printf("\n The Max Stress in timber is %.2f N/mm**2",sigma_max)
diff --git a/3772/CH5/EX5.15/Ex5_15.sce b/3772/CH5/EX5.15/Ex5_15.sce
new file mode 100644
index 000000000..6adcca091
--- /dev/null
+++ b/3772/CH5/EX5.15/Ex5_15.sce
@@ -0,0 +1,39 @@
+// Problem 5.15,Page no.137
+
+clc;clear;
+close;
+
+B=20 //cm //width of timber
+D=30 //cm //depth of timber
+d=25 //cm //depth of steel plate
+b=1.2 //cm //width of steel plate
+sigma_s=90 //N/mm**2 //Bending stress in steel
+sigma_t=6 //N/mm**2 //Bending stress in timber
+m=20 //Ratio of modulus of elasticity of of steel to timber
+
+//Calculation
+
+//Equivalent width of wood section,when 1.2 cm wide steel plate is replaced by steel plate is
+b_1=1.2*20 //cm
+d_1=25 //cm //depth of wood section
+y_1=d*2**-1 //cm //C.G of timber section
+y_2=D*2**-1 //cm //C.G of steel section
+
+Y_bar=(2*d*b_1*y_1+D*B*y_2)*(2*d*b_1+D*B)**-1 //cm //Distance of C.G from Bottom edge
+I=B*D**3*12**-1+B*D*(y_2-Y_bar)**2+2*(b_1*d_1**3*12**-1+b_1*d_1*(Y_bar-y_1)**2) //M.I of equivalent timber section about N.A
+Y=30-Y_bar //distance of C.G from top of equivalent wood section
+
+//Thus max stress will occur at top and that in steel will occur at bottom
+//sigma_s=m*Y_bar*Y**-1*sigma_t
+
+//After simplifying we get
+//sigma_s=15.99*sigma_t
+
+sigma_t=sigma_s*15.99**-1 //N/mm**2 //Max stress in Equivalent timber section
+
+Z_t=I*Y**-1 //Section modulus of equivalent section
+M=sigma_t*Z_t*10**-5*100 //Moment of resistance of beam
+
+//Result
+printf("Position of N.A is %.2f cm",Y_bar)
+printf("\n Moment of Resistance of beam is %.2f kN-m",M)
diff --git a/3772/CH5/EX5.2/Ex5_2.sce b/3772/CH5/EX5.2/Ex5_2.sce
new file mode 100644
index 000000000..bedd1f905
--- /dev/null
+++ b/3772/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,36 @@
+
+// Problem 5.2,Page no.122
+
+clc;clear;
+close;
+
+D=160 //mm //Overall Depth
+B=150 //mm //Width of Flange
+f_t=40 //mm //Flange thickness
+W_t=50 //mm //Web thickness
+sigma_t=20 //N/mm**2 //tension stress
+sigma_c=75 //N/mm**2 //compression stress
+
+//Calculations
+
+//Rectangle-1
+a_1=150*40 //mm**2 //Area of Rectangle-1
+y_1=40*2**-1 //mm //C.G of Rectangle-1
+
+//Rectangle-2
+a_2=120*50 //mm**2 //Area of Rectangle-2
+y_2=40+120*2**-1 //mm //C.G of Rectangle-2
+
+Y_bar=(a_1*y_1+a_2*y_2)*(a_1+a_2)**-1 //mm //Distance of C.G from the bottom flange
+I=1*12**-1*150*40**3+150*40*(60-40)**2+1*12**-1*50*120**3+50*120*(100-60)**2 //mm**4 //M.I of section about N.A
+y_t=60 //mm //Permissible tensile stress at the bottom face of flange from N.A
+y_c=100 //mm //Permissible tensile stress at the top face of flange from N.A
+
+//M=W*L*4**-1 //Max bending mooment at the centre
+
+//Using the relation M*I**-1=sigma_t*y_t**-1 we get
+W=(0.333*4*272*10**5)*(2.5*1000)**-1 //N //MAx central load
+
+//Result
+printf("The Max Bending Moment at the centre is %.2f N",W)
+//Answer is wrong in the textbook.
diff --git a/3772/CH5/EX5.3/Ex5_3.sce b/3772/CH5/EX5.3/Ex5_3.sce
new file mode 100644
index 000000000..1e8ba01c6
--- /dev/null
+++ b/3772/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,30 @@
+// Problem 5.3,Page no.123
+
+clc;clear;
+close;
+
+b=10 //cm //width of beam
+d=20 //cm depth of beam
+
+//Calculations
+
+//R_a and R_b are the reactions at A and B respectively.
+//Moment of all forces about A
+
+R_b=(4*4*4*2**-1-2*1.5)*(2)**-1 //KN
+//R_a+R_b=18
+R_a=18-R_b
+
+//Consider a section at a distance x from A
+//M_x=9.25*x-2(x-1.5)-4*x*x*2**-1=7.25*x+3-2*x**2
+
+//Taking derivative of above equation to find max value of M_x we get
+x=1.81 //m
+
+M=7.25*x+3-2*x**2 //kN*m
+I=b*d**3*12**-1 //cm**4 //M.I of the section
+y=10
+sigma=M*I**-1*y*10**8*(10**2)**-1 //Max bending stress
+
+//Result
+printf("The Max Bending stress is %.2f kN/m^2",sigma)
diff --git a/3772/CH5/EX5.4/Ex5_4.sce b/3772/CH5/EX5.4/Ex5_4.sce
new file mode 100644
index 000000000..8269d81ca
--- /dev/null
+++ b/3772/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,28 @@
+// Problem 5.4,Page no.124
+
+clc;clear;
+close;
+
+b=10 //cm //width of beam
+d=20 //cm depth of beam
+
+sigma=8 //N/mm**2 //Max bending stress
+W=5000 //N/m**2 //Load of floor
+A=450 //cm**2 //Area of joist
+L=5 //m //span of floor
+
+//Calculations
+//Let x be the centre to centre spacingof the joists
+
+//A_1=5*x**2 //m**2 //Area of floor between any two joists
+//W_1=5*x*W //N //total load supported by one interior joist
+//M=W_1*L*8**-1 //Max bending moment
+I=1*12**-1*b*(d*10**-2)**3*10**-2 //m**4 //M.I of joist
+y=0.15 //cm //Distance of of farthest fibre
+M=I*y**-1*sigma //N*m
+
+//Now equating to max bending moment we get
+x=(18000*8)*(25000*5)**-1
+
+//Result
+printf("The Max Bending Moment is %.2f m",x)
diff --git a/3772/CH5/EX5.8/Ex5_8.sce b/3772/CH5/EX5.8/Ex5_8.sce
new file mode 100644
index 000000000..189a866ea
--- /dev/null
+++ b/3772/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,32 @@
+// Problem 5.8,Page no.127
+
+clc;clear;
+close;
+
+L=3 //m //span of beam
+t=20 //mm //Thickness of steel
+D=200 //mm //overall depth
+B=140 //mm //overall width
+b=100 //mm //width of inner rectangle
+d=160 //mm //depth of inner rectangle
+w=77 //KN/mm**2
+sigma=100 //N/mm**2 //Bending stress
+//Calculations
+V=((D*10**-3*B*10**-3)-(d*10**-3*b*10**-3)) //m**3 //Volume of rectangular box
+W=V*3*w //KN //Weight of Beam
+I=(B*D**3-b*d**3)*12**-1 //mm**4 //M.I of beam section
+
+//Now using the relation,M*I**-1=sigma*y**-1
+
+y=200 //mm //distance from farthest fibre
+M=I*sigma*2*y**-1 //N*mm
+
+//M=3000*W+2772*3000*2**-1
+//After sub values in above equation we get
+
+W=((59.2*10**6-2772*3000*2**-1)*(3000)**-1)*10**-3 //KN //Max concentrated Load at free end
+
+F=W+2.772*2**-1 //KN //shear force at half length
+
+//Result
+printf("The shear force at half length is %.2f kN",F)
diff --git a/3772/CH5/EX5.9/Ex5_9.sce b/3772/CH5/EX5.9/Ex5_9.sce
new file mode 100644
index 000000000..390dc25fc
--- /dev/null
+++ b/3772/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,27 @@
+// Problem 5.9,Page no.128
+
+clc;clear;
+close;
+
+B=24 //mm //width of beam section
+D=21.7 //mm //depth of beam section
+E_1=11440 //MN/m**2 //Modulus of Elasticity parallel grain
+E_2=2860 //MN/m**2 ////Modulus of Elasticity perpendicular grain
+sigma_1=8.57 //MN/m**2
+sigma_2=2.14 //MN/m**2
+L=1.2 //m //span of beam
+
+//Calculations
+
+//Ratio of smaller modulus to larger modulus is E_2:E_1=1:4
+//Dimension of transformed Beam section
+b=18 //mm //width of Beam section
+d=3.1 //mm //depth of beam section
+
+I=(1*12**-1*B*10**-3*(D*10**-3)**3)-(3*(1*12**-1*b*10**-3*(d*10**-3)**3)) //m**4 //M.I of transformed section
+y=21.7*10**-3*2**-1
+M=I*sigma_1*10**6*y**-1 //N*m //Safe B.M
+P=4*M*L**-1 //N
+
+//Result
+printf("Safe value of Load is %.2f N",P)
generated by cgit (git 2.34.1) at 2024-12-22 09:40:37 +0000