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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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treedbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /698/CH4
parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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-rw-r--r--698/CH4/EX4.2/2_offset_bar.txt30
-rw-r--r--698/CH4/EX4.2/P2_offset_bar.sce74
-rw-r--r--698/CH4/EX4.3/3_determination_of_e.txt11
-rw-r--r--698/CH4/EX4.3/P3_determination_of_e.sce38
-rw-r--r--698/CH4/EX4.4/4_stress_in_spring_clip.txt12
-rw-r--r--698/CH4/EX4.4/P4_stress_in_spring_clip.sce42
-rw-r--r--698/CH4/EX4.5/5_curved_link_with_trapezium_cross_section.txt14
-rw-r--r--698/CH4/EX4.5/P5_curved_link_with_trapezium_cross_section.sce38
8 files changed, 259 insertions, 0 deletions
diff --git a/698/CH4/EX4.2/2_offset_bar.txt b/698/CH4/EX4.2/2_offset_bar.txt
new file mode 100644
index 000000000..28884930f
--- /dev/null
+++ b/698/CH4/EX4.2/2_offset_bar.txt
@@ -0,0 +1,30 @@
+(a) The bending moment at every section is=126 Nm
+
+(b) Refer to figure 4.3[A. Hall; Schaum's Outline on Machine Design]
+Tension occurs in upper fibres and compression in lower fibres.
+Maximum normal stress occurs at R=90mm and is compressive
+Location of maximum tensile stress is to be determined
+
+(c)For section R=90mm
+ Distance between centre of gravity and neutral axis e=0.000582 m
+
+(d)For section with R=90mm, stess at inner fibre
+ Si=M*hi/ A*e*ri
+ =26.65 MN/m^2 compression
+Stess at outer fibre
+ So=M*ho/ A*e*ro
+ =22.15 MN/m^2 tension
+
+For section R=115mm
+ Distance between centre of gravity and neutral axis e=0.0005 m
+
+(e)For section with R=115mm, stess at inner fibre
+ Si=M*hi/ A*e*ri
+ =23.60 MN/m^2 tension
+Stess at outer fibre
+ So=M*ho/ A*e*ro
+ =20.56 MN/m^2 compression
+
+(f) Maximum tension occurs at inner fibre of section R=115mm
+Maximum compression occurs at inner fibre of section R=90mm
+Maximum shear stress=half of greatest difference between any two extremes=13.3 MN/m^2
diff --git a/698/CH4/EX4.2/P2_offset_bar.sce b/698/CH4/EX4.2/P2_offset_bar.sce
new file mode 100644
index 000000000..220bd7f0c
--- /dev/null
+++ b/698/CH4/EX4.2/P2_offset_bar.sce
@@ -0,0 +1,74 @@
+clc
+//Example 4.2
+//Offset bar
+
+//------------------------------------------------------------------------------
+
+//Given data
+//Couple causing force
+F=900 // N
+//Distance between forces
+d=0.14 //m
+//Dimensions of bar
+h=0.025 // m
+R=0.09 // m
+t=0.05 // m
+
+res2=mopen(TMPDIR+'2_offset_bar.txt','wt')
+
+//Bending moment
+M=d*F
+mfprintf(res2,'(a) The bending moment at every section is=%d Nm\n\n',M)
+
+mfprintf(res2,'(b) Refer to figure 4.3[A. Hall; Schaum''s Outline on Machine Design]\n')
+mfprintf(res2,'Tension occurs in upper fibres and compression in lower fibres.\n')
+mfprintf(res2,'Maximum normal stress occurs at R=90mm and is compressive\n')
+mfprintf(res2,'Location of maximum tensile stress is to be determined\n\n')
+
+//Radii of curvature of fibres
+ri=R- h/2
+ro=R+ h/2
+rn=h/ log(ro/ri)
+
+//Distance between centre of gravity and neutral axis
+e=R-rn
+mfprintf(res2,'(c)For section R=90mm\n\tDistance between centre of gravity and neutral axis e=%0.6f m\n\n',e)
+
+//Checking for maximum tensile stress at R=90 mm
+Si=(M*0.01192)/(h*t*e*ri)
+mfprintf(res2,'(d)For section with R=90mm, stess at inner fibre\n\t')
+mfprintf(res2,'Si=M*hi/ A*e*ri\n')
+mfprintf(res2,'\t=%0.2f MN/m^2 compression\n',Si* 10^-6)
+s=Si/2
+So=(M*0.0131)/(h*t*e*ro)
+mfprintf(res2,'Stess at outer fibre\n\t')
+mfprintf(res2,'So=M*ho/ A*e*ro\n')
+mfprintf(res2,'\t=%0.2f MN/m^2 tension\n\n',So* 10^-6)
+
+//Checking for maximum tensile stress at R=115 mm
+R=0.115
+ri=R- h/2
+ro=R+ h/2
+rn=h/ log(ro/ri)
+e=R-rn
+e=ceil(e* 10^4)
+e=e* 10^-4
+mfprintf(res2,'For section R=115mm\n\tDistance between centre of gravity and neutral axis e=%0.4f m\n\n',e)
+Si=(M*0.012)/(h*t*e*ri)
+mfprintf(res2,'(e)For section with R=115mm, stess at inner fibre\n\t')
+mfprintf(res2,'Si=M*hi/ A*e*ri\n')
+mfprintf(res2,'\t=%0.2f MN/m^2 tension\n',Si* 10^-6)
+So=(M*0.013)/(h*t*e*ro)
+mfprintf(res2,'Stess at outer fibre\n\t')
+mfprintf(res2,'So=M*ho/ A*e*ro\n')
+mfprintf(res2,'\t=%0.2f MN/m^2 compression\n\n',So* 10^-6)
+
+//Maximum stresses
+mfprintf(res2,'(f) Maximum tension occurs at inner fibre of section R=115mm\n')
+mfprintf(res2,'Maximum compression occurs at inner fibre of section R=90mm\n')
+mfprintf(res2,'Maximum shear stress=half of greatest difference between any two extremes=%0.1f MN/m^2\n',s* 10^-6)
+
+mclose(res2)
+editor(TMPDIR+'2_offset_bar.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program----------------------------------- \ No newline at end of file
diff --git a/698/CH4/EX4.3/3_determination_of_e.txt b/698/CH4/EX4.3/3_determination_of_e.txt
new file mode 100644
index 000000000..d06f5fc54
--- /dev/null
+++ b/698/CH4/EX4.3/3_determination_of_e.txt
@@ -0,0 +1,11 @@
+(a) e=R-rn
+
+(b) Calculation of rn:
+ rn=(((bi-t)*ti)+((bo-t)*to)+(t*h))/((bi*log((ri+ti)/ri))+(t*log((ro-to)/(ri+ti)))+(bo*log(ro/(ro-to))))
+ rn=82.4473 mm
+
+(c)Calculation of R:
+ R=ri+ ((0.5* h^2 *t)+(0.5* ti^2 *(bi-t))+((bo-t)*to*(h- 0.5*to)))/(((bi-t)*ti)+((bo-t)*to)+(t*h))
+ R=88.1944 mm
+
+(d)e=R-rn= 5.7472 mm \ No newline at end of file
diff --git a/698/CH4/EX4.3/P3_determination_of_e.sce b/698/CH4/EX4.3/P3_determination_of_e.sce
new file mode 100644
index 000000000..aa7d3d1d4
--- /dev/null
+++ b/698/CH4/EX4.3/P3_determination_of_e.sce
@@ -0,0 +1,38 @@
+clc
+//Example 4.3
+//Determination of e
+
+//------------------------------------------------------------------------------
+
+//Given data
+//Dimensions of I section
+ro=0.125 //m
+ri=0.06 //m
+bi=0.06 //m
+bo=0.05 //m
+t=0.01 //m
+h=0.065 //m
+ti=0.015 //m
+to=0.01 //m
+
+res3=mopen(TMPDIR+'3_determination_of_e.txt','wt')
+mfprintf(res3,'(a)\te=R-rn\n\n')
+mfprintf(res3,'(b) Calculation of rn:\n')
+mfprintf(res3,'\trn=(((bi-t)*ti)+((bo-t)*to)+(t*h))/((bi*log((ri+ti)/ri))+(t*log((ro-to)/(ri+ti)))+(bo*log(ro/(ro-to))))\n')
+
+rn=(((bi-t)*ti)+((bo-t)*to)+(t*h))/((bi*log((ri+ti)/ri))+(t*log((ro-to)/(ri+ti)))+(bo*log(ro/(ro-to))))
+mfprintf(res3,'\trn=%0.4f mm\n\n',rn* 10^3)
+
+mfprintf(res3,'(c)Calculation of R:\n')
+mfprintf(res3,'\tR=ri+ ((0.5* h^2 *t)+(0.5* ti^2 *(bi-t))+((bo-t)*to*(h- 0.5*to)))/(((bi-t)*ti)+((bo-t)*to)+(t*h))\n')
+
+R=ri+ ((0.5* h^2 *t)+(0.5* ti^2 *(bi-t))+((bo-t)*to*(h- 0.5*to)))/(((bi-t)*ti)+((bo-t)*to)+(t*h))
+mfprintf(res3,'\tR=%0.4f mm\n\n',R* 10^3)
+
+e=R-rn
+mfprintf(res3,'(d)e=R-rn= %0.4f mm',e* 10^3)
+
+mclose(res3)
+editor(TMPDIR+'3_determination_of_e.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program----------------------------------- \ No newline at end of file
diff --git a/698/CH4/EX4.4/4_stress_in_spring_clip.txt b/698/CH4/EX4.4/4_stress_in_spring_clip.txt
new file mode 100644
index 000000000..40f0ed58f
--- /dev/null
+++ b/698/CH4/EX4.4/4_stress_in_spring_clip.txt
@@ -0,0 +1,12 @@
+(a) Maximum bending stress occurs at inside fibre of sections with
+ri=75mm (refer figure 4.5)
+
+(b) rn=(1/4)* (sqrt(ro) + sqrt(ri))^2
+ =0.08705 m
+
+(c) e=R-rn =0.0004 m
+
+(d) Si=(M*hi)/(A*e*ri) =41.0 MN/m^2
+
+(e)Maximum shear stress occurs at every point from A to B and from C to D
+Maximum shear stress=20.5 MN/m^2 \ No newline at end of file
diff --git a/698/CH4/EX4.4/P4_stress_in_spring_clip.sce b/698/CH4/EX4.4/P4_stress_in_spring_clip.sce
new file mode 100644
index 000000000..b744fe780
--- /dev/null
+++ b/698/CH4/EX4.4/P4_stress_in_spring_clip.sce
@@ -0,0 +1,42 @@
+clc
+//Example 4.4
+//Stress in spring clip
+
+//------------------------------------------------------------------------------
+
+//Given data
+//Diameter of spring clip
+d=0.025 // m
+//Area of cross section
+A=(%pi/4)* d^2
+//Force acting
+P=450 //N
+//Bending moment
+M=P*0.125 //N-m
+
+
+res4=mopen(TMPDIR+'4_stress_in_spring_clip.txt','wt')
+mfprintf(res4,'(a) Maximum bending stress occurs at inside fibre of sections with\nri=75mm (refer figure 4.5)\n\n')
+
+ro=0.1 //m
+ri=0.075 //m
+rn=(1/4)* (sqrt(ro) + sqrt(ri))^2
+mfprintf(res4,'(b)\trn=(1/4)* (sqrt(ro) + sqrt(ri))^2\n\t=%0.5f m\n\n',rn)
+
+R=ri+ d/2
+e=R-rn
+mfprintf(res4,'(c)\te=R-rn =%0.4f m\n\n',e)
+
+hi=0.01205
+Si=(M*hi)/(A*e*ri)
+mfprintf(res4,'(d)\tSi=(M*hi)/(A*e*ri) =%0.1f MN/m^2\n\n',Si* 10^-6)
+
+mfprintf(res4,'(e)Maximum shear stress occurs at every point from A to B and from C to D\n')
+//Maximum shear stress
+Smax=Si/2
+mfprintf(res4,'Maximum shear stress=%0.1f MN/m^2',Smax* 10^-6)
+
+mclose(res4)
+editor(TMPDIR+'4_stress_in_spring_clip.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program----------------------------------- \ No newline at end of file
diff --git a/698/CH4/EX4.5/5_curved_link_with_trapezium_cross_section.txt b/698/CH4/EX4.5/5_curved_link_with_trapezium_cross_section.txt
new file mode 100644
index 000000000..0f969c791
--- /dev/null
+++ b/698/CH4/EX4.5/5_curved_link_with_trapezium_cross_section.txt
@@ -0,0 +1,14 @@
+(a) Maximum stress occurs at internal radius of link
+ Si=(M*hi)/(A*e*ri)
+
+(b)To determine e:
+ rn=((bi+bo)/2)*h / ((((bi*ro - bo*ri)/h)*log(ro/ri)) - (bi-bo))
+ =69.4804 mm
+
+ R=ri+ (h*(bi+ 2*bo))/(3*(bi+bo))
+ =72.2222 mm
+
+ e=R-rn =2.7419 mm
+
+(c)Finding the stress
+ Si=82.100 Mpa \ No newline at end of file
diff --git a/698/CH4/EX4.5/P5_curved_link_with_trapezium_cross_section.sce b/698/CH4/EX4.5/P5_curved_link_with_trapezium_cross_section.sce
new file mode 100644
index 000000000..26f7f3b21
--- /dev/null
+++ b/698/CH4/EX4.5/P5_curved_link_with_trapezium_cross_section.sce
@@ -0,0 +1,38 @@
+clc
+//Example 4.5
+//Curved link with trapezium cross-section
+
+//------------------------------------------------------------------------------
+
+//Given data
+ro=0.1 //m
+ri=0.05 //m
+h=0.05 //m
+bo=0.025 //m
+bi=0.05 //m
+P=15000 //N
+A=(bi+bo)*h /2
+
+res5=mopen(TMPDIR+'5_curved_link_with_trapezium_cross_section.txt','wt')
+mfprintf(res5,'(a) Maximum stress occurs at internal radius of link\n\tSi=(M*hi)/(A*e*ri)\n\n')
+
+mfprintf(res5,'(b)To determine e:\n\t')
+mfprintf(res5,'rn=((bi+bo)/2)*h / ((((bi*ro - bo*ri)/h)*log(ro/ri)) - (bi-bo))\n')
+rn=((bi+bo)/2)*h / (((((bi*ro - bo*ri)/h)*log(ro/ri)) - (bi-bo)))
+mfprintf(res5,'\t=%0.4f mm\n\n',rn* 10^3)
+mfprintf(res5,'\tR=ri+ (h*(bi+ 2*bo))/(3*(bi+bo))\n')
+R=ri+ (h*(bi+ 2*bo))/(3*(bi+bo))
+mfprintf(res5,'\t=%0.4f mm\n\n',R* 10^3)
+e=R-rn
+mfprintf(res5,'\te=R-rn =%0.4f mm\n\n',e* 10^3)
+
+mfprintf(res5,'(c)Finding the stress\n')
+M=P*R
+hi=rn-ri
+Si=(M*hi)/(A*e*ri)
+mfprintf(res5,'\tSi=%0.3f Mpa',Si* 10^-6)
+
+mclose(res5)
+editor(TMPDIR+'5_curved_link_with_trapezium_cross_section.txt')
+//------------------------------------------------------------------------------
+//-----------------------------End of program----------------------------------- \ No newline at end of file