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"source": [
"# Chapter 1: Introduction—Concept of Stress\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Example 1.1, Page number 18 "
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Case(a): Shearing Stress in Pin A = 6790.6 psi\n",
"Case(b): Shearing Stress in Pin C = 7639 psi\n",
"Case(c): Largest Normal Stress in Link ABC = 2286 psi\n",
"Case(d): Average Shearing Stress at B = 171 psi\n",
"Case(e): Bearing Stress in Link at C = 6000 psi\n"
]
}
],
"source": [
"import math\n",
"\n",
"#Variable declaration\n",
"Fac = 750 #Force on rod AC(lb)\n",
"D = 0.375 #Diameter at the upper junction of rod ABC(in)\n",
"\n",
"\n",
"#Calculation \n",
"#Case(a)\n",
"A=(1/4)*((math.pi)*pow(D,2)) #Area at the upper junction of rod ABC(in^2) \n",
"tA=(Fac/A) #Shearing Stress in Pin A(psi) \n",
"#Case(b) \n",
"Ab=(1/4)*((math.pi)*pow(0.25,2)) #Area at the lower junction of rod ABC(in^2)\n",
"tC=(((1/2)*Fac)/Ab) #Shearing Stress in Pin C(psi)\n",
"#Case(c)\n",
"Anet=(3/8)*(1.25-0.375) #Area of cross section at A(in^2)\n",
"sA=(Fac/Anet) #Largest Normal Stress in Link ABC(psi)\n",
"#Case(d)\n",
"F1=750/2 #Force on each side(lb)\n",
"Ad=(1.25*1.75) #Area at junction B(in^2)\n",
"tB=(F1/Ad) #Average Shearing Stress at B\n",
"#Case(e)\n",
"Ae=0.25*0.25 #Area at point C(in^2)\n",
"sB=(F1/Ae) #Bearing Stress in Link at C\n",
"\n",
"\n",
"#Result\n",
"print('Case(a): Shearing Stress in Pin A = %.1f psi' %tA)\n",
"print('Case(b): Shearing Stress in Pin C = %.f psi' %tC)\n",
"print('Case(c): Largest Normal Stress in Link ABC = %.f psi' %sA)\n",
"print('Case(d): Average Shearing Stress at B = %.f psi' %tB)\n",
"print('Case(e): Bearing Stress in Link at C = %.f psi' %sB)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Example 1.2, Page number 19"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
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"outputs": [
{
"name": "stdout",
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"text": [
"Case(a): Diameter of the bolt = 28 mm\n",
"Case(a): Dimension b at Each End of the Bar = 62 mm\n",
"Case(a): Dimension h of the Bar = 34.300000 mm\n"
]
}
],
"source": [
"import math\n",
"\n",
"#Variable declaration\n",
"P = 120 #Maximum allowable tension force \n",
"s = 175 #Maximum allowable stress\n",
"t = 100 #Maximum allowable stress\n",
"Sb = 350 #Maximum allowable stress\n",
"\n",
"\n",
"#Calculation\n",
"#Case(a)\n",
"F1=P/2 #Current(A)\n",
"d=math.sqrt(((P/2)*1000)/((22/(4*7))*(100000000))) #Diameter of bolt(m)\n",
"d=d*1000 #Diameter of bolt(mm)\n",
"d=round(d,0) #Rounding of the value of diameter of bolt(mm)\n",
"Ad=(0.020*0.028) #Area of cross section of plate \n",
"tb=((P*1000)/Ad)/(1000000) #Stress between between the 20-mm-thick plate and the 28-mm-diameter bolt\n",
"tb=round(tb,0) #Rounding of the above calculated stress to check if it is less than 350\n",
"a=(P/2)/((0.02)*(175)) #Dimension of cross section of ring \n",
"a=round(a,2) #Rounding dimension of cross section of ring to two decimal places\n",
"b=28 + (2*(a)) #Dimension b at Each End of the Bar\n",
"b=round(b,2) #Rounding the dimension b to two decimal places \n",
"h=(P)/((0.020)*(175)) #Dimension h of the Bar\n",
"h=round(h,1) #Rounding dimension h of bar to 1 decimal place\n",
"\n",
"\n",
"#Result\n",
"print ('Case(a): Diameter of the bolt = %.f mm' %d)\n",
"print ('Case(a): Dimension b at Each End of the Bar = %.f mm' %b)\n",
"print ('Case(a): Dimension h of the Bar = %f mm' %h)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Example 1.3, Page number 34"
]
},
{
"cell_type": "code",
"execution_count": 45,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Case(a): Diameter of the bolt = 16.730000 mm\n",
"Case(a): Dimension b at Each End of the Bar = 22.000000 mm\n",
"Case(a): Dimension h of the Bar = 6.000000 mm\n"
]
}
],
"source": [
"import math\n",
"\n",
"#Variable declaration\n",
"Su = 600 #ultimate normal stress(MPa) \n",
"FS = 3.3 #Factor of safety with respect to failure\n",
"tU=350 #Ultimate shearing stress(MPa)\n",
"Cx=40 #X Component of reaction at C(kN)\n",
"Cy=65 #Y Component of reaction at C(kN)\n",
"Smax=300 #Allowable bearing stress of the steel \n",
"\n",
"#Calculation\n",
"C=math.sqrt((math.pow(40,2))+(math.pow(65,2)))\n",
"\n",
"#Case(a)\n",
"P=(15*0.6 + 50*0.3)/(0.6) #Allowable bearing stress of the steel(MPa)\n",
"Sall=(Su/FS) #Allowable Stress(MPa)\n",
"Sall=round(Sall,1) #Rounding Allowable stress to 1 decimal place(MPa)\n",
"Areqa=(P/(Sall*(1000))) #Cross Sectional area(m^2)\n",
"Areqa=round(Areqa,5) #Rounding cross sectional area to 5 decimal places(m^2)\n",
"dAB=math.sqrt(((Areqa)*(4))/(22/7)) #Diameter of AB(m)\n",
"dAB=dAB*1000 #Diameter of AB(mm)\n",
"dAB=round(dAB,2) #Rounding Diameter of AB(mm)\n",
"\n",
"#Case(b)\n",
"tALL=tU/FS #Stress(MPa)\n",
"tALL=round(tALL,1) #Rounding of Stress\n",
"AreqC=((C/2)/tALL) #Cross sectional area(m^2)\n",
"AreqC=AreqC*1000 \n",
"AreqC=round(AreqC,0) #Rounding the cross sectional area\n",
"dC=math.sqrt((4*AreqC)/(22/7)) #Diameter at point C\n",
"dC=round((dC+1),0) #Rounding of the diameter at C\n",
"\n",
"#Case(c)\n",
"\n",
"Areq=((C/2)/Smax) \n",
"Areq=Areq*1000 #Cross sectional area(mm^2)\n",
"t=(Areq/22) #Thickness of the bracket\n",
"t=round(t,0)\n",
"\n",
"#Result\n",
"print ('Case(a): Diameter of the bolt = % f mm' %dAB)\n",
"print ('Case(a): Dimension b at Each End of the Bar = % f mm' %dC)\n",
"print ('Case(a): Dimension h of the Bar = % f mm' %t)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"# Example 1.4, Page number 35"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Case(a): Control Rod = 5.263672 kips\n",
"Case(b): Bolt at B = 5.156250 kips\n",
"Case(c): Bolt at D = 6.865179 kips\n",
"Case(d): Bolt at C = 5.238095 kips\n",
"Summary. We have found separately four maximum allowable values of the force C. In order to satisfy all these criteria we must choose the smallest value, namely:= 5.156250 kips\n"
]
}
],
"source": [
"import math\n",
"\n",
"#Variable declaration\n",
"tU=40 #ultimate tensile stress\n",
"sU=60 #ultimate shearing stress\n",
"FS=3 #Mimnimum factor of safety\n",
"dA=(7/16) #Diameter of bolt at A(in)\n",
"dB=3/8 #Diameter of bolt at B(in) \n",
"dD=3/8 #Diameter of bolt at D(in)\n",
"dC=1/2 #Diameter of bolt at C(in)\n",
"\n",
"\n",
"#Calculation\n",
"Sall=(sU/FS) #Total tensile stress(kips)\n",
"B=Sall*((1/4)*(22/7)*(pow((7/16),2))) #Allowable force in the control rod(kips)\n",
"C1=1.75*(B) #Control Rod(kips)\n",
"tall=(tU/FS) #Total shearing stress\n",
"B=2*(tall*(1/4)*(22/7)*(3/8)*(3/8)) #Allowable magnitude of the force B exerted on the bolt\n",
"C2=1.75*B #Bolt at B(kips) \n",
"D=B #Bolt at D. Since this bolt is the same as bolt B, the allowable force is same(kips) \n",
"C3=2.33*D #Bolt at D(kips)\n",
"C4=2*(tall*(1/4)*(22/7)*(1/2)*(1/2)) #Bolt at C(kips) \n",
"list1=[C1,C2,C3,C4] #Adding all the maximum allowable forces on C(kips) \n",
"\n",
"\n",
"#Result\n",
"print ('Case(a): Control Rod = % f kips' %C1)\n",
"print ('Case(b): Bolt at B = % f kips' %C2)\n",
"print ('Case(c): Bolt at D = % f kips' %C3)\n",
"print ('Case(d): Bolt at C = % f kips' %C4)\n",
"print ('Summary. We have found separately four maximum allowable values of the force C. In order to satisfy all these criteria we must choose the smallest value, namely:= % f kips' %min(list1));"
]
}
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