{
"metadata": {
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2:Axial strains and Deformations in bars "
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.1 page number 77"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"l_ob = 2000 #mm - length of rod ob\n",
"l_bc = 1000 #mm - length of rod bc\n",
"l_cd = 1500 #mm - length of rod cd\n",
"p_ob = 100 #kN - Force in rods \n",
"p_bc = -150 #KN\n",
"p_cd = 50 #KN \n",
"A_ob = 1000 #mm2 - Area of rod ob\n",
"A_bc = 2000 #mm2 - Area of rod bc \n",
"A_cd = 1000 #mm2 - Area of rod cd\n",
"E = 200.0 #GPA \n",
"# the total deflection is algebraic sums of `deflection in each module \n",
"e_1 = p_ob*l_ob/(A_ob*E)\n",
"e_2 = p_bc*l_bc/(A_bc*E)\n",
"e_3 = p_cd*l_cd/(A_cd*E)\n",
"#All units are satisfied \n",
"e_total = e_1+ e_2 + e_3\n",
"print \"The total deflection is :\",round(e_total,3) ,\"mm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total deflection is : 1.0 mm\n"
]
}
],
"prompt_number": 77
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4 page number 80"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"p_app = 3 #Kips - applied force \n",
"P_A = 2.23 #Kips \n",
"p_B = -2.83 #kips - compressive force\n",
"l_ab = 6.71 #inch\n",
"l_bc = 8.29 #inch\n",
"s_ab = 17.8 #ksi - tensile stress\n",
"s_bc = -12.9 #ksi - compressive stress\n",
"E = 10.6 * pow(10,3) #ksi -youngs modulus \n",
"e_ab = s_ab*l_ab/E\n",
"\n",
"e_bc = s_bc*l_bc/E\n",
"x = e_ab/e_bc #the Ratio of cosines of the deflected angles \n",
"# t_1 and t_2 be deflected angles \n",
"#t_2 = 180-45-26.6-t_1 the sum of angles is 360\n",
"#t_1 = 52.2 degress\n",
"import math\n",
"e = e_ab/math.acos(math.radians(52.2)) #inch\n",
"k = p_app/e # kips/in vertical stiffness of the combination\n",
"print \"The vertical stiffness of the combination is\",round(k,3),\"kips/inch\"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"0.0112677358491\n",
"The vertical stiffness of the combination is 113.14 kips/inch\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.6 page number 83"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"dia = 50 #mm - diameter of aluminium \n",
"p = 100 # KN - instant force applid\n",
"dia_c = 0.1215 #mm- change in diameter \n",
"l_c = 0.219 #mm - change in length\n",
"l = 300 #mm - length \n",
"strain_dia = dia_c/dia # lateral strain \n",
"strain_l = l_c/l #longitudinal strain \n",
"po = strain_dia/strain_l # poission ratio \n",
"area = 3.14*dia*dia/4 #mm2 area\n",
"E = p*l/(area*l_c) #N/mm2 youngs modulus \n",
"print \"The lateral strain is:\",strain_dia,\"no units\"\n",
"print \"The longitudinal strain is:\",strain_l,\"no units\"\n",
"print \"The poissions ratio is:\",po,\"no units\"\n",
"print \"Youngs modulus:\",round(E,2),\"N/mm2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The lateral strain is: 0.00243 no units\n",
"The longitudinal strain is: 0.00073 no units\n",
"The poissions ratio is: 3.32876712329 no units\n",
"Youngs modulus: 69.8 N/mm2\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.7 page number 86"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"T = 12.9*pow(10,-6) #/F\n",
"t = 100.00 # F \n",
"e_ab = T*t*l_ab #in-elongation \n",
"e_bc = T*t*l_bc #in-elongation\n",
"k = e_ab/e_bc # ratio of cosines of deflected angles \n",
"# t_1 and t_2 be deflected angles \n",
"#t_2 = 180-45-26.6-t_1 the sum of angles is 360\n",
"t_1 = 26.6\n",
"import math\n",
"e = e_ab/math.acos(math.radians(26.6))\n",
"print \"The displacement in point B is :\",e ,\"in\"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The displacement in point B is : 0.00795578950395 in\n"
]
}
],
"prompt_number": 78
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.11 page number "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"mass = 4 #kg \n",
"dist = 1 #mt freely falling distance\n",
"l = 1500 #mm length of rod\n",
"d = 15 #mm diameter\n",
"E = 200 #GPA youngs modulus \n",
"k = 4.5 # N/mm stiffness costant\n",
"F = mass*9.81# The force applying\n",
"Area = 3.14*(d**2)/4 \n",
"# Two cases \n",
"#youngs modulus \n",
"e_y = F*l/(Area*E*pow(10,3))\n",
"# stiffness\n",
"e_f = F/k \n",
"#total\n",
"e = e_y +e_f\n",
"k = 1+(2/(e*pow(10,-3)))\n",
"stress_max_1 = F*(1+pow(k,0.5))/Area\n",
"print \"The maximum stress is:\",stress_max_1,\"Mpa\"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum stress is: 3.59377281766 Mpa\n"
]
}
],
"prompt_number": 56
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.12 page number 103"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"flex_a = 1#f\n",
"flex_b = 2#f\n",
"#removing lower support and solving FBD\n",
"e = -2 -(2+1)#fp\n",
"#e_1 = (2+1+1)*R\n",
"#e_1 = -e Making the elongations zero since the both ends are fixed\n",
"R = e/(2+1+1.0) #p\n",
"print \"The reactions at bottom is\",R,\"p\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The reactions at bottom is -1.25 p\n"
]
}
],
"prompt_number": 75
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.19 page number 113"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"l = 30 #in - The length of the rod\n",
"p_1 = 80 #kips - The Force on the end\n",
"p_2 = 125 #kips - The force on the other end\n",
"A_s = 0.5 #in2 - The crossection of the steel rod\n",
"A_a = 0.5 #in2 - The crossection of the aluminium \n",
"E_a = 10*(10**6) #psi - The youngs modulus of the aluminium \n",
"E_s = 30*(10**6) #psi - The youngs modulus of the steel\n",
"#Internally stastically indeterminant \n",
"p_a = p_1/4 #From solving we get p_s = 3*P_a\n",
"#From material properties point of view \n",
"#stress_steel = stress_aluminium\n",
"e = p_a*l*(10**3)/(A_a*E_a) #The end deflection \n",
"print \"The end deflection is\",e,\"in\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The end deflection is 0.12 in\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}