{
"metadata": {
"name": "",
"signature": "sha256:88c8946645e4f3f31d0d74d88b4d147ad878a660c1c0e5aff6962c72b58f187c"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"chapter 4:Shear stress in Beams and Related Problems "
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.1 page number 365"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"shear_v = 3000 #N - Transmitted vetical shear \n",
"shear_al = 700 #N - The maximum allowable \n",
"#We will divide this into two parts\n",
"l_1 = 50.0 #mm \n",
"l_2 = 200.0 #mm \n",
"b_1 = 200.0 #mm \n",
"b_2 = 50.0 #mm\n",
"A_1 = l_1* b_1 #mm2 - area of part_1\n",
"y_1 = 25.0 #mm com distance \n",
"A_2 =l_2*b_2 #mm2 - area of part_1\n",
"y_2 = 150.0 #in com distance \n",
"y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) #mm - The com of the whole system\n",
"c_max = (4-y_net) #mm - The maximum distace from com to end\n",
"c_min = y_net #mm - the minimum distance from com to end\n",
"I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n",
"I_2 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n",
"I_net = I_1 + I_2 #mm4 - the total moment of inertia\n",
"Q = A_1*(-y_1+y_net) #mm3\n",
"q = shear_v*Q/I_net #N/mm - Shear flow\n",
"d = shear_al/q # The space between the nails \n",
"print \"The minimal space between the nails \",round(d,0) ,\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimal space between the nails 42.0 mm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.2 pagenumber 365"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"l = 6 #m -length of the beam \n",
"p = 3 #KN-m _ the load applied\n",
"R_a = l*p/2 #KN -The reaction at a, Since the system is symmetry \n",
"R_b = l*p/2 #KN -The reaction at b \n",
"l_s = 10 #mm - The length of the screw \n",
"shear_al = 2 #KN - The maximum load the screw can take \n",
"I = 2.36*(10**9) #mm2 The moment of inertia of the whole system\n",
"#We will divide this into two parts\n",
"l_1 = 50.0 #mm \n",
"l_2 = 50.0 #mm \n",
"b_1 = 100.0 #mm \n",
"b_2 = 200.0 #mm\n",
"A_1 = l_1* b_1 #in2 - area of part_1\n",
"y_1 = 200.0 #mm com distance \n",
"A_2 =l_2*b_2 #mm2 - area of part_1\n",
"y_2 = 225.0 #in com distance\n",
"Q = 2*A_1*y_1 + A_2*y_2 # mm3 For the whole system\n",
"q = R_a*Q*(10**3)/I #N/mm The shear flow \n",
"d = shear_al*(10**3)/q #mm The space between the nails\n",
"print \"The minimal space between the nails \",round(d,0),\"mm\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimal space between the nails 123.0 mm\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.6 page number 376"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"#we will divide this into two equal parts and other part\n",
"l = 10.0 # in - The height \n",
"t = 0.1 # in - The width\n",
"b = 5.0 #mm- The width of the above part \n",
"A = t* b #in2 - area of part\n",
"y_net = l/2 # The com of the system \n",
"y_1 = l # The position of teh com of part_2\n",
"I_1 = t*(l**3)/12 #in4 The moment of inertia of part 1\n",
"I_2 = 2*A*((y_1-y_net)**2) #in4 The moment of inertia of part 2 \n",
"I = I_1 + I_2 #in4 The moment of inertia \n",
"e = (b**2)*(l**2)*t/(4*I) #in the formula of channels\n",
"l_sc = e - t/2 #in- The shear centre \n",
"print \"The shear centre from outside vertical face is \",l_sc ,\"in\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The shear centre from outside vertical face is 1.825 in\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.8 page number 387"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"dia = 10.0 #mm - The diameter of the cylinder \n",
"c = dia/2 #mm - the radius of the cylinder \n",
"A = 3.14*(c**2) #mm2 The area of the crossection \n",
"y = 4*c/(3*3.14) #mm The com of cylinder \n",
"I = 3.14*(c**4)/4 #mm4 - The moment of inertia of the cylinder\n",
"j = 3.14*(dia**4)/32 #mm4\n",
"T = 20.0 #N.m - The torque \n",
"V = 250.0 #N - The shear \n",
"M = 25.0 #N-m The bending moment \n",
"Q = A*y/2 #mm\n",
"stress_dmax = 4*V/(3*A) #V*Q/(I*d) #Mpa The direct maximum stress\n",
"stress_tmax = T*c*(10**3)/j #-Mpa The torsion maximum stress\n",
"stress_total = stress_dmax + stress_tmax #Mpa The total stress\n",
"print \"The direct maximum stress\",round(stress_dmax,2),\"Mpa\"\n",
"print \"The torsion maximum stress\",round(stress_tmax,2),\"Mpa\"\n",
"print \"The total stress\",round(stress_total,2),\"Mpa\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The direct maximum stress 4.25 Mpa\n",
"The torsion maximum stress 101.91 Mpa\n",
"The total stress 106.16 Mpa\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.9 page number 393"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given\n",
"dia = 15 #mm - The diameter of the rod\n",
"h = 0.5 #mt - The freely falling height \n",
"A = 3.14*(dia**2)/4 #mm2 The area of the crossection\n",
"E = 200 #Gpa -Youngs modulus\n",
"L = 750 #mm - The total length of the rod\n",
"G = 80 #gpa - Shear modulus \n",
"N = 10 #number of live coils\n",
"d = 5 #mm the diameter of live coil \n",
"m = 3 # the mass of freely falling body\n",
"H = 500 #mm -from mass to spring \n",
"F= m*9.81 #Kg the force due to that mass\n",
"#e = e_rod + e_spr\n",
"#e_rod\n",
"e_rod = p*L*(10**-3)/(A*E) #mm The elongation due to freely falling body\n",
"#e_spr\n",
"e_spr = 64*F*(dia**3)*N*(10**-3)/(G*(d**4)) #mm The elongation due to spring\n",
"e = e_rod + e_spr #mm The total elongation \n",
"p_dyn =F*(1+pow((1+(2*H/e)),0.5))\n",
"Stress_max = p_dyn/A #MPa - The maximum stress in the system \n",
"print \"The maximum stress in the system \",round(Stress_max,2),\"Mpa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum stress in the system 4.84 Mpa\n"
]
}
],
"prompt_number": 43
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}