{
"metadata": {
"name": "",
"signature": "sha256:6a2b99a338c7d2a71ec20bddeaf4c19ec8da980602f4c9834a7faee562a1e5cc"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6:Pure Bending and Bending with Axial force "
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.3 page number 293"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"#Entire area - hallow area\n",
"l_e = 60.0 #mm - length of the entire area\n",
"b_e = 40 #mm - width of the entire area\n",
"l_h = 30 #mm - length of the hallow area\n",
"b_h = 20 #mm - width of the hallow area\n",
"A_e = l_e*b_e #mm2 - The entire area\n",
"A_h = -l_h*b_h #mm2 - The hallow area '-' because its hallow\n",
"A_re = A_e + A_h #mm2 resultant area\n",
"y_e = l_e/2 # mm com from bottom \n",
"y_h = 20+l_h/2 #mm com from bottom \n",
"y_com = (A_e*y_e + A_h*y_h)/A_re \n",
"#moment of inertia caliculatins - bh3/12 +ad2\n",
"I_e = b_e*(l_e**3)/12 + A_e*((y_e-y_com)**2) #Parallel axis theorm\n",
"I_h = b_h*(l_h**3)/12 - A_h*((y_h-y_com)**2) #Parallel axis theorm\n",
"I_total = I_e - I_h\n",
"print \"The moment of inertia of total system is \",I_total,\"mm4\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The moment of inertia of total system is 655000.0 mm4\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.4 page number 295"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"l = 400 #mm - Length \n",
"b = 300 #mm - breath \n",
"F = 20 #KN _ the force applied on the beam \n",
"F_d = 0.75 #KN-m - The force distribution \n",
"d = 2 #mt - the point of interest from the free end\n",
"#caliculations \n",
"#From moment diagram\n",
"M = F*d - F_d*d*1\n",
"I = b*(l**3)/12 #mm4 - Bending moment diagram \n",
"c = l/2 # the stress max at this C\n",
"S = I/c #The maximum shear stress \n",
"shear_max = M*(10**6)/S #MPA - the maximum stress \n",
"print \"The maximum stress at 2 mt is\",round(shear_max,2),\"Mpa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum stress at 2 mt is 4.81 Mpa\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.5 pagr number 297"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"#We will divide this into three parts\n",
"F = 8 #k - force applied\n",
"d = 16 #inch -distance\n",
"l_1 = 1 #in \n",
"l_2 = 3 #in \n",
"b_1 = 4 #in \n",
"b_2 = 1 #in\n",
"A_1 = l_1* b_1 #in2 - area of part_1\n",
"y_1 = 0.5 #in com distance from ab\n",
"A_2 =l_2*b_2 #in2 - area of part_1\n",
"y_2 = 2.5 #in com distance from ab\n",
"A_3 = l_2*b_2 #in2 - area of part_1\n",
"y_3 = 2.5 #in com distance from ab\n",
"\n",
"y_net = (A_1*y_1 +A_2*y_2 + A_3*y_3)/(A_1+A_2+A_3) #in - The com of the whole system\n",
"c_max = (4-y_net) #in - The maximum distace from com to end\n",
"c_min = y_net #in - 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_3 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n",
"I_net = I_1 + I_2 + I_3 #in4 - the total moment of inertia\n",
"M_c = F*d*c_max \n",
"stress_cmax = M_c/I_net #Ksi - The maximum compressive stress\n",
"\n",
"M_t= F*d*c_min \n",
"stress_tmax = M_t/I_net #Ksi - The maximum tensile stress\n",
"print \"The maximum tensile stress\",stress_tmax ,\"Ksi\"\n",
"print \"The maximum compressive stress\",round(stress_cmax,1) ,\"Ksi\"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum tensile stress 16.0 Ksi\n",
"The maximum compressive stress 21.6 Ksi\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.8 page number 303"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"#Given \n",
"#We will divide this into two parts\n",
"E_w = 10.0 #Gpa - Youngs modulus of wood\n",
"E_s = 200.0 #Gpa - Youngs modulus of steel\n",
"M = 30.0 #K.N-m _ applied bending moment \n",
"n = E_s/E_w \n",
"l_1 = 250 #mm \n",
"l_2 = 10 #mm\n",
"b_1 = 150.0 #mm\n",
"b_2 = 150.0*n #mm\n",
"A_1 = l_1* b_1 #mm2 - area of part_1\n",
"y_1 = 125.0 #mm com distance from top\n",
"A_2 =l_2*b_2 #mm2 - area of part_1\n",
"y_2 = 255.0 #mm com distance from top\n",
"y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) #mm - The com of the whole system from top\n",
"I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n",
"I_2 = b_2*(l_2**3)/12.0 + A_2*((y_2-y_net)**2)\n",
"I_net = I_1 + I_2 #mm4 - the total moment of inertia\n",
"c_s= y_net # The maximum distance in steel \n",
"stress_steel = M*(10.0**6)*c_s/I_net #Mpa - The maximum stress in steel \n",
"\n",
"c_w= l_1+l_2-y_net # The maximum distance in wood \n",
"stress_wood = n*M*(10.0**6)*c_w/I_net #MPa - The maximum stress in wood \n",
"\n",
"print \"The maximum stress in steel \",round(stress_steel,2) ,\"Mpa\"\n",
"print \"The maximum stress in wood\",round(stress_wood,2) ,\"Mpa\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum stress in steel 11.49 Mpa\n",
"The maximum stress in wood 97.09 Mpa\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.9 page number 305"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"M = 50000 #ft-lb , positive bending moment applied\n",
"N = 9 # number of steel bars \n",
"n = 15 # The ratio of steel to concrete \n",
"A_s = 30 #in2 area of steel in concrete\n",
"#(10*y)*(y/2) = 30*(20-y)\n",
"#y**2 + 6*y -120\n",
"#solving quadractic equation \n",
"import math\n",
"\n",
"a = 1\n",
"b = 6\n",
"c = -120\n",
"# calculate the discriminant\n",
"d = (b**2) - (4*a*c)\n",
"\n",
"# find two solutions\n",
"sol1 = (-b-math.sqrt(d))/(2*a)\n",
"sol2 = (-b+math.sqrt(d))/(2*a)\n",
"y = sol2 # Nuetral axis is found\n",
"l_1 = y #in- the concrete below nuetral axis is not considered\n",
"b_1 = 10 #in - width\n",
"A_1 = l_1* b_1 #in2 - area of concrete\n",
"y_1 = y/2 #in com of the concrete \n",
"y_2 = 20-y #in com of the transformed steel \n",
"I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y)**2) #in4 parallel axis theorm\n",
"I_2 = A_s*((y_2)**2) #in4 first part is neglected\n",
"I_net = I_1 + I_2 #in4 - the total moment of inertia\n",
"c_c= y #in The maximum distance in concrete \n",
"stress_concrete = M*12*c_c/I_net #psi - The maximum stress in concrete \n",
"c_s= 20- y \n",
"stress_steel =n*M*12*c_s/I_net #psi - The maximum stress in concrete \n",
"print \"The maximum stress in concrete \",round(stress_concrete,2) ,\"psi\"\n",
"print \"The stress in steel\",round(stress_steel,2) ,\"psi\"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum stress in concrete 834.07 psi\n",
"The stress in steel 17427.61 psi\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example 6.10 page number 309"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"l = 50.0 #mm - the length of the beam \n",
"b = 50.0 #mm - the width of the beam\n",
"M = 2083 #Nm\n",
"A = l*b #mm2 - The area \n",
"#straight beam \n",
"I = b*(l**3)/12.0 #mm4 - The moment of inertia of the beam\n",
"c_1= l/2 # the distance where the stress is maximum \n",
"c_2 = -l/2 # the distance where the stress is maximum \n",
"s_1 = I/c_1\n",
"s_2 = I/c_2\n",
"stress_max_1 = M*(10**3)/s_1 #Mpa - the maximum strss recorded in the crossection\n",
"stress_max_2 = M*(10**3)/s_2 #Mpa - the maximum strss recorded in the crossection \n",
"print \"The maximum stress upward in straight case is\",stress_max_1,\"Mpa\"\n",
"print \"The maximum stress downward in straight case is\",stress_max_2,\"Mpa\"\n",
"\n",
"#curved beam \n",
"import math\n",
"r = 250.0 #mm Radius of beam curved \n",
"r_0 = r - l/2 # inner radius \n",
"r_1 = r + l/2 # outer radius\n",
"R = l/(math.log(r_1/r_0)) #mm \n",
"e = r - R \n",
"stressr_max_1 = M*(10**3)*(R-r_0)/(r_0*A*e)\n",
"stressr_max_2 = M*(10**3)*(R-r_1)/(r_1*A*e)\n",
"print \"The maximum stress upward in curved case is\",stressr_max_1,\"Mpa\"\n",
"print \"The maximum stress downward in curved case is\",stressr_max_2,\"Mpa\"\n",
"\n",
"#curved beam _2 \n",
"import math\n",
"r = 75.0 #mm Radius of beam curved \n",
"r_0 = r - l/2 # inner radius \n",
"r_1 = r + l/2 # outer radius\n",
"R = l/(math.log(r_1/r_0)) #mm \n",
"e = r - R \n",
"stressr_max_1 = M*(10**3)*(R-r_0)/(r_0*A*e)\n",
"stressr_max_2 = M*(10**3)*(R-r_1)/(r_1*A*e)\n",
"print \"The maximum stress upward in curved case2 is\",stressr_max_1,\"Mpa\"\n",
"print \"The maximum stress downward in curved case2 is\",stressr_max_2,\"Mpa\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum stress upward in straight case is 99.984 Mpa\n",
"The maximum stress downward in straight case is -99.984 Mpa\n",
"The maximum stress upward in curved case is 107.093207632 Mpa\n",
"The maximum stress downward in curved case is -93.6813516989 Mpa\n",
"The maximum stress upward in curved case2 is 128.733538525 Mpa\n",
"The maximum stress downward in curved case2 is -81.0307692623 Mpa\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Page number 6.14 page number 318"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given \n",
"#from example 6.9\n",
"St_ul = 2500 #psi - ultimate strength\n",
"st_yl = 40000 #psi _ yielding strength \n",
"b = 10 #in - width from example \n",
"A = 2 #in2 The area of the steel\n",
"d = 20 \n",
"t_ul = st_yl*A #ultimate capasity\n",
"y = t_ul/(St_ul*b*0.85) #in 0.85 because its customary\n",
"M_ul = t_ul*(d-y/2)/12 #ft-lb Plastic moment \n",
"print \"The plastic moment of the system is \",M_ul,\"ft-lb\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The plastic moment of the system is 120784.313725 ft-lb\n"
]
}
],
"prompt_number": 35
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.15 page number 231"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given \n",
"#From example 5.8 \n",
"W = 4.0 #N/m - The force distribution \n",
"L = 3 # m - The length of the force applied\n",
"M = W*L/8.0 # KN.m The moment due to force distribution\n",
"o = 30 # the angle of force applid to horizantal\n",
"l = 150.0 #mm length of the crossection \n",
"b = 100.0 #mm - width of the crossection \n",
"import math \n",
"M_z = M*(math.cos(3.14/6))\n",
"M_y = M*(math.sin(math.pi/6))\n",
"I_z = b*(l**3)/12.0\n",
"I_y = l*(b**3)/12.0\n",
"#tanb = I_z /I_y *tan30\n",
"b = math.atan(math.radians(I_z*math.tan(3.14/6.0)/I_y ))\n",
"print \"The angle at which nuetral axis locates is\",b,\"radians\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The angle at which nuetral axis locates is 0.0226547191205 radians\n"
]
}
],
"prompt_number": 50
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.16 pagenumber 323"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"M = 10 #KN.m - The moment applied\n",
"I_max = 23.95*(10**6) #mm4 - I_z The moment of inertia\n",
"I_min = 2.53*(10**6) #mm4 - I_y The moment of inertia\n",
"o = 14.34 # degress the principle axis rotated\n",
"#Coponents of M in Y,Z direction \n",
"M_z = M*(10**6)*math.cos(math.radians(o))\n",
"M_y = M*(10**6)*math.sin(math.radians(o))\n",
"#tanb = I_z /I_y *tan14.34\n",
"b = math.atan((I_max*math.tan(math.radians(o))/I_min ))\n",
"B = math.degrees(b) \n",
"y_p = 122.9 # mm - principle axis Y cordinate\n",
"z_p = -26.95 #mm - principle axis z cordinate\n",
"stress_B = - M_z*y_p/I_max + M_y*z_p/I_min #Mpa - Maximum tensile stress\n",
"y_f = -65.97 # mm - principle axis Y cordinate\n",
"z_f = 41.93 #mm - principle axis z cordinate\n",
"stress_f = - M_z*y_f/I_max + M_y*z_f/I_min #Mpa - Maximum compressive stress\n",
"print \"The maximum tensile stress\",round(stress_B,2) ,\"Mpa\"\n",
"print \"The maximum compressive stress\",round(stress_f,2),\"Mpa\"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum tensile stress -76.1 Mpa\n",
"The maximum compressive stress 67.73 Mpa\n"
]
}
],
"prompt_number": 83
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.18 page number 328"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"l = 50 #mm - The length of the beam \n",
"b = 50 #mm - The width of the beam \n",
"A = l*b #mm2 - The area of the beam \n",
"p = 8.33 #KN - The force applied on the beam \n",
"stress_max = p*(10**3)/A #Mpa After cutting section A--b\n",
"print \"The maximum stress in the beam\",stress_max ,\"Mpa \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum stress in the beam 3.332 Mpa \n"
]
}
],
"prompt_number": 51
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.24 page number 339"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"M = 10 #KN.m - The moment applied\n",
"I_max = 23.95*(10**6) #mm4 - I_z The moment of inertia\n",
"I_min = 2.53*(10**6) #mm4 - I_y The moment of inertia\n",
"o = 14.34 # degress the principle axis rotated\n",
"#Coponents of M in Y,Z direction \n",
"M_z = M*(10**6)*math.cos(math.radians(o))\n",
"M_y = M*(10**6)*math.sin(math.radians(o))\n",
"#tanb = I_z /I_y *tan14.34\n",
"b = math.atan((I_max*math.tan(math.radians(o))/I_min ))\n",
"B = math.degrees(b) \n",
"y_p = 122.9 # mm - principle axis Y cordinate\n",
"z_p = -26.95 #mm - principle axis z cordinate\n",
"stress_B = - M_z*y_p/I_max + M_y*z_p/I_min #Mpa - Maximum tensile stress\n",
"y_f = -65.97 # mm - principle axis Y cordinate\n",
"z_f = 41.93 #mm - principle axis z cordinate\n",
"stress_f = - M_z*y_f/I_max + M_y*z_f/I_min #Mpa - Maximum compressive stress\n",
"#location of nuetral axis To show these stresses are max and minimum \n",
"#tanB = MzI_z + MzI_yz/MyI_y +M_YI_yz\n",
"I_z = 22.64 *(10**6) #mm4 moment of inertia in Z direction\n",
"I_y = 3.84 *(10**6) #mm4 moment of inertia in Y direction\n",
"I_yz =5.14 *(10**6) #mm4 moment of inertia in YZ direction \n",
"M_y = M #KN.m bending moment in Y dorection \n",
"M_z = M #KN.m bending moment in Y dorection \n",
"B = math.atan(( M_z*I_yz)/(M_z*I_y )) #radians location on neutral axis\n",
"beta = math.degrees(B)\n",
"print \"By sketching the line with angle\",round(beta,1),\"degrees The farthest point associated with B and F\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"By sketching the line with angle 53.2 degrees The farthest point associated with B and F\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}