{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5: Stresses in Beams Basic Topics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.1, page no. 307"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"calculate radius of curvature, curvature & deflection of beam\n",
"\"\"\"\n",
"\n",
"import math\n",
"import numpy\n",
"\n",
"#initialisation\n",
"\n",
"L = 8.0 # length of beam in ft\n",
"h = 6.0 # Height of beam in inch\n",
"e = 0.00125 # elongation on the bottom surface of the beam\n",
"y = -3.0 # Dismath.tance of the bottom surface to the neutral surface of the beam in inch\n",
"\n",
"#Calculations\n",
"r = -(y/e) # Radius of curvature\n",
"print \"radius of curvature is\", round(r), \"inch\"\n",
"k = 1/r # curvature in in-1\n",
"print \"curvature\", round(k,5), \"ft-1\"\n",
"theta = numpy.degrees(numpy.arcsin(((L*12.0)/(2.0*r)))) # angle in degree\n",
"print \"Angle of twist\", round(theta,3), \"degree\"\n",
"my_del = r*(1-math.cos(math.radians(theta))) #Deflection in inch\n",
"print \"Deflection in the beam is \", round(my_del,4), \"inch\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"radius of curvature is 2400.0 inch\n",
"curvature 0.00042 ft-1\n",
"Angle of twist 1.146 degree\n",
"Deflection in the beam is 0.48 inch\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.2, page no. 315"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"Evaluate bending moment & maximum bending stress in the wire\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"d = 0.004 # thickness of wire in m\n",
"R0 = 0.5 # radius of cylinder in m\n",
"E = 200e09 # Modulus of elasticity of steel\n",
"s = 1200e06 # proportional limit of steel\n",
"\n",
"#calculation\n",
"\n",
"M = (math.pi*E*d**4)/(32*(2*R0+d)) # Bending moment in wire in N-m\n",
"print \"Bending moment in the wire is \", round(M,2), \"N-m\"\n",
"s_max = (E*d)/(2*R0+d) # Maximum bending stress in wire in Pa\n",
"print \"Maximum bending stress in the wire is %e\" %(s_max), \"Pa\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Bending moment in the wire is 5.01 N-m\n",
"Maximum bending stress in the wire is 7.968127e+08 Pa\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.3, page no. 316"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"find maximum tensile and compressive stress in the beam\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"L = 22 # Span of beam in ft\n",
"q = 1.5 # Uniform load intensity in k/ft\n",
"P = 12 # Concentrated in k\n",
"b = 8.75 # width of cross section of beam in inch\n",
"h = 27 # height of cross section of beam in inch\n",
"Ra = 23.59 # Reaction at point A\n",
"Rb = 21.41 # Reacyion at point B\n",
"Mmax = 151.6 # Maximum bending moment\n",
"\n",
"#calculation\n",
"\n",
"S = (b*h**2)/6 # Section modulus\n",
"s = (Mmax*12)/S # stress in k\n",
"st = s*1000 # Tensile stress\n",
"print \"Maximum tensile stress in the beam\", round(st), \"psi\"\n",
"sc = -s*1000 # Compressive stress\n",
"print \"Maximum compressive stress in the beam\", round(sc), \"psi\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum tensile stress in the beam 1711.0 psi\n",
"Maximum compressive stress in the beam -1711.0 psi\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.4, page no. 318"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"because of uniform load in the beam, calculate maximum tensile & compressive stress\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"q = 3200.0 # Uniform load intensity in N/m\n",
"b = 0.3 # width of beam in m\n",
"h = 0.08 # Height of the beam in m\n",
"t = 0.012 # thickness of beam in m\n",
"Ra = 3600.0 # Reaction at A in N\n",
"Rb = 10800.0 # Reaction at B in N\n",
"Mpos = 2025.0 # Moment in Nm\n",
"Mneg = -3600.0 # Moment in Nm\n",
"\n",
"#calculation\n",
"y1 = t/2.0\n",
"A1 = (b-2*t)*t \n",
"y2 = h/2\n",
"A2 = h*t \n",
"A3 = A2 \n",
"c1 = ((y1*A1)+(2*y2*A2))/((A1)+(2*A2))\n",
"c2 = h - c1 \n",
"Ic1 = (b-2*t)*(t**3)*(1.0/12.0)\n",
"d1 = c1-(t/2.0)\n",
"Iz1 = (Ic1)+(A1*(d1**2))\n",
"Iz2 = 956600e-12\n",
"Iz3 = Iz2 \n",
"Iz = Iz1 + Iz2 + Iz3 # Moment of inertia of the beam cross section\n",
"\n",
"# Section Modulli\n",
"S1 = Iz / c1 # for the top surface\n",
"S2 = Iz / c2 # for the bottom surface\n",
"\n",
"# Maximum stresses for the positive section\n",
"st = Mpos / S2 \n",
"print \"Maximum tensile stress in the beam in positive section is\", st, \"Pa\"\n",
"sc = -Mpos / S1 \n",
"print \"Maximum compressive stress in the beam in positive section is\", sc, \"Pa\"\n",
"\n",
"# Maximum stresses for the negative section\n",
"snt = -Mneg / S1 \n",
"print \"Maximum tensile stress in the beam in negative section is\", snt, \"Pa\"\n",
"snc = Mneg / S2 \n",
"print \"Maximum compressive stress in the beam in negative section is\", snc, \"Pa\"\n",
"\n",
"# Conclusion\n",
"st_max = st\n",
"sc_max = snc"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum tensile stress in the beam in positive section is 50468539.6422 Pa\n",
"Maximum compressive stress in the beam in positive section is -15157118.8248 Pa\n",
"Maximum tensile stress in the beam in negative section is 26945989.0219 Pa\n",
"Maximum compressive stress in the beam in negative section is -89721848.2528 Pa\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Exampe 5.5, page no. 325"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"choose a suitable size for the beam\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"L = 12 # Length of beam in ft\n",
"q = 420 # Uniform load intensity in lb/ft\n",
"s = 1800 # Allowable bending stress in psi\n",
"w = 35 # weight of wood in lb/ft3\n",
"\n",
"#calculation\n",
"M = (q*L**2*12)/8 # Bending moment in lb-in\n",
"S = M/s # Section Modulli in in3\n",
"\n",
"# From Appendix F\n",
"q1 = 426.8 # New uniform load intensity in lb/ft\n",
"S1 = S*(q1/q) # New section modulli in in3\n",
"\n",
"# From reference to appendix F, a beam of cross section 3*12 inch is selected\n",
"print (\"Beam of crosssection 3*12 is sufficient\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Beam of crosssection 3*12 is sufficient\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.6, page no. 326"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"find min. req. diameter of the woodpost & req. outer diameter of aluminum tube\n",
"\"\"\"\n",
"\n",
"import math\n",
"\n",
"#initialisation\n",
"P = 12000 # Lataeral load at the upper end in N\n",
"h = 2.5 # Height of post in m\n",
"Mmax = P*h # Maximum bending moment in Nm\n",
"\n",
"#calculation\n",
"# Part (a) : Wood Post\n",
"s1 = 15e06 # Maximum allowable stress in Pa\n",
"S1 = Mmax/s1 # Section Modulli in m3\n",
"d1 = ((32.0*S1)/math.pi)**(1.0/3.0) # diameter in m\n",
"print \"the minimum required diameter d1 of the wood post is\", round(d1,3), \"m\"\n",
"\n",
"# Part (b) : Alluminium tube\n",
"s2 = 50e06 # Maximum allowable stress in Pa\n",
"S2 = Mmax/s2 # Section Modulli in m3\n",
"d2 = (S2/0.06712)**(1.0/3.0) # diameter in meter.....(1) \n",
"print \"minimum required outer diameter d2 of the aluminum tube is\", round(d2,3),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the minimum required diameter d1 of the wood post is 0.273 m\n",
"minimum required outer diameter d2 of the aluminum tube is 0.208 m\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.7, page no. 326"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"evaluate and select a structural steel beam of wide-flange shape to support the loads\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"q = 2000.0 # Uniform load intensity in lb/ft\n",
"s = 18000.0 # Maximum allowable load in Psi\n",
"Ra = 18860.0 # Reaction at point A\n",
"Rb = 17140.0 # Reaction at point B\n",
"\n",
"#calculation\n",
"x1 = Ra/q # Distance in ft from left end to the point of zero shear\n",
"Mmax = (Ra*x1)-((q*(x1**2))/2.0) # Maximum bending moment in lb-ft\n",
"S = (Mmax*12.0)/s # Section Modulli in in3\n",
"\n",
"# Trial Beam\n",
"Ra_t = 19380.0 # Reaction at point A\n",
"Rb_t = 17670.0 # Reaction at point B\n",
"\n",
"#in Python the value for x1 differes by some points and hence the subsequent results differ\n",
"x1_t = Ra_t/q # Distance in ft from left end to the point of zero shear\n",
"Mmax_t = (Ra_t*x1_t)-((q*(x1_t**2))/2.0) # Maximum bending moment in lb-ft\n",
"S_t = (Mmax_t*12.0)/s # Section Modulli in in3\n",
"# From table E beam 12*50 is selected \n",
"print \"Beam of crosssection 12*50 is selected with section modulli\", round(S_t,1), \"in^3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Beam of crosssection 12*50 is selected with section modulli 62.6 in^3\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.8, page 329"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"find min. req. dimension of the posts\n",
"\"\"\"\n",
"\n",
"import math\n",
"\n",
"#initialisation \n",
"g = 9810 # Specific weight of water in N/m3\n",
"h = 2 # Height of dam in m\n",
"s = 0.8 # Dismath.tance between square cross section in m\n",
"sa = 8e06 # Maximum allowable stress in Pa\n",
"\n",
"#Calculations\n",
"b = ((g*(h**3)*s)/sa)**(1.0/3.0) # Dimension of croossection in m\n",
"\n",
"#Result\n",
"print \"the minimum required dimension b of the posts\", round(b,3), \"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the minimum required dimension b of the posts 0.199 m\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.11, page no. 341"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"find normal & shear stress at point C\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"L = 3 # Span of beam in ft\n",
"q = 160 # Uniform load intensity in lb/in\n",
"b = 1 # Width of cross section\n",
"h = 4 # Height of cross section\n",
"\n",
"# Calculations from chapter 4\n",
"Mc = 17920 # Bending moment in ld-in\n",
"Vc = -1600 # Loading in lb\n",
"I = (b*(h**3))/12.0 # Moment of inertia in in4\n",
"sc = -(Mc*1)/I # Compressive stress at point C in psi\n",
"Ac = 1*1 # Area of section C in inch2\n",
"yc = 1.5 # dismath.tance between midlayers od section C and cross section of beam\n",
"Qc = Ac*yc # First moment of C cross section in inch3\n",
"tc = (Vc*Qc)/(I*b) # Shear stress in Psi\n",
"print \"Normal stress at C\", sc, \"psi\"\n",
"print \"Shear stress at C\", tc, \"psi\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Normal stress at C -3360.0 psi\n",
"Shear stress at C -450.0 psi\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.12, page no. 342"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"calculate max. permissible value\n",
"\"\"\"\n",
"\n",
"import math\n",
"\n",
"#initialisation\n",
"s = 11e06 # allowable tensile stress in pa\n",
"t = 1.2e06 # allowable shear stress in pa\n",
"b = 0.1 # Width of cross section in m\n",
"h = 0.15 # Height of cross section in m\n",
"a = 0.5 # in m\n",
"\n",
"#Calculations\n",
"P_bending = (s*b*h**2)/(6.0*a) # Bending stress in N\n",
"P_shear = (2*t*b*h)/3.0 # shear stress in N\n",
"Pmax = P_bending # Because bending stress governs the design\n",
"\n",
"#Result\n",
"print \"the maximum permissible value Pmax of the loads\", Pmax, \"N\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the maximum permissible value Pmax of the loads 8250.0 N\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.13, page no. 345"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"evaluate max. shear stress in the pole & diameter of solid pole\n",
"\"\"\"\n",
"\n",
"import math\n",
"\n",
"#initialisation\n",
"d2 = 4 # Outer diameter in inch\n",
"d1 = 3.2 # Inner diameter in inch\n",
"r2 = d2/2 # Outer radius in inch\n",
"r1 = d1/2 # inner radius in inch\n",
"P = 1500 # Horizontal force in lb\n",
"\n",
"#calculation\n",
"# Part (a)\n",
"t_max = ((r2**2+(r2*r1)+r1**2)*4*P)/(3*math.pi*((r2**4)-(r1**4))) # Mximum shear stress in Psi\n",
"print \"Maximum shear stress in the pole is\", round(t_max), \"psi\"\n",
"\n",
"# Part (b)\n",
"d0 = math.sqrt((16*P)/(3*math.pi*t_max)) # Diameter of solid circular cross section in meter\n",
"print \"Diameter of solid circular cross section is \", round(d0,2), \"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum shear stress in the pole is 658.0 psi\n",
"Diameter of solid circular cross section is 1.97 m\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.14, page no. 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"maximum shear stress, minimum shear stress, and total shear force in the web.\n",
"\"\"\"\n",
"\n",
"import math\n",
"\n",
"#initialisation\n",
"b = 0.165 # in m\n",
"h = 0.320 # in m\n",
"h1 = 0.290 # in m\n",
"t = 0.0075 # in m\n",
"V = 45000.0 # Vertical force in N\n",
"\n",
"#calculation\n",
"I = (1.0/12.0)*((b*(h**3))-(b*(h1**3))+(t*(h1**3))) # Moment of inertia of the cros section\n",
"t_max = (V/(8.0*I*t))*((b*(h**2))-(b*(h1**2))+(t*(h1**2))) # Maximum shear stress in Pa\n",
"t_min = ((V*b)/(8*I*t))*(h**2-h1**2) # Minimum shear stress in Pa\n",
"T = ((t*h1)/3.0)*(2*t_max + t_min) # Total shear force in Pa\n",
"t_avg = V/(t*h1) # Average shear stress in Pa\n",
"\n",
"#Result\n",
"print \"Maximum shear stress in the web is\", round(t_max,2), \"Pa\"\n",
"print \"Minimum shear stress in the web is\", round(t_min,2), \"Pa\"\n",
"print \"Total shear stress in the web is\", round(T,2), \"N\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum shear stress in the web is 20985785.26 Pa\n",
"Minimum shear stress in the web is 17359517.46 Pa\n",
"Total shear stress in the web is 43015.04 N\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.15, page no. 352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"find shear stress at top of the web\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"V = 10000 # Vertical shear force in lb\n",
"b = 4 # in inch\n",
"t = 1 # in inch\n",
"h = 8 # in inch\n",
"h1 = 7 # in inch\n",
"\n",
"#calculation\n",
"A = b*(h-h1) + t*h1 # Area of cross section \n",
"Qaa = ((h+h1)/2.0)*b*(h-h1) + (h1/2.0)*(t*h1) # First moment of cross section\n",
"c2 = Qaa/A # Position of neutral axis in inch\n",
"c1 = h-c2 # Position of neutral axis in inch\n",
"Iaa = (b*h**3)/3.0 - ((b-t)*h1**3)/3.0 # Moment of inertia about the line aa\n",
"I = Iaa - A*c2**2 # Moment of inertia of crosssection\n",
"Q1 = b*(h-h1)*(c1-((h-h1)/2.0)) # First moment of area above the line nn\n",
"t1 = (V*Q1)/(I*t) # Shear stress at the top of web in Psi\n",
"Qmax = (t*c2)*(c2/2.0) # Maximum first moment of inertia below neutral axis\n",
"t_max = (V*Qmax)/(I*t) # Maximum Shear stress in Psi\n",
"\n",
"#Result\n",
"print \"Shear stress at the top of the web is\", round(t1), \"psi\"\n",
"print \"Maximum Shear stress in the web is\", round(t_max), \"Psi\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Shear stress at the top of the web is 1462.0 psi\n",
"Maximum Shear stress in the web is 1762.0 Psi\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.16, page no. 357"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"maximum permissible longitudinal spacing of the screws\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"\n",
"Af = 40*180 # Area of flange in mm2\n",
"V = 10500 # Shear force acting on cross section\n",
"F = 800 # Allowable load in shear\n",
"df = 120 # Dismath.tance between centroid of flange and neutral axis in mm\n",
"\n",
"#calculation\n",
"Q = Af*df # First moment of cross section of flange\n",
"I = (1.0/12.0)*(210*280**3) - (1.0/12.0)*(180*200**3) # Moment of inertia of entire cross section in mm4\n",
"f = (V*Q)/I # Shear flow\n",
"s = (2*F)/f # Spacing between the screw\n",
"\n",
"#Result\n",
"print \"The maximum permissible longitudinal spacing s of the screws is\", round(s,1), \"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum permissible longitudinal spacing s of the screws is 46.6 mm\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.17, page no. 362"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\"\"\"\n",
"maximum tensile and compressive stress in the beam\n",
"\"\"\"\n",
"\n",
"import math \n",
"\n",
"#initialisation\n",
"L = 60 # Length of beam in inch\n",
"d = 5.5 # dismath.tance from the point of application of the load P to the longitudinal axis of the tube in inch\n",
"b = 6 # Outer dimension of tube in inch\n",
"A = 20 # Area of cross section of tube in inch\n",
"I = 86.67 # Moment of inertia in in4\n",
"P = 1000 # in lb\n",
"theta = 60 # in degree\n",
"Ph = P*math.sin(math.radians(60)) # Horizontal component\n",
"Pv = P*math.cos(math.radians(60)) # Vertical component\n",
"\n",
"#Calculations\n",
"M0 = Ph*d # Moment in lb-in\n",
"y = -3 # Point at which maximum tensile stress occur in inch\n",
"N = Ph # Axial force\n",
"M = 9870 # Moment in lb-in\n",
"st_max = (N/A)-((M*y)/I) # Maximum tensile stress in Psi\n",
"yc = 3 # in inch\n",
"M1 = 5110 # moment in lb-in\n",
"sc_left = (N/A)-((M*yc)/I) # Stress at the left of point C in Psi\n",
"sc_right = -(M1*yc)/I # Stress at the right of point C in Psi\n",
"sc_max = min(sc_left,sc_right) # Because both are negative quantities\n",
"\n",
"#Result\n",
"print \"The maximum compressive stress in the beam is\", round(sc_max), \"psi\"\n",
"print \"The maximum tensile stress in the beam is\", round(st_max), \"psi\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum compressive stress in the beam is -298.0 psi\n",
"The maximum tensile stress in the beam is 385.0 psi\n"
]
}
],
"prompt_number": 26
}
],
"metadata": {}
}
]
}