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{
"metadata": {
"name": "",
"signature": "sha256:b60753ad5256b26ed900efc57b1a6de06dc4217fa7384fe1bb9afe99a185fc77"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter13-Elements of reinforced concrete"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1-486"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate The stress induced in the concrete and steel and \n",
"b = 10.;##inches\n",
"d = 1.;##inches\n",
"h = 20.;##inches\n",
"r = 2.;##inches\n",
"M_r = 500000.;##lb-inches\n",
"m = 15.;\n",
"A_r = 4.*0.25*math.pi*d**2;##in^2\n",
"h_eff = h-r;##inches\n",
"K = m*A_r/(b*h_eff);##inches\n",
"n1 = math.sqrt((K)**2+(2*K))-K;\n",
"n = n1*h_eff;##inches\n",
"a = h_eff-(n/3);##inches\n",
"c = 2.*M_r/(b*n*a);##lb/in^2\n",
"t = (h_eff-n)*m*c/n;##lb/in^2\n",
"print'%s %.d %s'%('The stress induced in the concrete and steel, t =',t,'lb/in^2');\n",
"\n",
"##there is a minute error in the answer given in textbook.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The stress induced in the concrete and steel, t = 10643 lb/in^2\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2-pg487"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate The distance of the N.A from the top edge and The safe moment of inertia and The safe moment of inertia\n",
"b = 8.;##inches\n",
"d = 7/8.;##inches\n",
"h = 18.;##inches\n",
"r = 2.;##inches\n",
"c = 750.;##lb/in^2\n",
"t_limit = 18000.;##lb/in^2\n",
"m = 8.;\n",
"h_eff = 16.;##inches\n",
"m = 18;\n",
"A_t = 3*0.25*math.pi*d**2;##in^2\n",
"K = m*A_t/(b*h_eff);##inches\n",
"n1 = math.sqrt((K)**2.+(2.*K))-K;\n",
"n = n1*h_eff;##inches\n",
"a = h_eff - (n/3.);##inches\n",
"t = m*c*(h_eff-n)/n;##lb/in^2\n",
"\n",
"if t<t_limit:\n",
" t = t;\n",
"else:\n",
" t = t_limit;\n",
"\n",
"M_r = t*A_t*a;## lb/inches\n",
"W = M_r*8./(12.*h_eff);##lb-wt\n",
"print'%s %.3f %s'%('The distance of the N.A from the top edge, n =',n,'inches.');\n",
"print'%s %.d %s'%('The safe moment of inertia is, t =',t,'lb/in^2.');\n",
"print'%s %.d %s %d %s'% ('Unifromly distributed load over the beam, W =',W,'lb-wt. or',W/16.011,'lb. per foot run');\n",
"\n",
"##there are calculation errors given in the answer in textbook.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The distance of the N.A from the top edge, n = 8.039 inches.\n",
"The safe moment of inertia is, t = 13368 lb/in^2.\n",
"Unifromly distributed load over the beam, W = 13385 lb-wt. or 836 lb. per foot run\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3-pg488"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate Area of steel reinforcement required and Corresponding stress in steel \n",
"import numpy\n",
"from numpy import roots\n",
"def quadratic(d,M_c,c_limit,b):\n",
" p = ([1 , -(d*3), + M_c*3/(0.5*c_limit*b)]);\n",
" Z = numpy.roots(p);\n",
" return Z;\n",
"\n",
"b = 12.;##inches\n",
"h = 22.;##inches\n",
"r = 2.;##inches\n",
"W = 1500.;##lb per foot run\n",
"d = h-r;##feet\n",
"l = 20.;##inches\n",
"c_limit = 700.;## lb/in^2\n",
"m = 15.;\n",
"M_c = W*20.*l*b/8.;## lb-inches\n",
"Z = quadratic(d,M_c,c_limit,b);\n",
"n = round(Z[1]);\n",
"t = m*c_limit*(d-n)/n;## lb/in62\n",
"A_t = 0.5*c_limit*b*n/t;## in^2\n",
"print'%s %.d %s'%('Area of steel reinforcement required is, A_t =',A_t,'in^2');\n",
"print'%s %.d %s'%(' Corresponding stress in steel is, t =',t,'lb/in^2');\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Area of steel reinforcement required is, A_t = 13 in^2\n",
" Corresponding stress in steel is, t = 4500 lb/in^2\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4-pg492"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate Effective deapth\n",
"m = 15.;\n",
"t = 18000.;## lb/in^2\n",
"c = 700.;## lb/in^2\n",
"b = 12.;## inches\n",
"M = 900000.;##bending moment lb/inches\n",
"k1 = 1./((t/(m*c))+1.);##k = n/d\n",
"k2 = 1.-k1/3.;##k2 = a/d\n",
"p = 0.5*c*k1/(t);\n",
"d = math.sqrt(M/(0.5*c*b*k1*k2));##inches\n",
"A_t = p*b*d;## sq.inches\n",
"A_t_previous = 0.25*math.pi*(7/8.)**2;##section area with diameter 7/8 inches\n",
"n = A_t/A_t_previous;\n",
"print'%s %.2f %s'%('Effective deapth is d =',d,'inches');\n",
"print'%s %.3f %s'%('A_t =',A_t,'sq.inches');\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Effective deapth is d = 25.75 inches\n",
"A_t = 2.214 sq.inches\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5-pg492"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate the effective death\n",
"l = 20.;##feet\n",
"W = 500.;## lb per foot run\n",
"c = 750.;## lb/in^2\n",
"t = 18000.;## lb/in^2\n",
"m = 15.;\n",
"BM_max = W*l*l*12./8. ;## lb-inches\n",
"##by making the effective deapth d twice the width b\n",
"d = (BM_max/(126.*0.5))**(1/3.);##inches\n",
"b = 0.5*d;##inches\n",
"##necessary reinforcement is 0.8% of concrete section\n",
"A_t = 0.008*b*d;## in^2\n",
"print'%s %.2f %s %.2f %s'%('d =',d,'inches ,b =',b,'inches');\n",
"print'%s %.3f %s'%('A_t =',A_t,'in^2');\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"d = 16.82 inches ,b = 8.41 inches\n",
"A_t = 1.132 in^2\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6-pg494"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate d1 and d2 and r1\n",
"W = 180.;## lb per sq.foot\n",
"l = 10.;## feet\n",
"b = 12.;##inches\n",
"c = 750.;## lb/in^2\n",
"m = 15.;\n",
"M = W*l*l*12./8.;##lb-inches\n",
"d_new = math.sqrt(M/(126.*b));##inches\n",
"A_t = 0.8*b*d_new/100.;##in^2\n",
"##using 3/8 inch rods \n",
"d1 = 3/8.;##inches\n",
"A1 = 0.25*math.pi*(d1)**2;##in^2 \n",
"r1 = A1*b/A_t;##inch\n",
"##using 1/2 inch rods \n",
"d2 = 1/2.;##inches\n",
"A2 = 0.25*math.pi*(d2)**2;##in^2 \n",
"r2 = A2*b/A_t;##inches\n",
"print'%s %.3f %s'%('d =',d_new,'inches');\n",
"print'%s %.3f %s'%('A_t =',A_t,'in^2');\n",
"print'%s %.3f %s %.2f %s'%('Using',d1,'inch rods, spacing centre to centre will be',r1,'inches');\n",
"print'%s %.2f %s %.1f %s'%('Using ',d2,' inch rods, spacing centre to centre will be ',r2,' inches');\n",
"##there are round-off errors in the answer given in textbook\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"d = 4.226 inches\n",
"A_t = 0.406 in^2\n",
"Using 0.375 inch rods, spacing centre to centre will be 3.27 inches\n",
"Using 0.50 inch rods, spacing centre to centre will be 5.8 inches\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7-pg495"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#calculate foot width of slab and spacing centre to centre \n",
"l = 12.;##feet\n",
"w = 150.;## lb per sq.foot\n",
"##Live load\n",
"LL = w*l;##lb-wt\n",
"##Dead Load assuming the slab thickness to be 6 inches\n",
"t = 6.;##inches\n",
"DL = t*l*12.;##lb-wt\n",
"##total load\n",
"W = LL+DL;##lb-wt\n",
"M = W*l*12./10.;##lb-inches\n",
"d = math.sqrt(M/(12.*126.));\n",
"print'%s %.4f %s'%('d =',d,'inches');\n",
"##With about an inch to cover the slab will be 6 inch thick\n",
"A_t = 0.8*l*d/100.;## in^2\n",
"##using 1/2 inch rods \n",
"d1 = 1/2.;##inches\n",
"A1 = 0.25*math.pi*(d1)**2;##in^2 \n",
"r1 = A1*l/A_t;##inches\n",
"print'%s %.4f %s'%(' Per foot width of slab, A_t =',A_t,'in^2');\n",
"print'%s %.2f %s %.3f %s'%('Using',d1,'inch rods, spacing centre to centre will be ',r1,' inches');\n",
"##there are minute calculation errors in the answer given in textbook.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"d = 5.0370 inches\n",
" Per foot width of slab, A_t = 0.4836 in^2\n",
"Using 0.50 inch rods, spacing centre to centre will be 4.873 inches\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}
|
