{ "metadata": { "name": "", "signature": "sha256:de92db499df3c69a3ee9a30a844e9f6a3e63f44e42060f2da976d486bd560283" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Safely Mounting a 40-lb Roll of Paper" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "By looking at the image and reading other redditer's comments, I'm assuming that the roll is 24\" wide. Using that as as a gage, we can estimate the spacing between the support screws at approximately 8\" apart. Neglecting the weight of the brackets and the support dowel we can draw the following free body diagram." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "If we add up all the forces in the vertical direction it must equal zero or the roll of paper would be accelerating downward ($F = m a$).\n", "\n", "$$\\sum F_y = R_y - 40 lb = 0$$\n", "$$R_y = 40 lb$$\n", "\n", "Next since we have a pair of equal an opposite forces offset from each other there would be a rotation unless some other force pair resisted it. This is know as a moment couple and since there is no rotation (while the roll isn't being unwound) we know that,\n", "\n", "$$\\sum \\tau = (40 lb)(8\") - R_{x2}(8\") = 0$$\n", "$$R_{x2} = 40 lb$$\n", "\n", "We also know that the roll isn't accelerating horizontally, so \n", "\n", "$$\\sum F_x = R_{x1} - 40 lb = 0$$\n", "$$R_{x1} = 40 lb$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We can now add in the values for the unknown forces to the drawing\n", "\n", "\n", "\n", "From this we can see that the four screws have to provide a vertical shear load of 40 lbs between them and the top two screws must resist a 40-lb withdraw load. OSHA requires a factor of safety of 4 when it comes to life safety so we can now determine the require with draw capacity of a top screw.\n", "\n", "* $FS = 4$ - Factor of safety\n", "* $n = 2$ - Number of top screws\n", "\n", "$$W_{req} = \\frac{F_{x1}}{n}\\times FS$$\n", "\n", "$$W_{req} = \\frac{40 lb}{2}\\times 4$$\n", "\n", "$$W_{req} = 80 lb$$" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Looking at Table 11.3.1 in the [NDS for Wood Construction, 2015 Edition](http://www.awc.org/standards/nds/2015.php) we can see that the allowable loading on screws is modified by load duration, temperature, moisture conditions, whether or not it is placed in the end grain of the wood, and if it is applied in a toe-nail configuration. In an indoor wall of a residence it is safe to assume temperature and moisture will be in normal ranges. Assuming standard construction of a wall and looking at the picture we can safely assume that the screw is not toe-nailed and is not in the end grain of a stud. This is also assuming that it is in a stud, which I believe is correct since there are two patched screw holes next to the top screw where it looks like someone tried to find a stud to screw into. This finally leaves us with the load duration adjustment factor found in Table 2.3.2. If we assume this roll will be left in place for more than two months then $C_D = 1.0$. This means we can use the values as written in table 12.2B." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Next we need to look at the withdraw capacity of screws based on the assumption that the wall is non-load baring and built using Spruce-Pine-Fir. This is a conservative assumption, but reasonable. The specific gravity of Spruce-Pine-Fir is found in Table 12.3.3A and equals 0.42. This can now be used in Table 12.2B to determine the withdraw capacity. Based on section 12.2.2.2 we know that these values are in pounds per inch of penetration into the stud (you can stop snickering now). For typical wall construction the drywall is $1/2\"$ thick. Also the bracket itself looks to be $1/8\"$ thick, so you would need an addition $5/8\"$ length added to the required penetration length. \n", "\n", "Based on these requirements we can calculate the minimum the top screw length for various sizes.\n", "\n", "$$L = \\frac{W_{req}}{W} + 5/8\"$$\n", "\n", "\\#6 Screw\n", "\n", "$$L = \\frac{80 lb}{67 lb} + 5/8\"$$\n", "\n", "$$L = 1.78\"$$\n", "\n", "or by automating the calculation," ] }, { "cell_type": "code", "collapsed": false, "input": [ "def Length(screw, W_req, W, extra):\n", " L = W_req/W + extra\n", " print('#{} Screws need a minimum length of {:.2f}\"'.format(screw,L))\n", "\n", "screws = [(6,69),\n", " (7,76),\n", " (8,82),\n", " (9,89),\n", " (10,95),\n", " (12,108)]\n", "\n", "for screw, W in screws:\n", " Length(screw,80,W,5/8)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "#6 Screws need a minimum length of 1.78\"\n", "#7 Screws need a minimum length of 1.68\"\n", "#8 Screws need a minimum length of 1.60\"\n", "#9 Screws need a minimum length of 1.52\"\n", "#10 Screws need a minimum length of 1.47\"\n", "#12 Screws need a minimum length of 1.37\"\n" ] } ], "prompt_number": 1 }, { "cell_type": "markdown", "metadata": {}, "source": [ "Based on experience, the \\#6 and \\#7 screws will be too likely to snap off during installation. Also a larger screw like a \\#10 or \\#12 would be too difficult to install without pre-drilling which would reduce their capacity if not expertly done. The \\#8 and \\#9 screw are not commonly sold in 1.75\" lengths so you would need to use the next larger size of 2\" to be safe." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We now need to check the adequacy of the screw for the combined shear and withdraw load. The shear capacity of a \\#8 screw in this configuration can be found in Table 12L. With the side member thickness of 5/8\" the capacity $Z= 63 lbs$ if the screw penetrates the stud at least 10 times its diameter. If it doesn't the this capacity needs to be reduced proportionally (based on note 3).\n", "\n", "* $p = 2\" - 5/8\" = 1.375\"$ - actual penetration length\n", "* $D = 0.164\"$ - screw diameter\n", "\n", "$$Z = 63 lb \\times \\frac{p}{10D}$$\n", "\n", "$$Z = 52.8 lb$$\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "We can use this to find the adjusted design value based on equation 12.4-1. \n", "\n", "In equation 12.4-1 we need to know the loading angle. This is the combination of the shear load and withdraw load. We know the withdraw load and the shear load is divided across 4 screws. \n", "\n", "$Z_{req} = \\frac{40 lb}{4} = 10 lb$ - shear load on an individual screw\n", "\n", "Applying a factor of safety of 4 we get $Z_{req} = 40 lb$.\n", "\n", "so the angle $\\alpha$ is," ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "$$\\alpha = tan^{-1}\\left(\\frac{W_{req}}{Z_{req}}\\right)$$\n", "\n", "$$\\alpha = tan^{-1}\\left(\\frac{80 lb}{40 lb}\\right)$$\n", "\n", "$$\\alpha = 63.43^o$$" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import arctan, degrees, cos, sin, sqrt\n", "alpha = arctan(80/40)\n", "print(\"Angle = {:.2f} degrees\".format(degrees(alpha)))" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Angle = 63.43 degrees\n" ] } ], "prompt_number": 2 }, { "cell_type": "markdown", "metadata": {}, "source": [ "Evaluating equation 12.4-1 we get,\n", "\n", "$$Z_{\\alpha}' = \\frac{(W' p)Z'}{(W' p)cos^2 \\alpha + Z' sin^2 \\alpha}$$\n", "\n", "\n", "Since the adjusted values of shear and withdraw are equal to the unadjusted values of shear and withdraw,\n", "\n", "$$Z_{\\alpha}' = \\frac{(112.75 lb \\times 1.375\")\\times 52.8 lb}{(112.75 lb \\times 1.375\" )cos^2 (63.43^o) + 52.8 lb \\times sin^2 (63.43^o)}$$\n", "\n", "$$Z_{\\alpha}' = 91.88 lb$$" ] }, { "cell_type": "code", "collapsed": false, "input": [ "Z_a = (82*1.375*52.8)/(82*1.375*cos(alpha)**2 + 52.8*sin(alpha)**2)\n", "print(\"The combined load resistance Z_a = {:.2f} lb\".format(Z_a))" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The combined load resistance Z_a = 91.88 lb\n" ] } ], "prompt_number": 3 }, { "cell_type": "markdown", "metadata": {}, "source": [ "The since the withdraw and shear loads are perpendicular to each other the combined loading can be calculated using the Pythagorean theorem.\n", "\n", "$$Z_{a,req} = \\sqrt{W_{req}^2 + Z_{req}^2}$$\n", "\n", "$$Z_{a,req} = \\sqrt{(80 lb)^2 + (40 lb)^2}$$\n", "\n", "$$Z_{a,req} = 89.4 lb$$" ] }, { "cell_type": "code", "collapsed": false, "input": [ "print(\"The required Z_a = {:.2f} lb\".format(sqrt(80**2+40**2)))" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The required Z_a = 89.44 lb\n" ] } ], "prompt_number": 4 }, { "cell_type": "markdown", "metadata": {}, "source": [ "So, if \\#8 screws are used, they must be at least 2\" long and screwed into the studs used to keep the roll of paper on the wall." ] } ], "metadata": {} } ] }