[Author Prev][Author Next][Thread Prev][Thread Next][Author Index][Thread Index]
Re: [ATM] Pressure per square mm
I've been thinking about this problem, too, and haven't been
too optimistic about getting realistic answers easily.
See the attached .rtf ("Rich Text") file, this was sent me by
a friend (engineer) for the simplified problem of a cantilevered
beam.
If you have trouble interpreting the format, let me know.
Dave
{\rtf1\ansi\ansicpg1252\uc1 \deff0\deflang1033\deflangfe1033{\fonttbl{\f0\froman\fcharset0\fprq2{\*\panose 02020603050405020304}Times New Roman;}{\f1\fswiss\fcharset0\fprq2{\*\panose 020b0604020202020204}Arial;}
{\f28\froman\fcharset238\fprq2 Times New Roman CE;}{\f29\froman\fcharset204\fprq2 Times New Roman Cyr;}{\f31\froman\fcharset161\fprq2 Times New Roman Greek;}{\f32\froman\fcharset162\fprq2 Times New Roman Tur;}
{\f33\froman\fcharset177\fprq2 Times New Roman (Hebrew);}{\f34\froman\fcharset178\fprq2 Times New Roman (Arabic);}{\f35\froman\fcharset186\fprq2 Times New Roman Baltic;}{\f36\fswiss\fcharset238\fprq2 Arial CE;}{\f37\fswiss\fcharset204\fprq2 Arial Cyr;}
{\f39\fswiss\fcharset161\fprq2 Arial Greek;}{\f40\fswiss\fcharset162\fprq2 Arial Tur;}{\f41\fswiss\fcharset177\fprq2 Arial (Hebrew);}{\f42\fswiss\fcharset178\fprq2 Arial (Arabic);}{\f43\fswiss\fcharset186\fprq2 Arial Baltic;}}
{\colortbl;\red0\green0\blue0;\red0\green0\blue255;\red0\green255\blue255;\red0\green255\blue0;\red255\green0\blue255;\red255\green0\blue0;\red255\green255\blue0;\red255\green255\blue255;\red0\green0\blue128;\red0\green128\blue128;\red0\green128\blue0;
\red128\green0\blue128;\red128\green0\blue0;\red128\green128\blue0;\red128\green128\blue128;\red192\green192\blue192;}{\stylesheet{\ql \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0
\fs24\lang1033\langfe1033\cgrid\langnp1033\langfenp1033 \snext0 Normal;}{\*\cs10 \additive Default Paragraph Font;}{\s15\qc \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 \fs24\ul\lang1033\langfe1033\cgrid\langnp1033\langfenp1033
\sbasedon0 \snext15 Title;}}{\*\listtable{\list\listtemplateid1471870664\listhybrid{\listlevel\levelnfc0\levelnfcn0\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698703\'02\'00.;}{\levelnumbers\'01;}
\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1\fbias0 \fi-360\li720\jclisttab\tx720 }{\listlevel\levelnfc0\levelnfcn0\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid-441522654
\'03(\'01);}{\levelnumbers\'02;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1\fbias0 \fi-390\li1470\jclisttab\tx1470 }{\listlevel\levelnfc0\levelnfcn0\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace360\levelindent0{\leveltext
\leveltemplateid67698703\'02\'02.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li2340\jclisttab\tx2340 }{\listlevel\levelnfc0\levelnfcn0\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext
\leveltemplateid67698703\'02\'03.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li2880\jclisttab\tx2880 }{\listlevel\levelnfc4\levelnfcn4\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext
\leveltemplateid67698713\'02\'04.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li3600\jclisttab\tx3600 }{\listlevel\levelnfc2\levelnfcn2\leveljc2\leveljcn2\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext
\leveltemplateid67698715\'02\'05.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-180\li4320\jclisttab\tx4320 }{\listlevel\levelnfc0\levelnfcn0\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext
\leveltemplateid67698703\'02\'06.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li5040\jclisttab\tx5040 }{\listlevel\levelnfc4\levelnfcn4\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext
\leveltemplateid67698713\'02\'07.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li5760\jclisttab\tx5760 }{\listlevel\levelnfc2\levelnfcn2\leveljc2\leveljcn2\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext
\leveltemplateid67698715\'02\'08.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-180\li6480\jclisttab\tx6480 }{\listname ;}\listid512039627}{\list\listtemplateid1462553878\listhybrid{\listlevel\levelnfc0\levelnfcn0\leveljc0
\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698703\'02\'00.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1\fbias0 \fi-360\li720\jclisttab\tx720 }{\listlevel\levelnfc4\levelnfcn4
\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698713\'02\'01.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li1440\jclisttab\tx1440 }{\listlevel\levelnfc2\levelnfcn2
\leveljc2\leveljcn2\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698715\'02\'02.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-180\li2160\jclisttab\tx2160 }{\listlevel\levelnfc0\levelnfcn0
\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698703\'02\'03.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li2880\jclisttab\tx2880 }{\listlevel\levelnfc4\levelnfcn4
\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698713\'02\'04.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li3600\jclisttab\tx3600 }{\listlevel\levelnfc2\levelnfcn2
\leveljc2\leveljcn2\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698715\'02\'05.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-180\li4320\jclisttab\tx4320 }{\listlevel\levelnfc0\levelnfcn0
\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698703\'02\'06.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li5040\jclisttab\tx5040 }{\listlevel\levelnfc4\levelnfcn4
\leveljc0\leveljcn0\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698713\'02\'07.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-360\li5760\jclisttab\tx5760 }{\listlevel\levelnfc2\levelnfcn2
\leveljc2\leveljcn2\levelfollow0\levelstartat1\levelspace0\levelindent0{\leveltext\leveltemplateid67698715\'02\'08.;}{\levelnumbers\'01;}\chbrdr\brdrnone\brdrcf1 \chshdng0\chcfpat1\chcbpat1 \fi-180\li6480\jclisttab\tx6480 }{\listname ;}\listid1660304817}}
{\*\listoverridetable{\listoverride\listid512039627\listoverridecount0\ls1}{\listoverride\listid1660304817\listoverridecount0\ls2}}{\info{\title Cantilevered Beam Problem}{\author KCC}{\operator KCC}{\creatim\yr2004\mo3\dy8\hr19\min1}
{\revtim\yr2004\mo3\dy8\hr19\min1}{\version2}{\edmins1}{\nofpages2}{\nofwords307}{\nofchars1752}{\nofcharsws2151}{\vern8247}}\widowctrl\ftnbj\aenddoc\noxlattoyen\expshrtn\noultrlspc\dntblnsbdb\nospaceforul\formshade\horzdoc\dgmargin\dghspace180
\dgvspace180\dghorigin1800\dgvorigin1440\dghshow1\dgvshow1\jexpand\viewkind1\viewscale125\pgbrdrhead\pgbrdrfoot\splytwnine\ftnlytwnine\htmautsp\nolnhtadjtbl\useltbaln\alntblind\lytcalctblwd\lyttblrtgr\lnbrkrule \fet0\sectd
\linex0\endnhere\sectlinegrid360\sectdefaultcl {\*\pnseclvl1\pnucrm\pnstart1\pnindent720\pnhang{\pntxta .}}{\*\pnseclvl2\pnucltr\pnstart1\pnindent720\pnhang{\pntxta .}}{\*\pnseclvl3\pndec\pnstart1\pnindent720\pnhang{\pntxta .}}{\*\pnseclvl4
\pnlcltr\pnstart1\pnindent720\pnhang{\pntxta )}}{\*\pnseclvl5\pndec\pnstart1\pnindent720\pnhang{\pntxtb (}{\pntxta )}}{\*\pnseclvl6\pnlcltr\pnstart1\pnindent720\pnhang{\pntxtb (}{\pntxta )}}{\*\pnseclvl7\pnlcrm\pnstart1\pnindent720\pnhang{\pntxtb (}
{\pntxta )}}{\*\pnseclvl8\pnlcltr\pnstart1\pnindent720\pnhang{\pntxtb (}{\pntxta )}}{\*\pnseclvl9\pnlcrm\pnstart1\pnindent720\pnhang{\pntxtb (}{\pntxta )}}\pard\plain \s15\qc \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0
\fs24\ul\lang1033\langfe1033\cgrid\langnp1033\langfenp1033 {Cantilevered Beam Problem
\par }\pard\plain \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 \fs24\lang1033\langfe1033\cgrid\langnp1033\langfenp1033 {
\par Statement of Problem:
\par
\par A bar of length L, width D, and weight W rests on a flat surface, with length L1 resting on the surface and L2 extending over the edge. Find the pressure p(x) on the surface.
\par
\par Assumptions:
\par
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 1.\tab}}\pard \qj \fi-360\li720\ri0\widctlpar\jclisttab\tx720\aspalpha\aspnum\faauto\ls1\adjustright\rin0\lin720\itap0 {The bar is homogeneous.
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 2.\tab}The bar is a rectangular right prism.
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par Solution:
\par
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 1.\tab}}\pard \qj \fi-360\li720\ri0\widctlpar\jclisttab\tx720\aspalpha\aspnum\faauto\ls2\adjustright\rin0\lin720\itap0 {
Let the origin be located at the edge of the surface, with the x-axis extending off the edge. Then the location of the unsupported end of the bar is x = L2, and the location of the supported end is x = -L1.
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 2.\tab}The weight W acts on the bar at its center of gravity, which is in the center of the bar, at x = (L2 \endash L1)/2.
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 3.\tab}Define r(x) = D*p(x) to be the differential of the reaction force supporting the bar. Then
\par }\pard \qj \li360\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin360\itap0 {\u8992\'28}{\f1\super 0
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 { \u9474\'a6 r(x) dx = W Force Balance Equation
\par \u8993\'29}{\f1\sub -L1
\par }{
\par }\pard \qj \li360\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin360\itap0 {\u8992\'28}{\f1\super 0
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 { \u9474\'a6 x r(x) dx = W (L2 \endash L1)/2 Moment Balance Equation
\par \u8993\'29}{\f1\sub -L1
\par }{
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 4.\tab}}\pard \qj \fi-360\li720\ri0\widctlpar\jclisttab\tx720\aspalpha\aspnum\faauto\ls2\adjustright\rin0\lin720\itap0 {Now assume that r(x) & the integral of [x r(x)] are polynomials in x,
\par }\pard \qj \li360\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin360\itap0 {
\par R(x) = a}{\f1\sub 0}{ + a}{\f1\sub 1}{x + a}{\f1\sub 2}{x}{\f1\super 2}{ + a}{\f1\sub 3}{x}{\f1\super 3}{
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 5.\tab}}\pard \qj \fi-360\li720\ri0\widctlpar\jclisttab\tx720\aspalpha\aspnum\faauto\ls2\adjustright\rin0\lin720\itap0 {Taking the derivative of R(x)
\par }\pard \qj \li360\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin360\itap0 {
\par x r(x) = a}{\f1\sub 1}{ + 2a}{\f1\sub 2}{x + 3a}{\f1\sub 3}{x}{\f1\super 2}{
\par
\par Since the first term is not divisible by x, a}{\f1\sub 1}{ must be zero, so
\par
\par r(x) = 2a}{\f1\sub 2}{ + 3a}{\f1\sub 3}{x
\par
\par And
\par
\par R(x) = a}{\f1\sub 0}{ + a}{\f1\sub 2}{x}{\f1\super 2}{ + a}{\f1\sub 3}{x}{\f1\super 3}{
\par
\par
\par
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 6.\tab}}\pard \qj \fi-360\li720\ri0\widctlpar\jclisttab\tx720\aspalpha\aspnum\faauto\ls2\adjustright\rin0\lin720\itap0 {Substituting this into the Moment Balance Equation, and evaluating:
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par }\pard \qj \li360\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin360\itap0 {( a}{\f1\sub 0}{ + a}{\f1\sub 2}{x}{\f1\super 2}{ + a}{\f1\sub 3}{x}{\f1\super 3}{ ) | }{\f1\sub (x=0)}{ - ( a}{\f1\sub 0}{ + a}{\f1\sub 2}{x}{\f1\super 2}{ + a}{
\f1\sub 3}{x}{\f1\super 3}{ ) | }{\f1\sub (x=-L1)}{ = W (L2 \endash L1) / 2
\par
\par - a}{\f1\sub 2}{ (L}{\f1\sub 1}{)}{\f1\super 2}{ + a}{\f1\sub 3}{ (L}{\f1\sub 1}{)}{\super 3}{ = W (L2 \endash L1) / 2
\par
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 7.\tab}}\pard \qj \fi-360\li720\ri0\widctlpar\jclisttab\tx720\aspalpha\aspnum\faauto\ls2\adjustright\rin0\lin720\itap0 {But we also have the Force Balance Equation, so
\par }\pard \qj \li360\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin360\itap0 {
\par \u8992\'28}{\f1\super 0
\par }{\u9474\'a6 ( 2a}{\f1\sub 2}{ + 3a}{\f1\sub 3}{x ) dx = W
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 { \u8993\'29}{\f1\sub -L1
\par }\pard \qj \li360\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin360\itap0 {
\par Integrating, we get
\par
\par [ 2a}{\f1\sub 2}{x + (3/2)a}{\f1\sub 3}{x}{\f1\super 2}{ ] | }{\f1\sub (x=0)}{ - [ 2a}{\f1\sub 2}{x + (3/2)a}{\f1\sub 3}{x}{\f1\super 2}{ ] | }{\f1\sub (x=-L1)}{ = W
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 8.\tab}}\pard \qj \fi-360\li720\ri0\widctlpar\jclisttab\tx720\aspalpha\aspnum\faauto\ls2\adjustright\rin0\lin720\itap0 {Solving the two equations from 6 & 7 simulataneously, we get:
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par }\pard \qj \fi360\li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {a}{\f1\sub 3}{ = 2 W L2 / L1}{\f1\super 3}{
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par }\pard \qj \fi360\li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {a}{\f1\sub 2}{ = (W/ 2L1) (1 + 3L2 / L1)
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 9.\tab}}\pard \qj \fi-360\li720\ri0\widctlpar\jclisttab\tx720\aspalpha\aspnum\faauto\ls2\adjustright\rin0\lin720\itap0 {Substituting into the definition of r(x), we have:
\par }\pard \qj \li360\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin360\itap0 {
\par r(x) = (W/L1)(1 + 3L2/L1) + 6xWL2/L1}{\f1\super 3}{
\par
\par p(x) = D*[(W/L1)(1 + 3L2/L1) + 6xWL2/L1}{\f1\super 3}{ ]
\par
\par
\par I have checked this for two cases:
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 (1)\tab}}\pard \qj \fi-390\li1470\ri0\widctlpar\jclisttab\tx1470\aspalpha\aspnum\faauto\ls1\ilvl1\adjustright\rin0\lin1470\itap0 {
In the case where L2 = 0 (no overhang) the reaction force must be evenly distributed over L1, so r(x) = W/L1.
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par {\listtext\pard\plain\hich\af0\dbch\af0\loch\f0 (2)\tab}}\pard \qj \fi-390\li1470\ri0\widctlpar\jclisttab\tx1470\aspalpha\aspnum\faauto\ls1\ilvl1\adjustright\rin0\lin1470\itap0 {In the case where L2 = L1, r(x) = 4W/L1 + 6xW/L1}{\f1\super 2}{
\par }\pard \qj \li1440\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin1440\itap0 {Substituting this into the Force and Moment Balance Equations also show that the equations are satisfied.
\par
\par
\par }\pard \qj \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par
\par
\par
\par
\par
\par }\pard \qc \li0\ri0\widctlpar\aspalpha\aspnum\faauto\adjustright\rin0\lin0\itap0 {
\par }}
_______________________________________________
ATM mailing list http://www.atmlist.net/