{"id":4532,"date":"2021-04-18T09:48:38","date_gmt":"2021-04-18T09:48:38","guid":{"rendered":"https:\/\/pickedshares.com\/?p=4532"},"modified":"2021-05-08T15:03:13","modified_gmt":"2021-05-08T15:03:13","slug":"engineering-mechanics-bending-line-of-a-beam-under-multiple-loads","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-bending-line-of-a-beam-under-multiple-loads\/","title":{"rendered":"Bending line of a beam under multiple loads"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">In this exercise, the bending line for a beam with partial line load, individual force and moment is calculated.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">A beam on a fixed bearing and a floating bearing is loaded over the length l by the line load q. Additionally the force F acts at 2l and the moment M on the floating bearing B. The bending line of the beam has to be determined.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"390\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-1-1024x390.png\" alt=\"Beam on two supports with line load, single load and moment\" class=\"wp-image-4502\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-1-1024x390.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-1-300x114.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-1-768x293.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-1-1536x586.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-1.png 1739w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Beam on two supports with line load, single load and moment<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">In the first step, the equations for the bearing reactions are determined. To determine the internal forces, the beam is divided into three sections. The horizontal forces are not considered further in the following equations, since they are obviously zero. <a href=\"https:\/\/pickedshares.com\/en\/right-hand-rules\/\">Left turning moments are positive<\/a>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Bearing loads<\/h3>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"424\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-2-1024x424.png\" alt=\"Bearing loads and beam sections\" class=\"wp-image-4508\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-2-1024x424.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-2-300x124.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-2-768x318.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-2-1536x636.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-2.png 1771w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Bearing loads and beam sections<\/figcaption><\/figure>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n<p>The sum of forces in z-direction gives<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{1} \\sum F_z = 0 = -F_{Az} - F_{Bz} + F + \\int_0^{l}{qdx} \\]<\/p>\n<p>\\[ \\tag{2} 0 = -F_{Az} - F_{Bz} + F + q \\cdot l \\]<\/p>\n<\/div>\n<p>The sum of moments regarding point A<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{3} \\sum M(A) = 0 = 3 \\cdot F_{Bz} \\cdot l - 2 \\cdot F \\cdot l - \\int_0^{l}{qxdx} + M \\]<\/p>\n<p>\\[ \\tag{4} 0 = 3 \\cdot F_{Bz} \\cdot l - 2 \\cdot F \\cdot l - \\frac{1}{2} q \\cdot l^2 + M \\]<\/p>\n<\/div>\n<p>Both bearing loads F<sub>Az<\/sub> and F<sub>Bz<\/sub> can now be solved. They are:<\/p>\n<p>\\[ \\tag{5} {F_{\\mathit{Bz}}}=\\frac{{{l}^{2}} q+4 F l-2 M}{6 l} \\]<\/p>\n<p>\\[ \\tag{6} {F_{\\mathit{Az}}}=\\frac{5 {{l}^{2}} q+2 F l+2 M}{6 l}\\]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The next step is to determine the internal forces in sections I to III. In order to be able to integrate the function of the line load from 0 to x, it is formulated as a function of the auxiliary coordinate \u03be. Normal forces are denoted by N, transverse forces by Q and bending moments by M<sub>b<\/sub>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Determination of the internal forces<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Section I<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image\"><img loading=\"lazy\" decoding=\"async\" width=\"929\" height=\"831\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-3.png\" alt=\"Internal forces and moments in section I\" class=\"wp-image-4510\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-3.png 929w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-3-300x268.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-3-768x687.png 768w\" sizes=\"auto, (max-width: 929px) 100vw, 929px\" \/><figcaption>Internal forces and moments in section I<\/figcaption><\/figure>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n<p>The sum of forces in z-direction gives<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{7} \\sum F_z = 0 = Q_I - F_{Az} + \\int_0^{x}{qd\\xi} \\]<\/p>\n<p>\\[ \\tag{8} {Q_I}=-\\frac{6 l q x-5 {{l}^{2}} q-2 F l-2 M}{6 l} \\]<\/p>\n<\/div>\n<p>The sum of moments regarding point x<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{9} \\sum M(x) = 0 = \\int_0^{x}{q (x-\\xi) d \\xi} - F_{Az} \\cdot x + M_{bI} \\]<\/p>\n<p>\\[ \\tag{10} {M_{\\mathit{bI}}}=-\\frac{3 l q\\, {{x}^{2}}+\\left( -5 {{l}^{2}} q-2 F l-2 M\\right)  x}{6 l}\\]<\/p>\n<\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Section II<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large custom-flex-image75\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"628\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-4-1024x628.png\" alt=\"Internal forces and moments in section II\" class=\"wp-image-4512\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-4-1024x628.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-4-300x184.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-4-768x471.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-4.png 1287w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Internal forces and moments in section II<\/figcaption><\/figure>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n<p>The sum of forces in z-direction gives:<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{11} \\sum F_z = 0 = Q_{II} - F_{Az} + \\int_0^{l}{q d \\xi} \\]<\/p>\n<p>\\[ \\tag{12} {Q_{\\mathit{II}}}=-\\frac{{{l}^{2}} q-2 F l-2 M}{6 l}\\]<\/p>\n<\/div>\n<p>The sum of moments regarding point x<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{13} \\sum M(x) = 0 = \\int_0^{l}{q (x - \\xi) d \\xi} - F_{Az} \\cdot x + M_{bII} \\]<\/p>    \n<p>\\[ \\tag{14} {M_{\\mathit{bII}}}=-\\frac{\\left( {{l}^{2}} q-2 F l-2 M\\right)  x-3 {{l}^{3}} q}{6 l}\\]<\/p>\n<\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Section III<\/h4>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"484\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-5-1024x484.png\" alt=\"Internal forces and moments in section III\" class=\"wp-image-4514\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-5-1024x484.png 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-5-300x142.png 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-5-768x363.png 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-5-1536x726.png 1536w, https:\/\/pickedshares.com\/wp-content\/uploads\/2021\/04\/tm2-22-5.png 1665w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Internal forces and moments in section III<\/figcaption><\/figure>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n<p>The sum of forces in z-direction gives<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{15} \\sum F_z = 0 = Q_{III} - F_{Az} + F + \\int_0^{l}{q d \\xi} \\]<\/p>\n<p>\\[ \\tag{16} {Q_{\\mathit{III}}}=-\\frac{{{l}^{2}} q+4 F l-2 M}{6 l}\\]<\/p>\n<\/div>\n<p>The sum of moments regarding point x<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{17} \\sum M(x) = 0 = \\int_0^{l}{q (x - \\xi) d \\xi} - F_{Az} \\cdot x + F \\cdot (x - 2 \\cdot l) + M_{bIII} \\]<\/p>\n<p>\\[ \\tag{18} {M_{\\mathit{bIII}}}=-\\frac{\\left( {{l}^{2}} q+4 F l-2 M\\right)  x-3 {{l}^{3}} q-12 F\\, {{l}^{2}}}{6 l}\\]<\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\">The next step is the determination of the bending line functions.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Bending line<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The function of the bending line is designated as w, or the first derivative as w' and the second derivative as w''. The modulus of elasticity is E and the area moment of inertia is I.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Section I<\/h4>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{19} {M_{\\mathit{bI}}}=-\\frac{3 l q\\, {{x}^{2}}+\\left( -5 {{l}^{2}} q-2 F l-2 M\\right)  x}{6 l} \\]<\/p>\n<p>\\[ \\tag{20}  w''_I = - \\frac{1}{E \\cdot I} M_{bI}   \\]<\/p>\n<p>\\[ \\tag{21} w''_I = -\\frac{-3 l q\\, {{x}^{2}}-\\left( -5 {{l}^{2}} q-2 F l-2 M\\right)  x}{6 E I l}\\]<\/p>\n<p>\\[ \\tag{22}  w'_I = {c_1}-\\frac{\\frac{\\left( 5 {{l}^{2}} q+2 F l+2 M\\right) \\, {{x}^{2}}}{2}-l q\\, {{x}^{3}}}{6 E I l}\\] <\/p>\n<p>\\[ \\tag{23}  w_I = -\\frac{\\frac{\\left( 5 {{l}^{2}} q+2 F l+2 M\\right) \\, {{x}^{3}}}{6}-\\frac{l q\\, {{x}^{4}}}{4}}{6 E I l}+{c_1} x+{c_2}\\] <\/p>\n<\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Section II<\/h4>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{24} {M_{\\mathit{bII}}}=-\\frac{\\left( {{l}^{2}} q-2 F l-2 M\\right)  x-3 {{l}^{3}} q}{6 l}\\]<\/p>\n<p>\\[ \\tag{25}  w''_{II} = - \\frac{1}{E \\cdot I} M_{bII}   \\]<\/p>\n<p>\\[ \\tag{26} w''_{II} = -\\frac{3 {{l}^{3}} q-\\left( {{l}^{2}} q-2 F l-2 M\\right)  x}{6 E I l}\\]<\/p>\n<p>\\[ \\tag{27}  w'_{II} = {c_3}-\\frac{\\frac{\\left( -{{l}^{2}} q+2 F l+2 M\\right) \\, {{x}^{2}}}{2}+3 {{l}^{3}} q x}{6 E I l} \\] <\/p>\n<p>\\[ \\tag{28}  w_{II} = -\\frac{\\frac{\\left( -{{l}^{2}} q+2 F l+2 M\\right) \\, {{x}^{3}}}{6}+\\frac{3 {{l}^{3}} q\\, {{x}^{2}}}{2}}{6 E I l}+{c_3} x+{c_4}\\] <\/p>\n<\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Section III<\/h4>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{29} {M_{\\mathit{bIII}}}= -\\frac{\\left( {{l}^{2}} q+4 F l-2 M\\right)  x-3 {{l}^{3}} q-12 F\\, {{l}^{2}}}{6 l}\\]<\/p>\n<p>\\[ \\tag{30}  w''_{III} = - \\frac{1}{E \\cdot I} M_{bIII}   \\]<\/p>\n<p>\\[ \\tag{31} w''_{III} = -\\frac{-\\left( {{l}^{2}} q+4 F l-2 M\\right)  x+3 {{l}^{3}} q+12 F\\, {{l}^{2}}}{6 E I l}\\]<\/p>\n<p>\\[ \\tag{32}  w'_{III} = {c_5}-\\frac{\\frac{\\left( -{{l}^{2}} q-4 F l+2 M\\right) \\, {{x}^{2}}}{2}+3 {{l}^{3}} q x+12 F\\, {{l}^{2}} x}{6 E I l}\\] <\/p>\n<p>\\[ \\tag{33}  w_{III} = -\\frac{\\frac{\\left( -{{l}^{2}} q-4 F l+2 M\\right) \\, {{x}^{3}}}{6}+\\frac{3 {{l}^{3}} q\\, {{x}^{2}}}{2}+6 F\\, {{l}^{2}}\\, {{x}^{2}}}{6 E I l}+{c_5} x+{c_6}\\] <\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\">The constants of integration are determined via the boundary and secondary conditions.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Boundary and secondary conditions<\/h4>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n<p>The deflection at x = 0 is zero.<\/p>\n<p>\\[ \\tag{34} w_I(x=0)=0=\\mathit{c_2} \\]<\/p>\n<p>The deflection at the point x = l is the same for bending line I and bending line II.<\/p>\n<p>\\[ \\tag{35} w_I(x=l)=w_{II}(x=l) \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{36} -\\frac{\\frac{{{l}^{3}}\\, \\left( 5 {{l}^{2}} q+2 F l+2 M\\right) }{6}-\\frac{{{l}^{5}} q}{4}}{6 E I l}+{c_1} l+{c_2}=-\\frac{\\frac{{{l}^{3}}\\, \\left( -{{l}^{2}} q+2 F l+2 M\\right) }{6}+\\frac{3 {{l}^{5}} q}{2}}{6 E I l}+{c_3} l+{c_4} \\]<\/p><\/div>\n\n<p>The inclination at the point x = l is the same for bending line I and bending line II.<\/p>\n<p>\\[ \\tag{37} w'_I(x=l)=w'_{II}(x=l) \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{38} {c_1}-\\frac{\\frac{{{l}^{2}}\\, \\left( 5 {{l}^{2}} q+2 F l+2 M\\right) }{2}-{{l}^{4}} q}{6 E I l}={c_3}-\\frac{\\frac{{{l}^{2}}\\, \\left( -{{l}^{2}} q+2 F l+2 M\\right) }{2}+3 {{l}^{4}} q}{6 E I l} \\]<\/p>\n<\/div>\n\n<p>The deflection at the point x = 2l is the same for bending line II and bending line III.<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{39} w_{II}(x=2l)=w_{III}(x=2l) \\]<\/p>\n<p>\\[ -\\frac{\\frac{4 {{l}^{3}}\\, \\left( -{{l}^{2}} q+2 F l+2 M\\right) }{3}+6 {{l}^{5}} q}{6 E I l}+2 {c_3} l+{c_4}=... \\]<\/p>\n<p>\\[ \\tag{40} ...=-\\frac{\\frac{4 {{l}^{3}}\\, \\left( -{{l}^{2}} q-4 F l+2 M\\right) }{3}+6 {{l}^{5}} q+24 F\\, {{l}^{4}}}{6 E I l}+2 {c_5} l+{c_6} \\]<\/p><\/div>\n\n\n\n<p>The inclination at the point x = 2l is the same for bending line II and bending line III.<\/p>\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{41} w'_{II}(x=2l)=w'_{III}(x=2l) \\]<\/p>\n<p>\\[ {c_3}-\\frac{2 {{l}^{2}}\\, \\left( -{{l}^{2}} q+2 F l+2 M\\right) +6 {{l}^{4}} q}{6 E I l}=... \\]<\/p>\n<p>\\[ \\tag{42} ...={c_5}-\\frac{2 {{l}^{2}}\\, \\left( -{{l}^{2}} q-4 F l+2 M\\right) +6 {{l}^{4}} q+24 F\\, {{l}^{3}}}{6 E I l} \\]<\/p><\/div>\n\n<p>The deflection at the point x = 3l is zero.<\/p>\n<p>\\[ \\tag{43} w_{III}(x=3l)=0 \\]<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{44} 0 = -\\frac{\\frac{9 {{l}^{3}}\\, \\left( -{{l}^{2}} q-4 F l+2 M\\right) }{2}+\\frac{27 {{l}^{5}} q}{2}+54 F\\, {{l}^{4}}}{6 E I l}+3 {c_5} l+{c_6} \\]<\/p><\/div>\n<p>The unknowns can now be resolved.<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{45} {c_1}=\\frac{25 {{l}^{3}} q+32 F\\, {{l}^{2}}+36 M l}{72 E I}\\]<\/p>\n<p>\\[ \\tag{46} {c_2}=0\\]<\/p>\n<p>\\[ \\tag{47} {c_3}=\\frac{37 {{l}^{3}} q+32 F\\, {{l}^{2}}+36 M l}{72 E I}\\]<\/p>\n<p>\\[ \\tag{48} {c_4}=-\\frac{{{l}^{4}} q}{24 E I}\\]<\/p>\n<p>\\[ \\tag{49} {c_5}=\\frac{37 {{l}^{3}} q+176 F\\, {{l}^{2}}+36 M l}{72 E I}\\]<\/p>\n<p>\\[ \\tag{50} {c_6}=-\\frac{{{l}^{4}} q+32 F\\, {{l}^{3}}}{24 E I}\\]<\/p>\n<\/div>\n<p>Finally, the three functions of the bending lines can now be described as follows:<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[ \\tag{51} w_I(x) =  \\frac{\\left( 25 {{l}^{3}} q+32 F\\, {{l}^{2}}+36 M l\\right)  x}{72 E I}-\\frac{\\frac{\\left( 5 {{l}^{2}} q+2 F l+2 M\\right) \\, {{x}^{3}}}{6}-\\frac{l q\\, {{x}^{4}}}{4}}{6 E I l}\\]<\/p>\n<p>\\[ \\tag{52} w_{II}(x) =  -\\frac{\\frac{\\left( -{{l}^{2}} q+2 F l+2 M\\right) \\, {{x}^{3}}}{6}+\\frac{3 {{l}^{3}} q\\, {{x}^{2}}}{2}}{6 E I l}+\\frac{\\left( 37 {{l}^{3}} q+32 F\\, {{l}^{2}}+36 M l\\right)  x}{72 E I}-\\frac{{{l}^{4}} q}{24 E I}\\]<\/p>\n<p>\\[ w_{III}(x) = -\\frac{\\frac{\\left( -{{l}^{2}} q-4 F l+2 M\\right) \\, {{x}^{3}}}{6}+\\frac{3 {{l}^{3}} q\\, {{x}^{2}}}{2}+6 F\\, {{l}^{2}}\\, {{x}^{2}}}{6 E I l}+\\frac{\\left( 37 {{l}^{3}} q+176 F\\, {{l}^{2}}+36 M l\\right)  x}{72 E I}... \\]<\/p>\n<p>\\[ \\tag{53} ...-\\frac{{{l}^{4}} q+32 F\\, {{l}^{3}}}{24 E I} \\]<\/p>\n<\/div>\n\n\n\n<p class=\"wp-block-paragraph\">The determination of the bending lines is now complete. It can be clearly seen that using the same coordinate origin in sections I to III results in very extensive formulas. A simplification can be achieved, for example, by introducing a substitute coordinate in the third area, which runs in the opposite direction to the original x-direction and starts at x = 3l.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Bending line of a beam under multiple loads\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-bending-line-of-a-beam-under-multiple-loads\/\" aria-label=\"Read more about Bending line of a beam under multiple loads\">Read 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