Stresses in Beams Part 4 W do people order doule ceese urgers,large fries, and a diet Coke. UNQUE VEW OF HSTORY FROM THE 6 t GRDE ncient Egpt was inaited mummies and te all wrote in draulics. Te lived in te Sara Dessert. Te climate of te Sara is suc tat all te inaitants e to live elsewere. Wen we did our design exercise te last class period, we actuall ceated We looked at te ending moment at te point on te eam and developed te axial force ased on tat moment ctuall tat moment was also developing some sear stress in te eam f we look at some differential element in a eam sujected to a distriuted load we e Saturda, Novemer 09, 00 Saturda, Novemer 09, 00 4 1
f we were to look at te moment diagram, etween te locations x and x+dx we would e some small cange in moment, dm Developing te sear at an dept in te eam on te left face we would e x Left M Saturda, Novemer 09, 00 5 Saturda, Novemer 09, 00 6 f we moved to te rigt face, we e canged te moment an amount dm so te sear on tat face is x Rgt ( M + ) dm Unless te dm is equal to 0, tere must e some oter force tat will off set te imalance in te axial stresses x Rgt ( M + ) dm Saturda, Novemer 09, 00 7 Saturda, Novemer 09, 00 8
f we take a section of te eam and look at te eam at some dept x Left M x Rgt ( M + ) dm Ten te sear stress acting on te top of te area identified as must offset te imalance in axial stresses x Left M x Rgt ( M + ) dm Saturda, Novemer 09, 00 9 Saturda, Novemer 09, 00 10 Visuall M ( + ) x Left xrgt M dm f we look at a differential slice of te cross section at an dept wit a tickness of d (rememer it as a dept of dx) We e a sear on te face parallel to ot te x and axis For te slice to e in equilirium, te sears must e equal x Left M x Rgt ( + ) M dm Saturda, Novemer 09, 00 11 Saturda, Novemer 09, 00 1
From te geometr of te section te area over wic te sear acts is dx (te area of te top face) Te force developed te sear at te face is equal to dx Saturda, Novemer 09, 00 1 Saturda, Novemer 09, 00 14 Now if we write our equilirium equation for te sum of te forces in te x-direction we e dx d + d 0 xleft x Rigt Sustituting our expressions for te sears on te rigt and left face we e ( + ) M M dm dx d+ d 0 Saturda, Novemer 09, 00 15 Saturda, Novemer 09, 00 16 4
Taking te values tat aren t a function of te area outside of te integrals we e ( M + dm ) M dx d d 0 + Expanding M M dm dx d d d 0 + + dm dx+ d 0 Saturda, Novemer 09, 00 17 Saturda, Novemer 09, 00 18 Solving for te erage sear stress we e dm d dm 1 dx dx d From te expression tat we developed in te sear and moment section we know tat dm/dx V so V d Saturda, Novemer 09, 00 19 Saturda, Novemer 09, 00 0 5
From te wa we developed te section, te axial forces were developed taking te summation of te axial forces acting on te area elow or te area V d Te integral term is te first moment of te area aout te origin so we can also use te definition to sustitute te centroid of times te distance to te centroid of to calculate te moment V ˆ ' ' Saturda, Novemer 09, 00 1 Saturda, Novemer 09, 00 Tis moment of te area ar is given te smol Q so we e Tis is a muc more complicated function tat we developed for te axial stress V ˆ ' ' V ˆ ' ' Saturda, Novemer 09, 00 Saturda, Novemer 09, 00 4 6
t is also not exactl correct and onl olds for certain sape properties ut for tis course, we will consider it as sufficient V ˆ ' ' We can start looking at te simplest cross section, a rectangular cross section, and seeing ow te sear varies on tat cross section Saturda, Novemer 09, 00 5 Saturda, Novemer 09, 00 6 Te complete cross section as a widt of and a dept of Te origin of te coordinate sstem is at te centroid of te section We want to find te erage sear stress at some distance along te -axis, x x ' Saturda, Novemer 09, 00 7 Saturda, Novemer 09, 00 8 7
We will set to e te top (or ase) of some section tat starts at te ase (or top) of te lock depending on if we are aove or elow te centroid of te eam s cross section We first need to locate te centroid of te new section, ar x Saturda, Novemer 09, 00 9 ' Saturda, Novemer 09, 00 0 ' ar ' Since we e a rectangular cross section, te centroid will e located at te midpoint of te section -' Saturda, Novemer 09, 00 1 ' ar ' So te distance to te centroid of te new section is ' ' ' ' ' + ' + + ar 4 4 -' Saturda, Novemer 09, 00 ' ar ' 8
Te area of te new section is Calculating te Q value at te dept ' ' ' ' Q ar ' + ' 4 ' ' ' Q + 4 4 16 4 -' ' ar ' ' Q 8 -' ' ar ' Saturda, Novemer 09, 00 Saturda, Novemer 09, 00 4 Sustituting tis value of Q into te expression for erage sear and using te moment of inertia of te original section aout te origin we e ' V 8 1 ' 1V 6 V ' 8 4 -' Saturda, Novemer 09, 00 5 ' ar ' So for a rectangular section wit a cross section Te erage sear at an dept is given 6 V ' 4 -' Saturda, Novemer 09, 00 6 ' ar ' 9
Notice tat since is squared, te erage sear will e te same at te same distance aove and elow te origin 6 V ' 4 -' Saturda, Novemer 09, 00 7 ' ar ' lso notice tat te maximum erage sear will e located wen is equal to 0 Tis is at te origin wic is on te centroid of te original face 6 V ' 4 -' Saturda, Novemer 09, 00 8 ' ar ' lso, notice tat te erage sear is a function of te square of so it will e a paraolic sape Prolem 8-6.8 6 V + ' 4 -' ' ar ' Saturda, Novemer 09, 00 9 Saturda, Novemer 09, 00 40 10
Homework 8-6. 8-6.15 8-6.16 Saturda, Novemer 09, 00 41 11