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Understanding Stresses in Beams — Transcript · YTtoTranscript

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Understanding Stresses in Beams

The Efficient Engineer · 2,230 words · 11 min read · EN-ORIG

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Below is the complete, readable transcript of Understanding Stresses in Beams by The Efficient Engineer on YouTube. Read the full text, copy any part you need, or generate a transcript for any video with our free tool.

if we apply a load to a beam it will deform by bending this generates internal stresses which can be represented by a shear force acting in the vertical direction and a bending moment the shear force is the resultant of vertical shear stresses which act parallel to the cross section and the bending moment is the resultant of

normal stresses called bending stresses which act perpendicular to the cross section it's important to have a good understanding of these stresses because any design or analysis of a beam will involve calculating them let's look at bending stresses first to keep things simple we'll consider a case of pure bending a section of a beam is said to

be in a state of pure bending when the shear force along it is equal to zero and so there is a constant bending moment along its length like there is for this beam loaded by two moments we also have a case of pure bending over the middle section of this beam where the bending moment is constant let's

look at how a beam deflects when it has a constant bending moment along its length if we imagine the beam as a collection of very small fibers as the beam deflects the fibers at the top of the beam gets shorter meaning that they are in compression and those at the bottom of the beam get longer so they

are in tension somewhere between the top and the bottom of the cross section there will be a surface containing fibers which stay the exact same length this is called the neutral surface it passes through the centroid of the cross section when looking at the beam in two dimensions we refer to it as the neutral

axis let's try and quantify the bending stresses that develop within the beam to resist these applied moments first let's calculate the strains in the beam this can be done quite easily just by considering the geometry of the deformation let's watch how a fiber at the neutral axis between points a and B and a fiber between Point C and D

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located at a distance Y from neutral axis deform since this is a case of pure bending we can see if the fibers Bend into a perfectly circular arc we'll call the center of the circle Oh before any deformation the fibers are all the same length after the deformation the length of the neutral

axis has stayed the same but the length of the fiber between points C and D has increased if theta is the angle of the arc and R is the radius of the arc to the neutral axis we can calculate the length of the arc between a and B like this and we can calculate the length of

the arc between C and D in the same way strain is defined as the change in length divided by the original length and so we can derive an equation for bending strain at any distance Y from the neutral axis

we define the distance Y as being positive downwards and so this equation will give us a positive strain for the bottom of the cross section which is in tension sometimes you'll see this equation written with a minus sign but that's because Y was defined as being positive upwards if we assume that stresses remain within the elastic

region of the stress-strain curve we can then apply Hookes law for uniaxial stress to calculate the bending stresses this gives us the equation for bending stress as a function of the radius of curvature R of the deformation but what we're really interested in is how the bending moment M affects the bending stress if we make an imaginary cut

through the beam we can expose the internal bending stresses represented here as a few discreet forces the resultant moment of these internal forces must be equal to the bending moment m and so we can calculate M by integration like this now we can plug in the equation for bending stress we just derived

when rearranged into this form we can notice that the integral on the right is the definition of the area moment of inertia this parameter which I've covered in detail in a separate video defines the resistance of a cross-section to bending due to its shape and is denoted using a letter I we can combine this equation for the

bending moment with the bending stress equation to obtain what is known as the flexure formula so what does it tell us bending stress increases linearly as the bending moment and the distance from the neutral axis increase and it decreases as the area moment of inertia increases the maximum stress occurs at the fibers

furthest from the neutral axis the term I over Y max depends only on the geometry of the cross-section and so it is called the section modulus and is denoted using the letter S you will often see the section modulus listed for a range of common beam cross sections and reference texts the i-beam

is a commonly used cross section because it has a large area moment of inertia which results in lower stresses here's how the bending stresses are distributed over an i-beam cross section they are zero at the neutral axis and reach a maximum and the outside surfaces of the flanges for a tee section the neutral

axis is shifted upwards and so the bending stress distribution looks like this so we've established how to calculate the bending stresses which are normal stresses for a case of pure bending most of the time we won't have pure bending is there will also be a shear force acting on the beam cross section like the beam we saw at the

start of the video it turns out that the presence of a shear force doesn't normally significantly affect the bending stresses and so luckily we can consider the flexure formula we derived earlier for pure bending to be valid for a more general case of bending the shear force V is the resultant of shear

stresses which act vertically parallel to the cross section we denote the shear stresses using the Greek letter tau to maintain equilibrium these vertical shear stresses have complimentary horizontal shear stresses which act between horizontal layers of the beam one way to visualize these horizontal stresses is to consider a beam made up of several planks of wood when a load is

applied there is a tendency for the planks to slide relative to one another now let's glue the planks together [Music] when the load is applied the planks cannot slide and so horizontal stresses develop between them if these shear stresses are larger than the shear strength of the glue bond the glue will fail these horizontal shear

stresses don't exist if we apply a moment instead of a force because that gives us a state of pure bending and so there is no tendency for the planks to slide relative to one another the presence of these horizontal shear stresses explains why wooden beams sometimes fail by splitting longitudinally this failure usually

occurs close to the neutral axis for reasons which will soon be obvious so how can we calculate the shear stresses we can calculate the average shear stress acting on the cross section as the shear force V divided by the cross-sectional area but the shear stresses aren't distributed uniformly across the beam cross section the shear

stress has to be zero at the free surfaces at the top and bottom of the beam so the average shear stress isn't very useful since it doesn't tell us the maximum shear stress instead we can use this equation to calculate the shear stresses acting on the cross section I won't cover the derivation of the

equation here but it's based on considering equilibrium of stresses acting on a small element within the beam the equation assumes that the shear stress is constant across the width B of the cross section so tau is a function of the distance along the beam X and the distance above the neutral axis Y V is

the shear force acting on the cross section which varies with the distance along the beam B is the width of the cross section it can vary with the distance Y from the neutral axis but in this case the cross section is rectangular so B is constant I is the area moment of inertia which is a

constant value calculated based on the shape of the cross section and Q is the first moment of area for the portion of the cross section of the location we want to calculate the shear stress for so it varies with the distance y above or below the neutral axis it is equal to the product of the

area above the location of interest in the distance between the centroid of that area and the neutral axis if the location of interest is below the neutral axis we consider the area below the axis instead of the area above it to calculate the first moment of area at this line which is at a distance Y from

the neutral axis we multiply the area of the blue rectangle above the line by the distance from the neutral axis to the centroid doing this calculation gives us an equation for Q as a function of the distance Y from the neutral axis for a rectangular cross-section and so we can obtain an equation which describes how

the shear stress varies with distance from the neutral axis the y term is squared and so the shear stress varies parabolically over the height of the cross-section with the maximum shear stress occurring at the neutral axis this is opposite to the bending stress which is zero at the neutral axis and explains why the horizontal shear

failure of a wooden beam we saw earlier occurs close to the neutral axis by setting Y to 0 in this equation we obtain an equation for the maximum shear stress in rectangular cross-sections [Music] it is equal to 1.5 times the average shear stress across the entire cross-section the derivation of this equation for shear stress makes a few

assumptions so we need to be careful with how we apply it first it assumes that the shear stresses are constant across the width of the cross section for rectangular cross sections this is a reasonable assumption if the rectangle is thin but for cross sections like this one the shear stresses can vary significantly over the

width and so in these cases the equation can really only give us the average shear stress across the width of the cross section it can't tell us what the maximum shear stress will be another assumption this equation makes is that the shear stresses are aligned with the y-axis shear stresses act and gently at

a free surface so for a circular cross section for example we can't strictly use this equation to get the distribution of shear stresses across the height of the cross section but we can still use it to estimate the shear stresses at the neutral axis because the shear stresses there are aligned with the y axis the equation for shear stress

at the neutral axis in a circular cross section is similar to the equation for a rectangular section where we have the average shear stress v over a multiplied by a constant the constant is 4 over 3 for a circular cross section and 3 over 2 for a rectangular one we can also use

the shear stress equation for thin walled sections like this I beam although things are a bit more complicated because the vertical shear stresses at the surface is shown in red must be 0 and because the flanges are very wide the vertical shear stress and the flanges is very small this means that the vertical shear stress is

distributed like this the web mostly carries the shear force and the flanges mostly carry the bending moment as we saw earlier you can see that the shear stresses are distributed quite evenly over the height of the web this is because the flanges contribute significantly to the first moment of area cue when calculating the shear

stresses in the web but they don't carry much of the vertical shear force since the web is thin the shear stresses are also distributed evenly across its width because of this we can easily calculate the approximate shear stress in the web like this more detailed analysis reveals that there are shear stresses in the

flanges but they are acting mainly in the horizontal direction the horizontal stresses on both sides of the flanges cancel each other out so the net shear force is still just a vertical force we can figure out the direction of the horizontal shear stresses based on the direction of the vertical shear stresses by imagining that the stresses are

flowing through the cross section

that's it for this review of bending and shear stresses and beams if you enjoyed the video and would like to see more content like this please consider supporting the channel on patreon thanks for watching