Strength of Materials: Analysis of shearing stress


In this session we want to discuss about shearing
stresses in beams. Let’s do a quick review of what we have discussed about shear so far.
Shear stresses arise from shear forces that are parallel to the cross section of a body.
I want to show this with an example. I like French baguette. It is very tasty and let
me grab one. If I want a slice of this bread, I take a
knife, hold the bread with one hand and then I am going apply shear force along the vertical
plane using this hand. Take a look at the free body diagram of this scenario. The force
applied by my hand to hold the bread in place and the force applied by the knife are shown.
Notice the shear plane in between the forces. As I slice this bread, the shear force is
applied parallel to this cross section. This shear force pushes one part of the bread in
one direction and the other part of the bread in the opposite direction. This causes one
surface to slide with respect to other, and the body shears. This is exactly what happens when you use
a scissor to cut a piece of cardboard. In both cases, you are applying shear forces
parallel to the cross section creating shear stress which breaks the parts into two. Average shear stress is equal to shear force
over area. But you know the average shear stress equation is not accurate. For example, the shear stress at the top or
the bottom of the body is equal to zero. You know this because there is no cross sectional
area right above the top, or, the bottom. So there is no shear stress at these locations.
But the equation indicates the stress is uniform from top to bottom. In reality the shearing
stress changes from zero from the top, or bottom, to a maximum value at the neutral
axis. So, in this session we want to have a better
understanding of what happens when shear force is applied, and develop a method for calculating
the shear stress in a beam. As you know, transverse forces are loads that
are applied perpendicular to the axis of the beam. Beams are typically subjected to shear
and bending. Pure bending causes deflection without sliding. But transverse shear causes
sliding of surfaces not in just one plane but also in a perpendicular plane. In order
for you to visually this, I want to show a demonstration. I have a few balsa wood pieces with me. I
want to make sure their edges are fully aligned. I am going to hold them together, and apply
a transverse force. As you can see this causes bending but you can also see that the pieces
of balsa wood also slide relative to each other. Take a closer look at how the edges
are out of alignment. This means that the each balsa wood piece is experiencing shear
stress in the longitudinal direction. This shows that the transvers force I applied
not only creates shear stress along the vertical plane but also along the longitudinal plane.
While this experiment shows that there is sliding in horizontal, longitudinal direction,
you won’t see this physically in a homogenous beam made of steel.
Even though you don’t see physical sliding in steel, the tendency to slide exists between
the surfaces. So shearing stresses occur in vertical, transverse plane as we well as on
horizontal, longitudinal plane. However, in material such as wood, you can see the effect
of this shear in the longitudinal direction. This pictures shows wood splitting along the
longitudinal axis. So our goal today is to develop a method to
calculate the shear stress in beams when subjected to transverse loads. Let’s get started.

82 Replies to “Strength of Materials: Analysis of shearing stress”

  1. Thanks, love the "real" examples! I hate when everything is explained all super bio-mechanical(y), using lots of scientific words and such.

  2. really your teaching power by showing simple examle is great i love your lecture style plz continue …i had never seen like you i love it.

  3. amaxing  explanation . i have never seen this kind of spoon feed explantion ever before with general life examples . simply superb

  4. what a wonderful lecture it is ! thank you very much sir.Plz upload such conceptual video. It would be very helpful for us.

  5. Helped me understand the directions of bending and shear stresses. Very informative. Thank you so much 😊

  6. I have a dumb question, you said that at the edges of the bread there is zero shear because there is zero area. so are you saying that we get max area at the neutral axis? don't we only get half the area at the neutral axis and the max area at the bottom end of the bread?

  7. Hello sir, am trying to define the bounds for shear modulus of composite with n phase isotropic materials but I finding it difficult to define the conditions for the correct solution. For the correct solution, is the shear strain constant or the shear stress. And also at for incorrect solution, is the shear stress constant for all directions? Your response will help me greatly in my project

  8. Amazingly awesome sir..!!.this is what I was searching…for..finally I found here ..where there is emphasis on building concepts..
    👌👌✌ worth sharing..so shared..

  9. Excellent video! Some more examples, diagrams along with your verbal explanations could further help. Did not quite fully understand why the shear at the edges of the baguette are zero. I get a rough idea why shear is maximum at the center–if a large diameter rod is extended under a weight, the middle part extends the most–just as you depicted in the "parabolic profile", but does that mean the shear is greatest at the center?? Is it because the area element at the center is the smallest while the area element at the sides is the greatest? Thanks in anticipation

  10. Sir sry to say that I hv one doubt how we appy shear force we will apply external transverse load …due to external load in opposite shear force will developed due to that shear force 1st shear stress will developed in horizontal direction…due to horizontal shear stress a complimentary vertical shear stress will developed…is there any wrong plz reply me??

  11. I still think I’m missing something. If sheer stress is maximum at the NA then why when we cut through steel does the material fracture from the edges (where the shear stress is zero) instead of fracturing from the middle?

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