# Kinematics of Euler Bernoulli and Timoshenko Beam Elements

 P: 660 Folks, Trying to get some appreciation for what is going on in the attached schematic of 1)Euler bernoulli and 2) Timoshenko beam elements. For the first one, ie the top picture, how was ##u- z \frac{dw}{dx}## arrived at? thanks Attached Thumbnails
 Engineering Sci Advisor HW Helper Thanks P: 7,177 dw/dx is the slope of the beam, which is assumed to be small. So dw/dx is also the angle the beam has rotated, in radians. The top picture (Euler beam theory) assumes that cross sections of the beam stay perpendicular to the neutral axis. So the angle between a cross section and the vertical is the same as the slope of the beam. The picture is (stupidly, IMHO) drawn with a "left handed" coordinate system (z and w positive downwards not upwards) which is where the minus signs come from. In the bottom picture (Timoshenko beam theory) plane sections of the beam do not stay perpendicular to the neutral axis, so there is an extra shear strain (measured by angle gamma) involved.
P: 660
 Quote by AlephZero dw/dx is the slope of the beam, which is assumed to be small. So dw/dx is also the angle the beam has rotated, in radians.
Ok

 Quote by AlephZero The top picture (Euler beam theory) assumes that cross sections of the beam stay perpendicular to the neutral axis. So the angle between a cross section and the vertical is the same as the slope of the beam.
I understand this.

 Quote by AlephZero The picture is (stupidly, IMHO) drawn with a "left handed" coordinate system (z and w positive downwards not upwards) which is where the minus signs come from.
Ok, how does the ##z\frac{dw}{dx}## come about? Is this equivalent to Z times the cos of the angle?

 Quote by AlephZero In the bottom picture (Timoshenko beam theory) plane sections of the beam do not stay perpendicular to the neutral axis, so there is an extra shear strain (measured by angle gamma) involved.
Thanks

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Kinematics of Euler Bernoulli and Timoshenko Beam Elements

 Quote by bugatti79 Ok, how does the ##z\frac{dw}{dx}## come about? Is this equivalent to Z times the cos of the angle?
dw/dx is the sine of the angle (sin θ = θ for small angles) but you are right about the basic idea.
 P: 660 What practical examples are there where one shouldn't use Euler-Bernouilli to track beam deflection etc. Would it for applications of plastic loading? Thanks
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 Quote by bugatti79 What practical examples are there where one shouldn't use Euler-Bernouilli to track beam deflection etc.
When the flexibility in shear is significant compared with the flexibility in pure bending.

For a rectangular section beam, Euler is OK when length/depth > 10 (some people say > 20).

For a more complicated criss sections, and/or composite beams made from several materials, you have to consider each case on its own merits.

With computer software like finite element analysis, you might as wel always use the Timoshenko formulation. Even if the correction is neglibile, it doesn't cause any numerical problems to include it.

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