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Why does a beam in shear like in this picture crack at 45 degrees instead of vertically? This is called "Punching shear", and requires concrete reinforcement at an angle:

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Assuming that the shear force $F$ acts vertically, and that the cross section area of the beam is $A$, the stress at 45 degrees is:

$$\sigma = \frac{F*cos(45)}{\frac{A}{cos(45)}}$$ $$\sigma = \frac{F}{A} * cos^2(45) < \frac{F}{A}$$

So the shear stress vertically should always be higher than at 45 degrees. So why is it normally assumed that the concrete cracks at 45 degrees?

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  • $\begingroup$ Typically that's where the stresses transition from tension to compression, making it the highest stressed area. $\endgroup$ – jko Mar 26 at 12:02
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In simple tension , the maximum resolved shear stress is 45 degrees to the tensile force . That is why a tensile test bar makes a "cup and cone" fracture face ; the cup edges are 45 degrees to the tensile force . This if for ductile materials that can deform in shear.

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Concrete crack is essentially caused by the diagonal tension stresses that acting normal to the crack plane. The tension stress is a maximum on a plane that is 45 degree to the analytical axes. You need to review the "Principal Stresses and Maximum Shear Stresses" in the engineer mechanics textbook.

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IMHO its a combination of many factors. First of all the loading.

The loading you are presenting on the top image will create bending load (i.e. normal loads) and shear loads. While the shear loads for a square cross-section can be though of as uniform, then normal loads (which are more significant and generate cracks) are greater away from the neutral axis. So you have a combination of stresses, and (in the majority of the cases) the normal stresses are more severe.

The other reason has to do with the type of failure. The Tresca criterion is one of the most basic failure theory suitable for ductile materials. It predicts that the material will fail, when the maximum shear stress exceeds a certain threshold. Because steels are ductile, the Tresca criterion is applicable. However the maximum shear stress is observed on the 45 deg plane, that is why most ductile materials exhibit necking prior to failure.

Even when you have reinforced concrete, the reincorcement is steel and its ductile. And although the concrete will fail in a brittle manner, the rebars will arrest the crack developement.

So, a crack will start to form at the location of the highest stress (near the vertical column because there the bending moment is greater), and it will progress in the direction of the 45o plane.

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If we try to imagine the path of tensile stress trajectories due to the moment they looks like the shape of a suspended bridge cable starting from the right top corner of the beam bending down passing near the bottom of the beam and bending back up to the left corner of the beam.

see the figure showing the trajectory of tension and compression in a beam. In the figure, there we have two small samples 1 and 2.

Sample one at the neutral axis carries pure shear and its principal stress angle is at 45 degrees.

$ \ \tau =V/A_n = \frac{P}{A}sin \theta cos \theta= \frac{P}{2A} sin2\theta $

which is max at 45 degrees.

The stresses in sample 2 are combined shear, v, and normal stress due to moment,f. and the tangent of the angle of principal stress is $tan \alpha = 2v/f, $ meaning the angle of the principal stress is inclined more sharply as shown in the figure.

This same pattern is the pattern of the shear crack.

The shear crack can and should be prevented by shear ties.

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trajectory of tension

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