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In the example you are presenting the shear stresses on cross-section of beam AB are as in the image below. with $\color{green}{green}$ is the shear stress $\tau_v$ which has a single direction (parallel to the shear force) with $\color{red}{red}$ is the torsional stress $\tau_T$, which has a changing direction (see the following graph) if you notice at ...


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Strain is the ratio of change in length of a layer of fiber in the beam with respect to its original length. And it is a function of displacement (elongation/shortening) in direction of the longitudinal axis, x, at time t (for the changes, length and sign, are dynamic in nature).


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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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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 ...


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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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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. ...


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