| Shear Stresses in Beams |
| Rectangular Beams |
| Built-up Members/Beams |
| Shear Stresses in Thin-Walled Members |
Transverse Shear
Shear stress in beams
Shear stress due to transverse loading creates corresponding longitudinal shear stresses which will act along longitudinal planes of the beam
When a shear force is applied, it tends to cause warping of the cross section. Therefore, when a beam is subject to moments and shear forces, the cross section will not remain plane as assumed in the derivation of the flexural formula. However, we can assume that the warping due to shear is small enough to be neglected, which is particularly true for slender beams.
Transverse loading applied to a beam results in normal and shearing stresses in transverse sections
Distribution of normal and shearing stresses satisfies
\[ \begin{align*} &F_x = \int\sigma_x dA = 0 &M_x &= \int(y\tau_{xz} - z\tau_{xy})dA=0 \\ &F_y = \int\tau_{xy} dA = -V &M_y &= \int z\sigma_{x} dA = 0 \\ &F_z = \int\tau_{xz} dA = 0 &M_z &= \int(-y\sigma_{x})dA = M \\ \end{align*} \]
When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces
Longitudinal shearing stresses must exist in any member subjected to transverse loading.
Symmetry of stress: transverse xy stress implies longitudinal yx stress
Rectangular beam cross section: Shear Stresses
Anisotropic material: wood fails at locations of maximum longitudinal stress
Built-up Members/Beams: Shear Flow
\[dF = \frac{VQ}{I}dx\]
Where shear flow is: $$q = \frac{dF}{dx} = \frac{VQ}{I} $$
Shear Stresses in Thin-Walled Members
The variation of shear flow across the section depends only on the variation of the first moment.
\[q = \tau t = \frac{VQ}{I}\]
For a box beam, q grows smoothly from zero at A to a maximum at C and C’ and then decreases back to zero at E.
For a wide-flange beam, the shear flow increases symmetrically from zero at A and A’, reaches a maximum at C and then decreases to zero at E and E’.
