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    Transverse Shear

    Transverse loading creates corresponding longitudinal shear stresses which will act along longitudinal planes of the beam

    Pure Transverse Shear Tensor

    For pure transverse shear loading conditions, the stress tensor can be simplified as shown below, depending on whether shearing occurs along the y or z axis.

    Transverse shear stress tensor. #ts-tsr
    $$ \begin{align} T= \begin{bmatrix} 0 & \frac{V_yQ_y}{I_yt} & 0 \\ \frac{V_yQ_y}{I_yt} & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ T = \begin{bmatrix} 0 & 0 & \frac{V_zQ_z}{I_zt} \\ 0 & 0 & 0 \\ \frac{V_zQ_z}{I_zt} & 0 & 0 \end{bmatrix} \ \end{align} $$

    Shear Stress in Beams

    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 (\( L>5d \)).
    Transverse loading applied to a beam results in normal and shearing stresses in transverse sections, as well as a bending moment. The distribution of normal and shearing stresses satisfies the following where \( dA \) is the cross sectional area.
    Equilibrium.
    $$ \begin{align*} &F_x = \int\sigma_x dA = 0 &M_x &= \int(y\tau_{xz} - z\tau_{xy})dA=0 \\ &F_y = \int\tau_{xy} dA = -V &M_y &= \int z\sigma_{x} dA = 0 \\ &F_z = \int\tau_{xz} dA = 0 &M_z &= \int(-y\sigma_{x})dA = M \\ \end{align*} $$
    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.

    Symmetry.
    $$ \tau_{xy} = \tau_{yx}\ $$

    Average Shear Stress

    Shear stress varies throughout the cross-section of the beam where the maximum shear stress is at the neutral axis and is zero at the edges. Shear stress can be represented as an average across the beam.
    Avg. shear stress.
    $$ \tau = \frac{V}{A}\ $$
    Shear stress can also be evaluated at any given point (\( y_1 \)) from the z-axis.
    shear stress at a point. #shr-pts
    $$ \tau = \frac{VQ}{It}\ $$
    (A) An element of width \( dx \) in a bending beam has a FBD of the bending moment stress distribution.

    (B) An element distance \( y \) away from the neutral axis has a FBD of the bending moment stress distribution For this element to be in equilibrium, \( \tau_{xy} \) must be present.

    $$ \sigma_{x_1} \neq \sigma_{x_2}\ $$
    (C) Simplified FBD. The width of this element is \( t(y) \)
    $$ F_x = \int\sigma_x dA = 0\ $$
    $$ \Sigma F_x = F_{x_1} - F_{x_2} + \Delta H\ $$
    $$ \int_{y}^{y_{top}} \sigma_{x_1} t(y) dy - \int_{y}^{y_{top}} \sigma_{x_2} t(y) dy + \tau_{xy} t(y) dx = 0\ $$
    $$ \int_{y}^{y_{top}} \frac{M(x)y}{I} t(y) dy - \int_{y}^{y_{top}} \frac{(M(x)+dM(x))y}{I} t(y) dy + \tau_{xy} t(y) dx = 0\ $$
    $$ \tau_{xy} = \frac{dM(x)}{dx}\frac{1}{I t(y)} \int_{y}^{y_{top}} y t(y) dy\ $$
    $$ V(x) = \frac{dM(x)}{dx}\ $$
    $$ \tau_{xy} = \frac{V(x)}{I t(y)} \int_{y}^{y_{top}} y t(y) dy\ $$
    $$ \tau_{xy} = \frac{V(x) Q(y)}{I t(y)}\ $$
    • \( V \) : shear force
    • \( Q \) : First area moment of inertia of cut section ( \( Q = \Sigma \bar{y}'A' \) )
    • \( I \) : Second area moment of inertia of whole cross-sectional area
    • \( t \) : beam thickness

    Rectangular Beam Cross-Section

    Shear stress at a given point (y) in a rectangular beam. #shr-rct
    $$ \begin{align*} &\tau_{cut} = \frac{3}{2} \frac{V}{bh}(1-\frac{y^2}{c^2}) \\ \text{where} \\ &c = \frac{1}{2}h \\ \text{or} \\ &\tau_{cut} = \frac{6V}{bh^3}(\frac{h^2}{4}-y^2) \end{align*} $$
    $$ I_z = \frac{1}{12} bh^3\ $$
    $$ t=b\ $$
    $$ \begin{aligned}Q &= A\bar{y}\\ &= b(\frac{1}{2}h-y)[y+\frac{1}{2}(\frac{1}{2}h-y)]\\ &= \frac{1}{2}b(\frac{1}{4}h^2-y^2)\end{aligned} $$
    Maximum shear stress occurs at the neutral axis.
    Max. shear stress in rectangular beam.
    $$ \tau_{max} = 1.5 \frac{V}{A}\ $$

    I-Shaped Beam Cross-Section

    The neutral axis of an I-beam is the center. To find the I, use the parallel axis theorem. When determining the stress distribution, break the beam up into sub-sections of constant thickness for easier calculations. The stress distribution is a discontinuous parabola.

    Built-up Members/Beams: Shear Flow

    In some cases, it is more useful to look at shear flow (\( q \)) through a structure than internal shear stress.
    L: Shear flow direction. R: Shear flow magnitude
    For an I-beam, the shear flow increases symmetrically on top from the end to the neutral axis in the y-direction, and on the side increases from the flanges to the neutral axes. Shear flow is the a measure of force per length.
    Shear flow (q). #shr-flw
    $$ \begin{align*} &dF = \frac{VQ}{I}dx \\ \text{where} \\ &q = \frac{dF}{dx} = \frac{VQ}{I} \\ \end{align*} $$
    $$ F_x = \int\sigma_x dA = 0\ $$
    $$ \Sigma F_x = F_{x_1} - F_{x_2} + \Delta H\ $$
    $$ \Delta H + \int \frac{M_{x_1} y}{I}dA - \int \frac{M_{x_2} y}{I}dA = 0\ $$
    $$ \Delta H = \frac{M_{x_1}-M_{x_2}}{I} \int ydA = 0\ $$
    $$ \Delta H = \frac{\Delta M}{I} Q = 0\ $$
    $$ \frac{\Delta H}{\Delta x} = \frac{\Delta M}{\Delta x} \frac{Q}{I} = 0\ $$
    $$ \lim_{x\to 0} \frac{\Delta H}{\Delta x} = \frac{dH}{dx} = q\ $$
    $$ q = \frac{dM}{dx} \frac{Q}{I}\ $$
    Shear stress using shear flow.
    $$ \tau = \frac{q}{t} $$
    Shear flow is most applicable in design situations such as built-up beams. Built-up beams are beams that have been put together with either an adhesive (glue, epoxy, etc) or fasteners (nails, bolts, etc). Calculating shear flow allows for determining how much adhesive or how often a fastener is needed.
    For uniformly spaced fasteners:
    $$ F = q\Delta s $$

    Adhesives supply resistance along the length of the contacting parts. Determine the minimum shear strength at these contacting/weak points using the shear stress equation.

    Example Problem: The following I-beam is composed of 4 boards glued together #blt-adh

    The vertical shear is is \( V = 100 \text{ } N \). Find the minumum required shear strength for the glue (\( \tau_{glue} \)).

    $$ \begin{align*} &\tau_{AB} = \frac{V Q_A}{I t_A},\ &\tau_{BD} = \frac{V Q_D}{I t_D}\ \end{align*} $$
    \( V \) : shear force
    $$ V = 100 \text{ } N $$
    \( Q \) : First area moment of inertia of cut section ( \( Q = \Sigma \bar{y}'A' \) )
    $$ Q_A = A \bar{y_A} = 3.13*10^4 \text{ } mm^3 $$
    $$ Q_D = A \bar{y_D} = 9.38*10^4 \text{ } mm^3 $$
    \( I \) : Second area moment of inertia of whole cross-sectional area
    $$ I = I_1 + 2 I_2 = 1.04*10^7 \text{ } mm^4 $$
    $$ I_1 = \frac{1}{12}b_1 h_1^3 = 8.79*10^5 \text{ } mm^4 $$
    $$ I_2 = \frac{1}{12}b_2 h_2^3 +A_2d_2^2 = 47.85*10^5 \text{ } mm^4 $$
    \( t \) : length of connected region
    $$ t_A = 25 \text{ } mm $$
    $$ t_D = 25 \text{ } mm $$
    Solve.
    $$ \tau_{AB} = 12.0 \text{ } kPa $$
    $$ \tau_{BD} = 36.1 \text{ } kPa $$
    $$ \begin{align*} &\tau_{AB} < \tau_{BD} \ &\tau_{glue} = 36.1 \text{ } kPa\ \end{align*} $$

    Fasteners supply resistance at fixed intervals. Use the shear flow formula to analyze the beam.

    Example Problem: The following beam is composed of 2 boards connected by 6 nails. #blt-fst

    The nails are at an interval of of \( \Delta s = 50 \text{ } mm \). The shear force is \( V = 100 \text{ } kN \). Find the shear flow (\( q \)) and the load at each fastener (\( f_n \)).

    $$ \begin{align*} &q = \frac{VQ}{I},\ &f_n = \frac{q*s_{total}}{4}\ \end{align*} $$
    Shear flow.
    $$ Q = (b*h)\bar{y} = (300*50)*25 = 3.75*10^5 \text{ } mm^3 $$
    $$ I = \frac{1}{12}b h^3 = \frac{1}{12}*300*100^3 = 2.5*10^7 \text{ } mm^4 $$
    $$ q = 1.5 \text{ } \frac{kN}{mm} $$
    Force in the nail.

    Since the nails are uniformally distributed, each nail is assumed to supply a shearing force \( f_n \) that resists the shear flow in the beam. Along a single row of nails (refer to the free-body diagram above), the total force is:

    $$ \Sigma F_{total} = \frac{1}{2}f_n + f_n + \frac{1}{2}f_n = 2f_n $$

    Because there are two rows of nails, the total resistance load supplied by the nails is:

    $$ \Sigma F_{total} = 2*F_{total} = 2*2f_n = 4f_n $$

    To find the shearing force \( f_n \) in each nail, we equate the total force supplied by the nails with the total shear flow acting across their total spacing, \( q*s_{total} = 2q\Delta s \), to obtain:

    $$ s_{total} = 2\Delta s = 100 \text{ } mm^4 $$
    $$ f_n = 37.5 \text{ } kN $$

    Heads up!

    shear stresses in thin-walled members builds on this content.

    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 symmetrically. From the edge to where the beam becomes hollow, the flow increases linearly and from where it becomes hollow to the neutral axis, the flow increases parabolically.
    For a u-channel beam, the shear flow increases linearly on the sides from the edge to the base, and on the base it increases from the sides to the neutral axis.