10.4 Shear center

\[\begin{equation*} q\left(s\right)=q(s=0)-\frac{\left(\int_{s_{0}}^{s}{t\left(s\right) \,x\left(s\right)\;ds}\right)\,S_x}{{I_{yy}}} \end{equation*}\]

For different sections of our cross section, these have constant values

  • \[\begin{equation*}\small \begin{pmatrix}t\left(s_1\right)=t \\ x\left(s_1\right)=-\frac{b}{2} \\ q_1(s_1=0)=0 \\ \end{pmatrix} \end{equation*}\]
  • \[\begin{equation*}\small \begin{pmatrix}t\left(s_2\right)=t \\ x\left(s_2\right)=s_2-\frac{b}{2} \\ q_2(s_2=0)=q_1\left(s_1=h\right) \\ \end{pmatrix} \end{equation*}\]
  • \[\begin{equation*}\small \begin{pmatrix}t\left(s_3\right)=t \\ x\left(s_3\right)=\frac{b}{2} \\ q_3(s_3=0)=q_2\left(s_2=b\right) \\ \end{pmatrix} \end{equation*}\]

We can perform the integrations to obtain the shear flows: \[\begin{equation*} q_1\left(s_1\right)= \frac{S_x}{{I_{yy}}} \left[ \frac{b\,s_1\,t}{2} \right] \end{equation*}\]

\[\begin{equation*} q_2\left(s_2\right)= \frac{S_x}{{I_{yy}}} \left[ \frac{b\,h\,t}{2}-\frac{ \left(s_2^2-b\,s_2\right)\,t}{2} \right] \end{equation*}\]

\[\begin{equation*} q_3\left(s_3\right)= \frac{S_x}{{I_{yy}}} \left[ \frac{b\,h\,t}{2}-\frac{b \,s_3\,t}{2} \right] \end{equation*}\]

We can perform the integrations again to obtain the resultant forces: \[\begin{equation*} F=\int_{s_0}^{s_1}{q\left(s \right)\;ds} \end{equation*}\]

For sub-section 1: \[\begin{equation*} F_1=\int_{s_1=0}^{s_1=h}{q_1\left(s_1 \right)\;ds_1} \end{equation*}\]

\[\begin{equation*} F_1= \frac{S_x}{{I_{yy}}} \left. \frac{b\,\left(\frac{s_1^2}{2} \right)\,t}{2} \right|_0^h \end{equation*}\]

\[\begin{equation*} F_1=\frac{S_x}{{I_{yy}}} \left[\frac{b\,h^2\,t}{4}\right] \end{equation*}\]

For sub-section 2: \[\begin{equation*} F_2=\int_{s_2=0}^{s_2=b}{q_2\left(s_2 \right)\;ds_2} \end{equation*}\]

\[\begin{equation*} F_2= \left. \frac{S_x}{{I_{yy}}} \frac{\left(-2\,s_2^3+3\,b\,s_2^2-6\,b\,h\,s_2\right)\,t}{12} \right|_0^b \end{equation*}\]

\[\begin{equation*} F_2=\frac{S_x}{{I_{yy}}} \frac{\left(6\,b^2\,h+b^3\right)\,t}{12} \end{equation*}\]

For sub-section 3: \[\begin{equation*} F_3=\int_{s_3=0}^{s_3=h}{q_3\left(s_3 \right)\;ds_3} \end{equation*}\]

\[\begin{equation*} F_3= \frac{S_x}{{I_{yy}}} \left. \frac{\left(b\,s_3^2-2\,b\,h\, s_3\right)\,t}{4} \right|_0^h \end{equation*}\]

\[\begin{equation*} F_3=\frac{S_x}{{I_{yy}}} \frac{b\,h^2\,t}{4} \end{equation*}\]

The moment of inertia about the \(y\) axis is (by inspection using parallel axis theorem): \[\begin{equation*} {I_{yy}}=\frac{t\,b^3}{12}+2\,h\,t\,\left(\frac{b}{2}\right)^2 + 2 \frac{h t^3}{12} \end{equation*}\]

\[\begin{equation*} {I_{yy}}=\frac{t\,b^3}{12}+2\,h\,t\,\left(\frac{b}{2}\right)^2 + 0 \end{equation*}\]

Now: \[\begin{equation*} F_1=S_x \frac{3\,h^2}{6\,b\,h+b^2} \end{equation*}\]

\[\begin{equation*} F_2=S_x \end{equation*}\]

\[\begin{equation*} F_3=S_x \frac{3\,h^2}{6\,b\,h+b^2} \end{equation*}\]

The resultant forces must be the static equivalent of the shear forces applied through the shear center.

  • Shear Center
    • The point at which shear loads must be applied in order to no have twisting of the beam to occur
      • Location depends on geometry, not loads
      • Always located on a line of symmetry

Moments about the shear center are zero, shear center assumed to be below the channel.

\[\begin{equation*} \frac{b\,F_1}{2}-F_2\,e_y+\frac{F_3\,b}{2}=0 \end{equation*}\]

\[\begin{equation*} e_y = \frac{3\,h^2}{6\,h+b} \end{equation*}\]

Moments about the shear center are zero, shear center assumed to be above the channel \[\begin{equation*} F_2\,e_y+\frac{F_3 \,b}{2}+\frac{F_1\,b}{2}=0 \end{equation*}\]

\[\begin{equation*} e_y=-\frac{3\,h^2}{6\,h+b} \end{equation*}\]

Moments about corner A is equivalent to resultant shear moment, shear center is assumed below the channel \[\begin{equation*} F_1\,b=e_y\,S_x \end{equation*}\]

\[\begin{equation*} e_y=\frac{3\,h^2}{6\,h+b} \end{equation*}\]

Moments about corner A is equivalent to resultant shear moment, shear center is assumed above the channel \[\begin{equation*} F_1\,b=-e_y\,S_x \end{equation*}\]

\[\begin{equation*} e_y=-\frac{3\,h^2}{6\,h+b} \end{equation*}\]

10.4.1 Steps to take to find the shear center

  1. Assume a shear load is applied at the shear center
    • If symmetry, assume the load lies on the line of symmetry
    • Assume a convenient location, it is often easiest to assume the load is outside the geometry.
  2. Compute the shears flows and resultant shear forces
  3. Sum the moments about a convenient point.
    • If the point is the shear center, the moments need to sum to zero.
    • If the point is off the shear center, the moments must sum to the same moment as the resultant force would impart.

Remember: the moment from the shear flows must be the same as that of the resultant shear vector that is assumed. The computed and assumed resultants are statically equivalent! If they are not, you have made an error.