Observations of beam physical deflection

• Bending deflection

• Shear deflection

Definition of a beam

• Assumptions that underlie beam theory
  • A beam is much longer than its characteristic cross section
  • A beam carries:
    • Axial forces (P)
    • Shear forces (S)
    • Distributed loads (p)
    • Torques (T)
    • Bending moments (M)
  • The reference line for a beam is the line of the centroids
  • The reference line is often assumed to be straight

Kinematics of an Euler-Bernoulli beam

Assumed Euler-Bernoulli beam bending kinematics

• An Euler-Bernoulli beam also assumes:
  • \[w(x,y,z) = w_0(z) + \alpha(z) x + \beta(z) y\]
  • Plane sections remain plane under load
  • Plane sections remain perpendicular to the reference line
• Therefore, the axial deformation is a linear function of the position in the plane of the cross section

Therefore: \[w(x,y,z) = w_0(z) - x {u {}_{,z}} - y {v {}_{,z}}\]

Now, we can write strain:

\[\begin{align} {\varepsilon_{zz}}=& \, {\frac{\partial w}{\partial z}} \\ {\gamma_{zy}}=& \, {\frac{\partial v}{\partial z}} + {\frac{\partial w}{\partial y}} = {\frac{\partial v_0}{\partial z}} + \beta = 0\\ {\gamma_{zx}}=& \, {\frac{\partial u}{\partial z}} + {\frac{\partial w}{\partial x}} = {\frac{\partial u_0}{\partial z}} + \alpha = 0\\ {\varepsilon_{zz}}=& \, {w_0 {}_{,z}} - x {\alpha {}_{,z}} - y {\beta {}_{,z}} \\ \end{align}\]

• When the assumptions of an Euler-Bernoulli beam hold, shear deformation is assumed negligible: \[\begin{align} {\gamma_{zy}}=& \, 0 = {\frac{\partial v_0}{\partial z}} + \beta \\ {\gamma_{zx}}=& \, 0 = {\frac{\partial u_0}{\partial z}} + \alpha \\ \end{align}\]

• thus: \[\begin{align} \alpha(z) =& \, - {u {}_{,z}} \\ \beta(z) =& \, - {v {}_{,z}} \\ \end{align}\] \[\begin{align} {\varepsilon_{zz}}=& \, {w_0 {}_{,z}} - x {\alpha {}_{,z}} - y {\beta {}_{,z}} \\ \end{align}\]

Strain in a beam

\[\begin{align} {\varepsilon_{zz}}=& \, {w_0 {}_{,z}} - x {u {}_{,zz}} - y {v {}_{,zz}}\\ {\varepsilon_{zz}}=& \, {\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}} \end{align}\]

• The symbol \({\varepsilon_{zz 0}}\) is the axial strain of the reference line.
• The symbols \({u {}_{,zz}}\) and \({v {}_{,zz}}\) are the curvatures of the reference line in the \(x-z\) and \(y-z\) planes.

Stress in a beam

The stress can now be written: \[\begin{align} {\sigma_{zz}}=& \, E {\varepsilon_{zz}}\\ {\sigma_{zz}}=& \, E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \end{align}\]

Beam loads

Axial load in a beam

• Therefore: \[\begin{align} P =& \, E \int_A \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \, dA\\ =& \, E \left({\varepsilon_{zz 0}}\int_A \, dA - {u {}_{,zz}} \int_A x \, dA - {v {}_{,zz}} \int_A y \, dA \right) \end{align}\]

Definitions

• Lets define some terms: \[\begin{align} \int_A \, dA =& \, A \\ \int_A x \, dA =& \, x_c A \\ \int_A y \, dA =& \, y_c A \\ \end{align}\]

• Now we have: \[\begin{align} P =& \, E \left(A {\varepsilon_{zz 0}}- x_c A {u {}_{,zz}} - y_c A {v {}_{,zz}} \right) \\ \end{align}\]

Choice of coordinate system

• We choose our coordinate system so that: \[\begin{align} x_c = 0 \\ y_c = 0 \\ \end{align}\]

• Which leads to: \[\begin{align} P =& \, E A \, {\varepsilon_{zz 0}}\\ \end{align}\]

• This means that the axial force in the beam is due to the elongation of the centroidal axis and is not related to the bending of the beam.

Bending moments about \(x\)

\[\begin{align} {\sigma_{zz}}=& \, E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \end{align}\]

• The resultant bending moment about the centroid is: \[M_x = \int_A y \cdot {\sigma_{zz}}\, dA\]

• \[M_x = E \left( {\varepsilon_{zz 0}}\cdot \int_A y \, dA - {u {}_{,zz}} \cdot \int_A x y \, dA - {v {}_{,zz}} \cdot \int_A y^2 \, dA\right)\]

• Recall:

• \(\int_A y \, dA = 0\) : By choice of coordinate system
• \(\int_A x y \, dA = {I_{xy}}\) : Area product of inertia
• \(\int_A y^2 \, dA = {I_{xx}}\) : Area moment of inertia about \(x\)-axis
• Note: \({I_{xy}}= 0\) if there is symmetry about the \(x\) or \(y\) axis. (If there is symmetry about another axis, the computation can be transformed to be about the principal axis of the cross section.)
• If \(E=E(x,y)\), then it must remain inside the integral and we do not have a clean separation between material and cross-section properties. (See ME6520 Composite Materials)

Bending moments about \(y\)

\[\begin{align} {\sigma_{zz}}=& E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \end{align}\]

• The resultant bending moment about the centroid is: \[M_y = - \int_A x \cdot {\sigma_{zz}}\, dA\]

\[M_y = - E \left( {\varepsilon_{zz 0}}\cdot \int_A x \, dA - {u {}_{,zz}} \cdot \int_A x^2 \, dA - {v {}_{,zz}} \cdot \int_A x y \, dA\right)\]

• Recall:

• \(\int_A x \, dA = 0\) : By choice of coordinate system
• \(\int_A x y \, dA = {I_{xy}}\) : Area product of inertia
• \(\int_A x^2 \, dA = {I_{yy}}\) : Area moment of inertia about \(y\)-axis
• Note: \({I_{xy}}= 0\) if there is symmetry about the \(x\) or \(y\) axis (If there is symmetry about another axis, the computation can be transformed to be about the principal axis of the cross section.)

\[\begin{align} M_x =& \, - E{I_{xy}}{u {}_{,zz}} - E{I_{xx}}{v {}_{,zz}} \\ M_y =& \, + E{I_{yy}}{u {}_{,zz}} + E{I_{xy}}{v {}_{,zz}} \\ \frac{P}{A} =& \, E {\varepsilon_{zz 0}}= {\sigma_{zz}}^{\mathrm{ave}} \\ \end{align}\]

• These equations are uncoupled due to the centroidal \({z}\)-axis
• Note also we can invert and solve

Differential equations of static equilibrium

Axial static equilbrium

• In the axial direction (\(z\)):
  • \(P = P(z)\) – axial force resultant
  • \(p_z = p_z(z)\) – distributed axial load [force/length]
    • e.g., \(p_z = \rho A g_z\)
    • Note \(\rho g_z\) is related to the stress equilibrium equations (body force per unit volume) as \(b_z = \rho g_z\)
  • \(\displaystyle\sum F_z = 0 = P(z + dz) - P(z) + p_z dz\)

Hence:

• \(0 = \frac{P(z + dz) - P(z)}{dz} + p_z\)
• \(\frac{d P}{d z} = - p_z(z)\)

• For \(P = {\sigma_{zz}}\cdot A = E A \, {\varepsilon_{zz 0}}= E A \, {{w}_0 {}_{,z}}\)
• \({\frac{d P(z)}{d z}} = {\frac{d }{d z}} \left[ E A {\frac{d w_0}{d z}} \right] = - p_z(z)\)
• For constant \(EA\)
  • \(E A \, {w_0 {}_{,zz}} = - p_z(z)\)
• Note, when \(p_z = 0\), the axial force does not vary (for small deflections)

Bending static equilibrium

Beam bending: In the lateral directions (\(x\) & \(y\)):

• (\(S_x\) ; \(S_y\)) - shear force resultants
• (\(p_x\) ; \(p_y\)) - distributed lateral loads [force/length]
• (\(M_x\) ; \(M_y\)) - bending moment resultants

• Summing forces, we find: \[\displaystyle \sum F_x = S_x(z+dz) -S_x(z) + p_x dz = 0\] \[{\frac{d S_x(z)}{d z}} = -p_x\]

• Using similar arguments, we can find: \[{\frac{d S_y(z)}{d z}} = -p_y\]

• Now use moment equilibrium and including bending moments:

• Consider first the counterclockwise moments about \(y\) axis (using the the center as reference point): \[\begin{align} \left(M_y(z+dz) - M_y(z)\right)& \\ + \left(S_x(z+dz) +S_x(z)\right)& \frac{d z}{2} = \; 0\\ & d M_y = -(2 S_x +d S_x)\frac{d z}{2} \end{align}\]

• In the limit as \(d z\) goes to zero. \[\begin{align} \frac{dM_y}{dz} =& -S_x \\ \end{align}\]

• The same can be applied to moments about the \(x\) axis:

• Note carefully the difference in the convention for direction of the moments. \[\begin{align} \frac{dM_x}{dz} =& +S_y \\ \end{align}\]

• In summary, equilibrium arguments require: \[\begin{align} {\frac{d S_x(z)}{d z}} =& -p_x \\ {\frac{d S_y(z)}{d z}} =& -p_y \\ \frac{d M_x}{dz} =& +S_y \\ \frac{d M_y}{dz} =& -S_x \\ \end{align}\]

• Combining these leads to: \[\begin{align} \frac{d^2 M_x}{dz^2} =& -p_y \\ \frac{d^2 M_y}{dz^2} =& +p_x \\ \end{align}\]

• Recall also: \[\begin{align} M_x =& \, - E{I_{xy}}{u {}_{,zz}} - E{I_{xx}}{v {}_{,zz}} \\ M_y =& \, + E{I_{yy}}{u {}_{,zz}} + E{I_{xy}}{v {}_{,zz}} \\ \end{align}\]

• In matrix form these can be written as: \[\begin{align} \left\{ \begin{array}{c} M_x \\ M_y\\ \end{array} \right\} = \left[ \begin{array}{cc} -EI_{xy} & -EI_{xx} \\ EI_{yy} & EI_{xy} \\ \end{array} \right] \left\{ \begin{array}{c} {u {}_{,zz}} \\ {v {}_{,zz}} \\ \end{array} \right\} \end{align}\]

• Recall inversion: \[\begin{pmatrix}a & b \\ c & d \\ \end{pmatrix}^{-1} = \begin{pmatrix}\frac{d}{a\,d-b\,c} & -\frac{b}{a\,d-b\,c} \\ - \frac{c}{a\,d-b\,c} & \frac{a}{a\,d-b\,c} \\ \end{pmatrix}\]

• Simple inversion yields the following: \[\begin{align} \left\{ \begin{array}{c} {u {}_{,zz}} \\ {v {}_{,zz}} \\ \end{array} \right\} = \frac{1}{\det EI} \left[ \begin{array}{cc} EI_{xy} & EI_{xx} \\ -EI_{yy} & -EI_{xy} \\ \end{array} \right] \left\{ \begin{array}{c} M_x \\ M_y\\ \end{array} \right\} \end{align}\] where: \[{\det EI} = + EI_{xx} EI_{yy} - (EI_{xy})^2\]

• Now, lets examine the stress equation again: \[\begin{equation} \begin{split} {\sigma_{zz}}=& E \left( -{u {}_{,zz}} x - {v {}_{,zz}} y \right) \\ =& \frac{E}{\det EI} \left[\right. \\ & -x \left(EI_{xy} M_x + EI_{xx} M_y \right) \\ & \left. + y \left(EI_{yy} M_x + EI_{xy} M_y\right) \right] \\ \end{split} \end{equation}\]

• It is these equations which allow us to compute deflections, strains, and stresses from geometry, material, and loads.

Summary of beam equations

• Thus: \[\begin{align} \frac{d^2}{d z^2} \left[ E {I_{yy}}{u {}_{,zz}} + E {I_{xy}}{v {}_{,zz}} \right] =& \, p_x \\ \frac{d^2}{d z^2} \left[ E {I_{xy}}{u {}_{,zz}} + E {I_{xx}}{v {}_{,zz}} \right] =& \, p_y \\ \end{align}\]

• If symmetry exists across the \(x\) or \(y\) axis, the bending equations are uncoupled. \[\begin{align} \frac{d^2}{d z^2} \left[ E {I_{yy}}{u {}_{,zz}} \right] =& \; p_x \\ \frac{d^2}{d z^2} \left[ E {I_{xx}}{v {}_{,zz}} \right] =& \; p_y \\ M_x =& \, - E{I_{xx}}{v {}_{,zz}} \\ M_y =& \, + E{I_{yy}}{u {}_{,zz}} \\ \end{align}\]

Example: Uniform beam

With uniform loading

With uniform loading

\[\begin{align} E I_{yy} \; {u {}_{,zzzz}} =& \; p_x \\ E I_{yy} \; {u {}_{,zzz}} =& \; p_x z + C_1 \\ E I_{yy} \; {u {}_{,zz}} =& \; p_x \frac{z^2}{2} + C_1 z + C_2\\ E I_{yy} \; {u {}_{,z}} =& \; p_x \frac{z^3}{6} + C_1 \frac{z^2}{2} + C_2 z +C_3\\ E I_{yy} \; u(z) =& \; p_x \frac{z^4}{24} + C_1 \frac{z^3}{6} + C_2 \frac{z^2}{2} +C_3 z +C_4\\ \end{align}\]

• What are some boundary conditions we can use? \[\begin{align} u(0) =& \; 0 \Longrightarrow C_4 = 0\\ u'(0) =& \; 0 \Longrightarrow C_3 = 0\\ \end{align}\] \[\begin{align} M(0) =& \; E I_{yy} {u {}_{,zz}} = \frac{p_x l^2}{2} \Longrightarrow C_2 = \frac{p_x l^2}{2}\\ M(l) =& \; E I_{yy} {u {}_{,zz}} = 0 \Longrightarrow C_1 = - l p_x\\ \end{align}\]

\[E I_{yy} u(z) = p_x \left(\frac{z^4}{24}-\frac{l\,z^3}{6}+\frac{l^2\,z^2}{4}\right)\]

Example of a uniform beam with off-centroid point load

This example is likely not given in lecture but is provided for reference

\[\begin{align} M_z =& \; 2F \cdot \left(\frac{3}{2}\right) = 3000 \; [{\; \mathrm{in\cdot lb} \;}] \\ \end{align}\]

• For a boundary condition, compute moments at \(z=L\) due to \(F\) about its centroid: \[\begin{align} M_{Fo} =& \; (3 \, \hat{j}+1.5 \, \hat{i}) \times F \, \hat{k} \\ =& \; 3F \, \hat{i} - 1.5F \, \hat{j} \\ =& \; 3000 \, \hat{i} - 1500 \, \hat{j} \; [{\; \mathrm{in\cdot lb} \;}] \\ \end{align}\]

- Thus, the components are: \[\begin{align} M_x =& \; 3000 \; [{\; \mathrm{in\cdot lb} \;}] \\ M_y =& \; -1500 \; [{\; \mathrm{in\cdot lb} \;}] \\ \end{align}\]

Summarizing at the end \(z=L:\)

• Forces \[\begin{align} P&=1000, [{\; \mathrm{lb} \;}] \\ S_x&=0, [{\; \mathrm{lb} \;}] \\ S_y&=2000 [{\; \mathrm{lb} \;}] \\ \end{align}\]

• Moments \[\begin{align} M_x&=+3000, \; [{\; \mathrm{in\cdot lb} \;}] \\ M_y&=-1500, \; [{\; \mathrm{in\cdot lb} \;}] \\ M_z&=+3000 \; [{\; \mathrm{in\cdot lb} \;}] \\ \end{align}\]

• Distributed loads \[\begin{align} p_x &= \, 0, \\ p_y &= \, 0, \\ p_z &= \, 0 \end{align}\]

• Integrate the distributed load to get shear: \[\begin{align} {\frac{d S_x}{d z}} =& \; - p_x \\ S_x =& \; \mathrm{constant} = 0 \\ {\frac{d S_y}{d z}} =& \; - p_y \\ S_y =& \; \mathrm{constant} = 2000 {\; \mathrm{lb} \;}\\ {\frac{d P}{d z}} =& \; - p_z \\ \end{align}\]
• Because there are no distributed axial loads or torques: \[\begin{align} P =& \; \mathrm{constant} = 1000 {\; \mathrm{lb} \;}\\ M_z =& \; 3000 {\; \mathrm{in\cdot lb} \;}\\ \end{align}\]

• Integrate shear (\(S\)) to get moment:
• \(M_x\) \[{\frac{d M_x}{d z}} =S_y = 2000 \Longrightarrow M_x = 2000 \cdot z + C\]
  • Apply boundary condition at \(z=100 [in]\): \[\begin{align} M_x =& \; 3000 = 2000 \cdot 100 + C \\ C =& \; - 197,000 \\ M_x(z) =& \; 2000 \cdot z -197,000 \; [{\; \mathrm{in\cdot lb} \;}] \\ \end{align}\]
• \(M_y\) \[{\frac{d M_y}{d z}} =-S_x =0\] \[M_y = 0 \cdot z + C\]
  • Apply boundary condition at \(z=100 [in]\): \[\begin{align} M_y =& \; -1500 = C \\ M_y(z) =& \; -1500 {\; \mathrm{in\cdot lb} \;}\\ \end{align}\]

• Find stress and strain: \[\begin{align} \sigma_{zz} =& \; EA(\varepsilon_{zo} - x \, {u {}_{,zz}} - y \, {v {}_{,zz}}) \\ \varepsilon_{zz} =& \; \varepsilon_{zo} - {u {}_{,zz}} \cdot x - {v {}_{,zz}} \cdot y \\ \varepsilon_{zo} =& \; \frac{P}{EA} = \frac{1000}{30 \cdot 10^6 \cdot 6 \cdot 3} = \frac{1}{540000} \\ \end{align}\]
• For a symmetric cross-section: \[\begin{align} M_x =& \; - EI_{\mathrm{xx}} {v {}_{,zz}} \\ M_y =& \; + EI_{\mathrm{yy}} {u {}_{,zz}} \\ \end{align}\]

Hence:

• \[\begin{align} {u {}_{,zz}} =& \; \frac{M_y}{EI_{\mathrm{yy}}} = \frac{-1500}{[30 \cdot 10^6 \cdot (6 \cdot 3^3 /12)]} \\ =& \; -\frac{1}{270000} [{\; \mathrm{in}^{-1} \;}] \\ \end{align}\]
• \[\begin{align} {v {}_{,zz}} =& \; -\frac{M_x}{EI_{\mathrm{xx}}} \\ =& \; \frac{-(2000 \cdot z -197,000)}{[30 \cdot 10^6 \cdot (3 \cdot 6^3 /12)]} \\ =& \; \frac{(197 - 2 \cdot z)}{1620000} [{\; \mathrm{in}^{-1} \;}] \\ \end{align}\]

• Calculate stress: \[\begin{align} \sigma_{zz} =& \; EA(\varepsilon_{zo} - x {u {}_{,zz}} - y {v {}_{,zz}}) \\ \end{align}\]

• Recall: \[\begin{align} \varepsilon_{zo} =& \; \frac{1}{540000} \\ {u {}_{,zz}} =& \; -\frac{1}{270000} [{\; \mathrm{in}^{-1} \;}] \\ {v {}_{,zz}} =& \; \frac{(197 - 2 \cdot z)}{1620000} [{\; \mathrm{in}^{-1} \;}] \\ \end{align}\]

• Thus: \[\begin{align} \sigma_{zz}(1.5,+3,0) =& \; -10722 [\mathrm{psi}] \\ \sigma_{zz}(1.5,-3,0) =& \; +11167 [\mathrm{psi}] \\ \end{align}\]

• To obtain the displacements / deflections of the reference line: \[\varepsilon_{zo} = {w_o {}_{,z}} = \frac{1}{540000} \Longrightarrow w_o = \frac{z}{540000} + C\]

Boundary conditions:

• At \(z=0\):
  • \[\begin{align} wo =& \; 0 \\ \Longrightarrow wo =& \; \frac{z}{540000} \\ \end{align}\]
  • \[\begin{align} {u {}_{,zz}} =& \; -\frac{1}{270000} \Longrightarrow {u {}_{,z}} = -\frac{z}{270000} + C \\ \end{align}\]
• At \(z=0\)
  • \[\begin{align} {u {}_{,z}} =& \; 0 \Longrightarrow {u {}_{,z}} = -\frac{z}{270000} \\ {u {}_{,z}} =& \; -\frac{z}{270000} \Longrightarrow u = -\frac{z^2}{540000} + C \\ \end{align}\]

• At \(z=0\):

  • \[\begin{align} u =& \; 0 \Longrightarrow u = -\frac{z^2}{540000} \\ {v {}_{,zz}} =& \; \frac{(197 - 2 \cdot z)}{1620000} \\ {v {}_{,z}} =& \; \frac{(197 \cdot z - z^2)}{1620000} \\ v =& \; \left(\frac{197}{2} - \frac{z}{3}\right) \cdot \frac{z^2}{1620000} \\ \end{align}\]

• Also at \(z=0\) \[\begin{split} {u {}_{,z}} =& \; 0 \Longrightarrow {u {}_{,z}} = -\frac{z}{270000} \\ {u {}_{,z}} =& \; -\frac{z}{270000} \Longrightarrow u = -\frac{z^2}{540000} + C \\ \end{split}\]

• At \(z=0\): \[\begin{split} u =& \; 0 \Longrightarrow u = -\frac{z^2}{540000} \\ {v {}_{,zz}} =& \; \frac{(197 - 2 \cdot z)}{1620000} \\ {v {}_{,z}} =& \; \frac{(197 \cdot z - z^2)}{1620000} \\ v =& \; \left(\frac{197}{2} - \frac{z}{3}\right) \cdot \frac{z^2}{1620000} \\ \end{split}\]