8.8 Example: Uniform beam

8.8.1 With uniform loading

Cantilever beam uniformly loaded

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

8.8.2 With uniform loading

Cantilever beam uniformly loaded

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

Cantilever beam uniformly loaded

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

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

8.8.3 Example of a uniform beam with off-centroid point load

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

@AriGur2007

  • Given:
    • L=100
    • E=30 MPsi
    • F=1000
  • Required:
    • \(P(z)\)
    • \(S_x(z)\)
    • \(M_x(z), M_y(z), M_z(z)\)
    • \({\varepsilon_{zz}}(x,y,z)\)
    • \({\sigma_{zz}}(x,y,z)\)

@AriGur2007

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

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

Due to overlapping material and to provide continuity, the lectures slides for Lecture 1 are included in the materials for Lecture 0.

Due to overlapping material and to provide continuity, the lectures slides for Lecture 1 are included in the materials for Lecture -1.