11.3: Force and Moment Balances
- Page ID
- 7852
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Force Balance
The fact that the two layers are in fact bonded together must now be taken into account. Assuming first that the layers remain flat, there is clearly a requirement that both must end up with the same length (in the x-direction). This will require a tensile stress to be created in one layer (the one that is initially shorter) and a compressive stress in the other. The forces acting in each layer (in the x-direction) must add up to the externally applied force (ie a force balance must apply). Since there is no applied force in the case we are currently considering, the forces in the two layers must sum to zero. In addition, the misfit strain must be partitioned in some way between the two layers. There are thus two simultaneous equations, allowing the solution to be found. This is illustrated in the figure below. (The subscripts d and s represent “deposit” and “substrate” - ie the upper and lower layers.)
Moment Balance
The force balance is relatively straightforward, but the situation depicted in the figure above does not represent complete static equilibrium. This is because the forces being exerted by the stresses acting in the two layers produce a bending moment that tends to create curvature in the x-y plane - for the case shown, the top surface would become convex. The distributions of stress and strain must therefore change, creating an internal balancing moment (and also creating curvature), while maintaining a force balance.
Using the moment balance to find the curvature (and the associated stress and strain distributions) is slightly more complex than applying the force balance. Derivation of the expression for the curvature is shown here. The outcome is:
\[\kappa=\frac{6 E_{d} E_{s}(h+H) h H \Delta \varepsilon}{E_{d}^{2} h^{4}+4 E_{d} E_{s} h^{3} H+6 E_{d} E_{s} h^{2} H^{2}+4 E_{d} E_{s} h H^{3}+E_{s}^{2} H^{4}}\]
It’s important to recognize that the curvature of a beam is equal to the through-thickness gradient of the associated (in-plane) strain distribution. The other key concept here is that of the neutral axis of the beam. This is the location - strictly, it’s a plane (in 3-D), rather than an axis - where no (in-plane) strains arise from the bending (adoption of curvature). Its location, for a 2-layer system, is derived here. The result is:
\[\delta=\frac{\left(h^{2} E_{d}-H^{2} E_{s}\right)}{2\left(h E_{d}+H E_{s}\right)}\]
Using these equations, the final outcome (of imposing a misfit strain) can be obtained. This is illustrated in the figure below. It should be noted that adoption of curvature does NOT lead to zero strain at the neutral axis, but rather to NO CHANGE there - ie it remains at the value resulting from the force balance. (If we simply applied an external bending moment, rather than an internal misfit strain, then there would be no strain at the neutral axis.)
