Mechanical Systems V - Statics Flashcards

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• Force and Moment: 3D Reminders

Rectangular Components (3D)

A force vector $\mathbf{F}$ can be expressed in terms of its rectangular components:

• Vector form: $\mathbf{F} = \mathbf{F}_x + \mathbf{F}_y + \mathbf{F}_z$

• Unit vector notation: $\mathbf{F} = F_x \mathbf{i} + F_y \mathbf{j} + F_z \mathbf{k}$

Determination of the projection of $\mathbf{F}$ on each axis using direction cosines:

• $F_x = F \cos(\theta_x)$

• $F_y = F \cos(\theta_y)$

• $F_z = F \cos(\theta_z)$

Magnitude calculation:

• $F = \sqrt{F_x^2 + F_y^2 + F_z^2}$

Moment (3D)

While scalar multiplication (moment-arm rule) is convenient for 2D, 3D analysis requires a vector approach involving cross-product multiplication because determining perpendicular distances between lines of action and points is computationally tedious.

Evaluation of the cross product at point $O$:

• $\mathbf{M}_0 = \mathbf{OA} \times \mathbf{F} = \begin{pmatrix} r_x \\ r_y \\ r_z \end{pmatrix} \times \begin{pmatrix} F_x \\ F_y \\ F_z \end{pmatrix} = \begin{pmatrix} M_x \\ M_y \\ M_z \end{pmatrix}$

• Expansion of components:

  • $M_x = r_y F_z - r_z F_y$

  • $M_y = r_z F_x - r_x F_z$

  • $M_z = r_x F_y - r_y F_x$

Couple (3D)

A couple magnitude is calculated as follows:

• $\mathbf{M} = \mathbf{r}_A \times \mathbf{F} + \mathbf{r}_B \times (-\mathbf{F}) = (\mathbf{r}_A - \mathbf{r}_B) \times \mathbf{F}$

• Letting $\mathbf{r} = \mathbf{r}_A - \mathbf{r}_B$, the formula simplified becomes: $\mathbf{M} = \mathbf{r} \times \mathbf{F}$

• Key Properties of Couples:

  • The couple is identical about all points; reference to the moment center $O$ disappears.

  • A couple is a free vector in 3D.

  • In contrast, the moment of a force about a point is a sliding vector directed along the axis through that point.

Resultants (3D)

• Force Resultant: The simplest force combination that can replace original forces without changing the external effect on a rigid body. Equilibrium occurs when the resultant of all forces acting on a body is zero.

  • $\mathbf{R} = \sum \mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 + \dots$

• Principle of Moments: The moment resultant is the sum of all individual moments at a specific point (e.g., point $O$).

  • $\mathbf{M}_0 = \sum (\mathbf{r}_i \times \mathbf{F}_i) = \mathbf{M}_1 + \mathbf{M}_2 + \mathbf{M}_3 + \dots$

Representation of Distributed Loads

Forces can be distributed across a surface or length in three common ways:

1. Uniformly distributed load: Represented as a constant rectangular profile.

2. Uniformly varying load: Represented as a triangular profile.

3. Uniformly distributed and varying load: A trapezoidal combination of constant and varying profiles.

Mechanical Actions

Link Diagrams and System Isolation

• Physical System: A set of materials with constant mass over time.

• System Boundary: To study a solid in statics, it must be isolated via an imaginary border separating the internal part ($\Sigma$) from the external part ($\bar{\Sigma}$).

• Mechanical Action Diagram: Visualizes external actions. Example with solids $0$, $1$, and $2$:

  • If isolating solid $1$: The balance includes external actions on $1$ (ext $\rightarrow 1$), connection action from $2$ ($2 \rightarrow 1$), and connection action from $0$ ($0 \rightarrow 1$).

  • If isolating system $A = \{1 + 2\}$: The balance includes ext $\rightarrow 1$, ext $\rightarrow 2$, $0 \rightarrow 1$, and $0 \rightarrow 2$. Note that $2 \rightarrow 1$ becomes an internal mechanical action and is excluded from the balance.

Mechanical Action Torsor

A mechanical action of solid $S1$ on $S2$ is defined by two vectors reducing to a specific point (reduction point):

1. Resultant (Pushing action): $\mathbf{R}$.

2. Moment (Twisting action): $\mathbf{M}_O$.

Representation of the Torsor:

• $\{\tau_{S1 \rightarrow S2}\} = \begin{cases} X_{12} & L_{12} \\ Y_{12} & M_{12} \\ Z_{12} & N_{12} \end{cases}_{(O, \mathbf{x}, \mathbf{y}, \mathbf{z})}$

Summary of Kinematic Joints

Each joint is defined by its degrees of freedom ($n_c$) and degrees of connection ($n_s$). Below are the details for various joints:

• Fixed Joint ($n_c = 0$, $n_s = 6$):

  • Kinematic Torsor: All elements are zero.

  • Action Torsor: $\begin{cases} X_{12} & L_{12} \\ Y_{12} & M_{12} \\ Z_{12} & N_{12} \end{cases}$.

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Power and Perfect Joints

Developed Power

The power developed by an action $j$ on solid $S$ at time $t$ in reference frame $(R)$ using torsor $\tau$ at point $M$:

• $P_{R}(\{\tau_{S/j}\}) = \{\tau_{S/j}\} \cdot \{\mathcal{V}_{S/R}\}$

• Expanded form: $P_{R}(\{\tau_{S/j}\}) = \mathbf{R}(\{\tau_{S/j}\}_M) \cdot \mathbf{v}(\{\mathcal{V}_{S/R}\}_M) + \mathbf{M}(\{\tau_{S/j}\}_M) \cdot \mathbf{\omega}(\{\mathcal{V}_{S/R}\}_M)$

Perfect Kinematic Joint

A joint is considered perfect if there is no gap and no friction. In this case, the power developed between the two solids is zero:

• $P(\{\tau_{S2/S1}\}) = \{\tau_{S2/S1}\} \cdot \{\mathcal{V}_{S1/S2}\} = 0$

• Example (Hinge of axis $\mathbf{x}$): The rotation exists only around $x$, but the action torsor has a zero moment component $L_{12}$ along $x$, resulting in zero power.

Fundamental Principle of Statics (FPS)

Theoretical Theorems

• General Theorem: If a physical system is in static equilibrium, the total external mechanical action torsor applied on $\Sigma$ is null:

  • $\{\tau_{\Sigma \rightarrow \bar{\Sigma}}\} = \{0\}$

• Theorem of Static Resultant: In equilibrium, the resultant of the external mechanical action torsor is zero:

  • $\mathbf{R}(\{\tau_{\Sigma \rightarrow \bar{\Sigma}}\}) = \mathbf{0}$

• Theorem of Static Moment: In equilibrium, the moment of the external mechanical action torsor calculated at any point $A$ is zero:

  • $\mathbf{M}_A(\{\tau_{\Sigma \rightarrow \bar{\Sigma}}\}) = \mathbf{0}$

• Theorem of Reciprocal Actions: (Newton's Third Law) For systems in equilibrium, the torsor of $\Sigma_1$ on $\Sigma_2$ is the negative of the torsor of $\Sigma_2$ on $\Sigma_1$:

  • $\{\tau_{\Sigma_2 \rightarrow \Sigma_1}\} = -\{\tau_{\Sigma_1 \rightarrow \Sigma_2}\}$

Problem Solving Approach

1. Step 1: Draw the link diagram.

2. Step 2: Isolate the specific solid(s) for study.

3. Step 3: Determine all mechanical action torsors acting on the isolated system(s).

4. Step 4: Solve for unknowns by applying the FPS.

Beams

Determinacy and Stability

• Determinate Structure: Unknown external reactions or internal members can be found using only equilibrium conditions.

• Indeterminate Structure: Unknowns cannot be determined by statics alone; compatibility conditions of structural parts are required.

• Mathematical Conditions:

  • Let $R$ = number of support reactions.

  • Let $Fi$ = number of reaction forces from internal hinges.

  • Let $m$ = number of members.

  • Statically Unstable: $R + Fi < 3m$

  • Statically Determinate and Stable: $R + Fi = 3m$

  • Statically Indeterminate: $R + Fi > 3m$

Free-Body Diagram (FBD)

The FBD is a simple visual representation of the mechanical action torsor components applied to the beam.

• Example Components: Reactions at supports (e.g., $A_x, A_y, M_A$ at point $A$, $B_y$ at point $B$), beam length $L$, and external point loads $P$.

Beam Problem Solving Approach

1. Step 1: Draw the FBD, replacing distributed loads with simplified representations and identifying all torsor components.

2. Step 2: Check stability and determinacy. If the structure is unstable or indeterminate, isolate multiple specific beams to solve.

3. Step 3: Solve for components of the mechanical action torsor using FPS (decomposing into resultants and moments).