Problem 1.1 Translational mechanical power is the dot product of force and the velocity of the point of application of the force, \(\mathsf{\mathbb{P}}=\mathbf{F}\cdot \mathbf{v}\) When the force and velocity of the point of application of the force are colinear, the scalar equation \(\mathsf{\mathbb{P}}=F\,v\) can be used. A mechanical system was de-energized before the force shown in Fig. P1.1 a with the colinear velocity shown in Fig. P1.1 b acted on it.

Fig. P1.1 a Force F (t ) acting on a mechanical system. b Colinear velocity v (t )

Problem 1.2 DC Electrical power is product of current times voltage, \(\mathsf{\mathbb{P}}=i\,v\) A DC motor and the system it drives was de-energized, before the DC motor’s power source provided it the current i at the voltage v , shown in Fig. P.1.2 a and b.

Fig. P1.2 a DC current i (t ) driving a motor. b Voltage v (t ) of the power supply

Problem 1.3 A piping system is shown in Fig. P1.3 . The fluid is modeled as incompressible. Consequently, the continuity equations can be written in terms of volume flow rate, Q, rather than in terms of mass flow rate. The branches are identified by letter, and pressure nodes between the branches are numbered.

Fig. P1.3 Piping system schematic. The fluid is modeled as incompressible. There is fluid resistance in each branch, which decreases pressure in the direction of the fluid flow

1.3.a Orient the flow in each branch. The positive direction for flow through pump is from node 5 to node 1.

1.3.b Write a complete set of independent compatibility equations in the form of path equations.

1.3.c Write a complete set of independent continuity equations for volume flow rate.

Problem 1. 4 A piping system is shown in Fig. P1.4 . The fluid is modeled as incompressible. Consequently, the continuity equations can be written in terms of volume flow rate, Q, rather than in terms of mass flow rate. The branches are identified by letter, and pressure nodes between the branches are numbered.

Fig. P1.4 Piping system schematic. The fluid is modeled as incompressible. There is fluid resistance in each branch, which decreases pressure in the direction of the fluid flow

1.4.a Orient the flow in each branch. The positive direction for flow through pump is from node 5 to node 1.

1.4.b Write a complete set of independent compatibility equations in the form of path equations.

1.4.c Write a complete set of independent continuity equations in terms of volume flow rate, Q .

Problem 1.5 The stress vs. strain plot of a specimen tested to rupture is approximated by the elastic-perfectly plastic model shown in Fig. P1.5 .

Fig. P1.5 Stress–strain plot of a specimen tested to rupture. The specimen was 0.5 in. in diameter and 3 in. long. The yield stress, yield strain, and rupture strain are\({{\sigma}_{yield}}=90\,\text{ksi}\quad{{\varepsilon}_{yield}}=0.18\,{\%}\) \({{\varepsilon }_{rupture}}=1{.}02\,\text{\%}\)

Problem 1.6 The force vs. displacement plot of a load test is shown in Fig. P1.6 .

Fig. P1.6 Force in pounds vs. displacement in inches. The test results are approximated by

$$ F\left( t \right)=\left( 7\times {{10}^{4}}\,\text{lb} \right)\left( 1-{{\text{e}}^{-\,\frac{x}{\text{in}}}} \right). $$

1.6.a Calculate the energy transferred to the specimen for the displacement, \( x=4\,\text{in,} \) in foot-pounds and joules.

1.6.b Calculate the coenergy for the displacement, \( x=4\,\text{in}\text{.} \)

Problem 1.7 An electric circuit consisting of a voltage source, represented as a battery, a switch, a resistor, an inductor, and a capacitor, is shown in Fig. P1.7 . The energetic equations of this circuit are listed. Use elimination by substitution to derive the system equation for the input voltage and output variable indicated:

Fig. P1.7 Electric circuit consisting of a battery, resistor, inductor, and capacitor in series, annotated with voltage nodes and the positive directions of current through the elements

i. Input Variable: v _{1g} , Output Variable: v _{12}

ii. Input Variable: v _{1g} , Output Variable: v _{23}

iii. Input Variable: v _{1g} , Output Variable: v _{3g}

iv. Input Variable: v _{1g} , Output Variable: i _{L}

and check its units in terms of the power variables, current, voltage and time.

Energetic Equations:

Continuity (Conservation of Charge), Node Eqs:

$$\begin{aligned} & \text{Node}1:{{i}_{source}}={{i}_{R}} \\ & \text{Node}\text{2: }{{i}_{R}}={{i}_{L}} \\ & \text{Node}3:{{i}_{L}}={{i}_{C}} \\ \end{aligned}$$

Compatibility of Voltage Drops, Path Eq:

$${{v}_{1\text{g}}}={{v}_{12}}+{{v}_{23}}+{{v}_{3\text{g}}}$$

Element Eqs:

Energy Eqs:

$$\begin{aligned} & \text{System:}\ {{\mathbb{E}}_{sys}}={{\mathbb{E}}_{L}}+{{\mathbb{E}}_{C}}\ \\ & \text{Inductor:}\ {{\mathbb{E}}_{L}}=\frac{1}{2}Li_{L}^{2}\ \\ & \text{Capacitor:}\ {{\mathbb{E}}_{C}}=\frac{1}{2}Cv_{3g}^{2} \\ \end{aligned}$$

Problem 1.8 An electric circuit consisting of a voltage source, a resistor, an inductor, and a capacitor, annotated with nodes of distinct values of voltage and arrows indicating the positive direction of current through each element, is shown in Fig. P1.8 . The energetic equations of this circuit are listed. Use elimination by substitution to derive the system equation for the input and output variable indicated:

Fig. P1.8 Electric circuit consisting of a voltage source, resistor, inductor, and capacitor in series, annotated with voltage nodes and the positive directions of current through the elements

i. Input Variable: v _{1g} , Output Variable: i _{ R }

ii. Input Variable: v _{1g} , Output Variable: v _{2g}

iii. Input Variable: v _{1g} , Output Variable: i _{ L }

iv. Input Variable: v _{1g} , Output Variable: v _{12}

v. Input Variable: v _{1g} , Output Variable: i _{C}

and check its units in terms of the power variables and time.

Energetic Equations:

Continuity (Conservation of Charge), Node Eqs:

$$\begin{aligned} & \text{Node}\ \text{1:}\ {{i}_{source}}={{i}_{R}}\ \\ & \text{Node}\ \text{2:}\ {{i}_{R}}={{i}_{L}}+{{i}_{C}} \\ \end{aligned}$$

Compatibility of Voltage Drops, Path Equations:

$${{v}_{1\text{g}}}={{v}_{12}}+{{v}_{2g}}\ \ \ \ \ \ \ \ \ \ \ \ {{v}_{2g}}={{v}_{2g}}$$

Element Eqs:

$$\begin{aligned} & \text{Resistor:}\ {{v}_{12}}=R{{i}_{R}}\ \\ & \text{Inductor:}\ {{v}_{2g}}=L\frac{d{{i}_{L}}}{dt} \\ & \text{Capacitor:}\ {{i}_{C}}=C\frac{d{{v}_{2g}}}{dt} \\ \end{aligned}$$

Energy Eqs:

$$\begin{aligned} & \text{System}:\ {{\mathbb{E}}_{sys}}={{\mathbb{E}}_{L}}+{{\mathbb{E}}_{C}}\ \\ & \text{Inductor:}\ {{\mathbb{E}}_{L}}=\frac{1}{2}Li_{L}^{2}\ \\ & \text{Capacitor:}\ {{\mathbb{E}}_{C}}=\frac{1}{2}Cv_{2g}^{2} \\ \end{aligned}$$

Problem 1.9 An electric circuit consisting of a voltage source, a resistor, an inductor, and a capacitor is shown in the schematic Fig. P1.9 . The energetic equations of this circuit are listed. Use elimination by substitution to derive the system equation for the input and output variable indicated:

Fig. P1.9 Electric circuit consisting of a battery, resistor, inductor, and capacitor in series, annotated with voltage nodes and the positive directions of current through the elements

i. Input Variable: v _{1g} , Output Variable: v _{12}

ii. Input Variable: v _{1g} , Output Variable: v _{2g}

iii. Input Variable: v _{1g} , Output Variable: i _{ L }

iv. Input Variable: v _{1g} , Output Variable: i _{ R }

v. Input Variable: v _{1g} , Output Variable: i_{ C }

and check its units in terms of the power variables, current, voltage and time.

Energetic Equations:

Continuity (Conservation of Charge), Node Eqs:

$$\begin{aligned} & \text{Node}\ 1:\ {{i}_{source}}={{i}_{L}}\ \\ & \text{Node}\ 2:\ {{i}_{L}}={{i}_{R}}+{{i}_{C}} \\ \end{aligned}$$

Compatibility of Voltage Drops, Path Eqs:

$${{v}_{1\text{g}}}={{v}_{12}}+{{v}_{2g}}\,\,\,\,\,\,\,\,\,\,\,\,{{v}_{2g}}={{v}_{2g}}$$

Element Eqs:

$$\begin{aligned} & \text{Resistor:}\ {{v}_{2g}}=R{{i}_{R}}\ \\ & \text{Inductor:}\ {{v}_{12}}=L\frac{d{{i}_{L}}}{dt}\ \\ & \text{Capacitor:}\ {{i}_{C}}=C\frac{d{{v}_{2g}}}{dt} \\ \end{aligned}$$

Energy Eqs:

$$ \begin{aligned} & \text{System:}\ {{\mathbb{E}}_{sys}}={{\mathbb{E}}_{L}}+{{\mathbb{E}}_{C}}\ \ \ \\ & \text{Inductor: }{{\mathbb{E}}_{L}}=\frac{1}{2}Li_{L}^{2}\ \ \ \\ & Capacitor:{{\ }_{C}}=\frac{1}{2}Cv_{2g}^{2} \\ \end{aligned} $$

Problem 1.10 A translational mechanical system consisting of a mass M sliding on a lubricating fluid film with damping b , and a spring K attached between the mass and ground is shown in Fig. P1.10a . The linear graph of this energetic system, analogous to an electric circuit diagram is Fig. P1.10b . The energetic equations are listed. Use elimination by substitution to derive the system equation for the input and output variable indicated:

Fig. P1.10 a Force F(t) acts on mass M which slides on a lubricating film with damping b against spring K which is connected to ground. b The linear graph of the system

i. Input Variable: F(t) , Output Variable: F _{ b }

ii. Input Variable: F(t) , Output Variable: F _{ M }

iii. Input Variable: F(t) , Output Variable: F _{ K }

iv. Input Variable: F(t) , Output Variable: v _{1g}

and check the its units in terms of the power variables, force, velocity and time.

Energetic Equations:

Continuity (Force Equilibrium) Node Eqs: \( F(t)={{F}_{b}}+{{F}_{M}}+{{F}_{K}} \)

Compatibility of Velocity, Path Eq: \( {{v}_{1\text{g}}}={{v}_{1\text{g}}} \)

Element Eqs: \( {{F}_{b}}=b{{v}_{1g}}\quad {{F}_{M}}=M\frac{d{{v}_{1g}}}{dt}\quad \frac{d{{F}_{K}}}{dt}=K{{v}_{1g}} \)

Energy Eqs:

$$\begin{aligned} & \text{System:}\ {{\mathbb{E}}_{sys}}={{\mathbb{E}}_{M}}+{{\mathbb{E}}_{K}}\ \ \ \\ & \text{Mass:}\ {{\mathbb{E}}_{M}}=\frac{1}{2}Mv_{1g}^{2} \\ & \text{Spring:}\ {{\mathbb{E}}_{K}}=\frac{F_{K}^{2}}{2K} \\ \end{aligned}$$

Problem 1.11 A schematic of a hydraulic system is shown in Fig. P1.11a . The pump, modeled as a pressure source p (t ), discharges fluid into a hydraulic circuit consisting of two fluid resistances, R _{1} and R_{2} , a fluid inertance I , which stores kinetic energy, and a fluid accumulator with capacitance C , which stores energy by compressing a spring or nitrogen-filled bladder. Figure P1.11b is the linear graph of the system, analogous to an electric circuit. The energetic equations are listed. Use elimination by substitution to derive the system equation for the input and output variable indicated:

Fig. P1.11a Fluid system modeled as a pressure source, two fluid resistances, a fluid “inertance,” and a fluid capacitance

Fig. P1.11b Linear graph of the hydraulic system

i. Input Variable: p (t ), Output Variable: p _{12}

ii. Input Variable: p (t ), Output Variable: p _{23}

iii. Input Variable: p (t ), Output Variable: p _{3g}

iv. Input Variable: p (t ), Output Variable: Q _{ L }

v. Input Variable: p (t ), Output Variable: Q _{ C }

vi. Input Variable: p (t ), Output Variable: Q _{ R2}

and check its units in terms of the power variables, pressure, volume flow rate Q, and time.

Energetic Equations:

Continuity (Conservation of Volume of Incompressible Fluid), Node Eqs:

$$ \begin{aligned} & \text{Node}\ \text{1:}\ Q={{Q}_{{{R}_{1}}}} \\ & \text{Node}\ \text{2:}\ {{Q}_{R}}={{Q}_{I}}\ \ \ \\ & \text{Node}\ \text{3:}\ {{Q}_{I}}={{Q}_{{{R}_{2}}}}+{{Q}_{C}} \\ \end{aligned} $$

Compatibility of Pressure Drops, Path Equations: \( {{p}_{1\text{g}}}={{p}_{12}}+{{p}_{23}}+{{p}_{3\text{g}}}\quad {{p}_{3g}}={{p}_{3g}} \)

Element Eqs:

$$ \begin{aligned} \text{Fluid Resistance }{{{R}}_{{1}}}\text{:}{{p}_{12}}={{R}_{1}}{{Q}_{{{R}_{1}}}}\quad \quad \text{Fluid Resistance }{{{R}}_{{2}}}\text{:}\ {{p}_{3g}}={{{R}}_{{2}}}{{Q}_{{{{R}}_{{2}}}}}\\ \text{Fluid Inertance I:}{{p}_{23}}=I\frac{d{{Q}_{I}}}{dt}\quad \quad \text{Fluid Capacitance C:}\ {{Q}_{\text{C}}}=\text{C}\frac{d{{p}_{3\text{g}}}}{dt}\\\end{aligned} $$

Energy Eqs:

$$ \begin{aligned} & \text{System}:\ {{\mathbb{E}}_{sys}}={{\mathbb{E}}_{I}}+{{\mathbb{E}}_{C}} \\ & \text{Inertance:}\ {{\mathbb{E}}_{I}}=\frac{1}{2}IQ_{I}^{2} \\ & \text{Capacitance:}\ {{\mathbb{E}}_{C}}=\frac{1}{2}Cp_{3g}^{2} \\ \end{aligned} $$