Statics

This chapter covers the following:

• Force, resolution, and resultant (two dimensions)
• Moments (couples)
• Systems of forces and equilibrium requirements
• Centroids of masses, areas, lengths, and volumes
• Moment of inertia
• Parallel axis theorem
• Radius of gyration
• Product of inertia
• Screw thread
• Belt friction
• Statically determinate truss
• Plane truss: method of joints
• Plane truss: method of sections
• Concurrent forces

Force, resolution, and resultant (two dimensions)

Force

Definitions

A force is a vector quantity having both magnitude and direction.

Example:

A box is pulled with a force of 100 N along the ground. The force is directed at 30° above the horizontal. You can study its effect by separating it into horizontal and vertical components.

Resolution of a force

Resolving a force means replacing one force with two perpendicular components that have the same overall effect.

If a force $F$ acts at an angle $\theta$ from the horizontal, its components are:

$$\begin{gathered} F_x=F\cos \theta \\ {F}_y=F\sin \theta \end{gathered}$$

Example:

A force of 200 N is applied at an angle of 60°.

• $F_x=200\cos 60°=100$ N
• $F_y=200\sin 60°\approx 173.2$ N

Resultant of forces

When multiple forces act at a point, you can find a single equivalent force (the resultant) by adding components.

First add components in each direction:

$$\begin{gathered} R_x=\sum F_x,R_y=\sum F_y \\ R=\sqrt{R_x^2+R_y^2},\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right) \end{gathered}$$

Example:

Two forces act at a point: 50 N at 0° and 30 N at 90°.

$$\begin{gathered} R_x=50,R_y=30 \\ R=\sqrt{50^2+30^2}=\sqrt{3400}\approx 58.3\text{ N} \end{gathered}$$

Moments (couples)

Moment of a force

The moment measures the turning effect of a force about a point or axis.

$$M=F\cdot d$$

Where:

• $F$: Force
• $d$: Perpendicular distance to the axis

Example:

A 20 N force is applied at the end of a 0.5 m wrench.

$M=20\cdot 0.5=10$ N·m.

Couple

Definitions

A couple consists of two equal and opposite forces separated by a distance, producing a pure moment.

A couple produces rotation without a net force. For two equal and opposite forces separated by a distance $d$:

$$M_{\text{couple}}=F\cdot d$$

Example:

Two forces of 100 N each act in opposite directions, 0.3 m apart.

$M_{\text{couple}}=100\cdot 0.3=30$ N·m.

Systems of forces and equilibrium requirements

Types of force systems

• Concurrent
• Parallel
• Coplanar
• Non-concurrent

Example:

Three forces of 40 N, 30 N, and 20 N act at a common point but in different directions. This is a concurrent force system because their lines of action meet at one point.

Equilibrium conditions

A body is in equilibrium when there is no net force and no net moment.

$$\sum F_x=0,\sum F_y=0,\sum M=0$$

Example:

A ladder rests against a wall with forces due to its weight and the reactions at the floor and wall. You apply the equilibrium equations to check whether the ladder can remain at rest.

Centroids of masses, areas, lengths, and volumes

For composite areas:

For a composite area made of parts $A_i$ located at coordinates $(x_i,y_i)$, the centroid coordinates are:

$$\bar{x}=\frac{\sum x_i{A}_i}{\sum A_i},\bar{y}=\frac{\sum y_i{A}_i}{\sum A_i}$$

Example:

• Area 1 $=20c{m}^2$ at $x_1=2$
• Area 2 $=30c{m}^2$ at $x_2=6$

$$\bar{x}=\frac{20\cdot 2+30\cdot 6}{20+30}=\frac{220}{50}=4.4\text{ cm}$$

Moment of inertia

Area moment of inertia

The area moment of inertia describes how area is distributed about an axis. About the x and y axes:

$$I_x=\int y^{2}dA,I_y=\int x^{2}dA$$

Example:

For a rectangle of width $b=4$ cm and height $h=6$ cm, about the x-axis:

$I_x=\frac{b{h}^3}{12}=\frac{4\cdot 6^3}{12}=72{\text{cm}}^4$.

Parallel axis theorem

The parallel axis theorem lets you shift a moment of inertia from a centroidal axis to a parallel axis a distance $d$ away.

$$I=I_{\text{centroid}}+A{d}^2$$

Where:

• $I$: Moment of inertia about new axis
• $I_{\text{centroid}}$: About centroidal axis
• $d$: Distance between axes
• $A$: Area

Example:

Rectangle $b=4$ cm, $h=6$ cm, centroidal $I_x=72{\text{cm}}^4$. If the new axis is 3 cm away:

$I=72+(24)(3^2)=72+216=288{\text{cm}}^4$.

Radius of gyration

The radius of gyration $k$ is a way to express the moment of inertia in a length-like form.

$$k=\sqrt{\frac{I}{A}}$$

Where $I$: Moment of inertia, $A$: Area

Example:

$I=288{\text{cm}}^4$, $A=24{\text{cm}}^2$:

$k=\sqrt{288/24}=\sqrt{12}\approx 3.46$ cm.

Product of inertia

The product of inertia is used in principal axis calculations.

$$I_{xy}=\int xydA$$

Used in principal axis calculations.

Example:

For a rectangle symmetric about both axes, $I_{xy}=0$ due to symmetry.

Friction

Coulomb’s law of friction

Coulomb’s law relates the frictional force to the normal force.

$$F_f=\mu N$$

Where:

• $F_f$: Frictional force
• $\mu$: Coefficient of friction
• $N$: Normal force

Example:

A block of weight 100 N rests on a horizontal surface with $\mu =0.3$.

$F_f=0.3\cdot 100=30$ N.

Screw thread

Torque to raise a load:

$$T=\frac{W{d}_m}{2}\left(\frac{l+\pi \mu d_m}{\pi d_m-\mu l}\right)$$

Where:

• $T$: Torque
• $W$: Load
• $d_m$: Mean diameter
• $l$: Lead
• $\mu$: Coefficient of friction

Example:

$W=500$ N, $d_m=20$ mm, $l=5$ mm, $\mu =0.15$:

$T=\frac{500\cdot 20}{2}\cdot \frac{5+\pi (0.15)(20)}{\pi (20)-0.15(5)}\approx 3140$ N·mm.

Belt friction

$$\frac{T_1}{T_2}=e^{\mu \theta }$$

Where:

• $T_1$: Tension on tight side
• $T_2$: Tension on slack side
• $\mu$: Coefficient of friction
• $\theta$: Angle of contact in radians

Example:

$\mu =0.25$, $\theta =\pi$ rad, $T_2=100$ N:

$T_1=100e^{0.25\pi }\approx 221.4$ N.

Statically determinate truss

A truss is statically determinate if the number of unknowns matches the number of available equilibrium equations.

$$m+r=2j$$

Where:

• $m$: Members
• $r$: Reactions
• $j$: Joints

Example:

A simple triangular truss: $m=3$, $r=3$, $j=3$.

$3+3=6$, $2\cdot 3=6$. Hence, statically determinate.

Plane truss: method of joints

At each joint, treat the joint as a particle in equilibrium and apply:

$$\sum F_x=0,\sum F_y=0$$

Solve two equations per joint to find forces in members.

Example:

At a joint with two unknown member forces and one known external load, the equilibrium equations give the member forces.

Plane truss: method of sections

To find forces in selected members, cut the truss and apply equilibrium to one side of the cut:

$$\sum F_x=0,\sum F_y=0,\sum M=0$$

Solve for up to three unknowns in the cut section.

Example:

For a Pratt truss cut through 3 members, use moment equilibrium about a joint to directly solve one member force, then apply $\sum F_x=0$ and $\sum F_y=0$.

Concurrent forces

Definitions

Concurrent forces are forces whose lines of action intersect at a common point.

Equilibrium conditions:

For concurrent forces in a plane, equilibrium requires:

$$\sum F_x=0,\sum F_y=0$$

Example:

Three concurrent forces: 100 N east, 80 N north, and 128 N southwest. Their vector sum can satisfy $\sum F_x=0$, $\sum F_y=0$ if in equilibrium.