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

Exam tip: The FE Reference Handbook (available during the exam) includes the formulas for moments of inertia, the parallel axis theorem, screw-thread torque, and belt friction shown in this chapter - look them up there instead of memorizing them. When you substitute numbers, keep every quantity in one unit system (don’t mix mm with m or N with lbf), and carry full precision through intermediate steps, rounding only at the final answer.

Force, resolution, and resultant (two dimensions)

Force

Definitions

Force

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

Couple

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.

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

Watch out: Pick a sign convention for forces and moments before you write the equilibrium equations - for example, forces up and to the right positive, counterclockwise moments positive - and hold that convention for the entire problem. Switching conventions partway through is a common source of sign errors.

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.

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

Example:

$I_{xy}=0$ whenever either axis is an axis of symmetry: every element of area has a mirror image across that axis, and their contributions cancel. A rectangle, symmetric about both axes, has $I_{xy}=0$, and so does a channel, which is symmetric about only one.

Friction

Coulomb’s law of friction

Coulomb’s law sets the largest friction force a surface can supply. Static friction takes whatever value equilibrium needs, from zero up to that limit, and reaches the limit only when sliding impends; once sliding starts, the kinetic coefficient applies instead.

$$F_f\le {\mu }_{s}N$$

Where:

• $F_f$: Friction force
• ${\mu }_s$: Coefficient of static friction
• $N$: Normal force

Example:

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

The largest friction force the surface can supply is $F_{f,max}=0.3\cdot 100=30$ N. With no horizontal push, the actual friction force is zero. A horizontal push of 20 N is resisted by 20 N of friction, and the block starts to slide only when the push exceeds 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

This is the power-screw form with collar friction neglected. The FE Reference Handbook’s statics section gives the equivalent form $M=Pr\tan (\alpha \pm \varphi )$, where $r$ is the mean thread radius, $\alpha$ is the lead angle ($\tan \alpha =l/(\pi d_m)$), $\varphi$ is the friction angle ($\tan \varphi =\mu$), and the plus sign applies to raising the load.

Example:

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

$T=\frac{500\cdot 20}{2}\cdot \frac{5+\pi (0.15)(20)}{\pi (20)-0.15(5)}\approx 1162$ N·mm.

Belt friction

When a belt is on the verge of slipping, the tensions on either side of the pulley are related by the formula below. The FE Reference Handbook writes it as $F_1=F_2{e}^{\mu \theta }$, with $F_1$ the force in the direction of impending motion. Short of slip, the ratio can be anything up to this value.

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

Where:

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

Example:

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

$T_1=100e^{0.25\pi }\approx 219.3$ 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:

Take the triangular truss above: pin support at $A=(0,0)$, roller at $B=(4,0)$ m, apex $C=(2,3)$ m, with a $100$ N downward load applied at $C$. By symmetry, each support carries a $50$ N vertical reaction.

At joint $C$, members $AC$ and $BC$ meet the load at the same angle from horizontal, $\theta ={\tan }^{-1}(3/2)\approx 56.3°$. Both members push up on the joint (compression) to balance the $100$ N load, and their horizontal components cancel by symmetry:

$$2F\sin \theta =100⟹F=\frac{100}{2\sin 56.3°}\approx 60.1\text{ N (compression)}$$

Answer: Members $AC$ and $BC$ each carry about $60.1$ N in compression.

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

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

Because all forces pass through the same point, the moment equation is automatically satisfied - equilibrium only requires $\sum F_x=0$ and $\sum F_y=0$ from the equilibrium conditions above.

Example:

Three concurrent forces: $100$ N east, $100$ N north, and $141.4$ N southwest.

$$\begin{gathered} \sum F_x=100-141.4\cos 45°=100-100=0 \\ \sum F_y=100-141.4\sin 45°=100-100=0 \end{gathered}$$

Answer: The forces satisfy both equilibrium equations, so the system is in equilibrium.