Mechanics: Statics formulary

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Unit 1Basic Definitions and Vectors

A vector perpendicular to both, with magnitude $V_1 V_2 \sin\theta$ .

A scalar; zero when the vectors are perpendicular.

Force is mass times acceleration; in statics the net force is zero, so $\vec{a} = 0$ .

A downward force; $g \approx 9.81\ \mathrm{m/s^2}$ .

Unit 2Geometry and Trigonometry

The cosines satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ .

For resolving force triangles that are not right-angled.

Unit 3Forces, Moments and Centroids

Replace $dA$ with $dw$ , $dm$ , or $dV$ for weight, mass, or volume centroids.

Split a shape into known pieces; subtract holes with a negative area.

Centroid sits one quarter of the height up from the base.

$a$ semi-major, $b$ semi-minor axis.

Build the unit vector from $A$ to $B$ , then scale it by the force magnitude.

$\bar{y}$ measured from the flat circular face.

Stiffer (larger $EA$ ) or shorter members stretch less under the same force.

The expansion of $\vec{r} \times \vec{F}$ .

$d$ is the perpendicular lever arm. Summing moments about a smart point removes unknowns.

Seen in Practice exam 3483.

$\vec{r}$ runs from the point to any point on the force line.

Apply a unit virtual load to find a single displacement.

Centroid distances measured from the right-angle corner.

$\bar{y}$ measured from the flat edge.

$b$ members, $n$ joints. Satisfied means statically determinate.

A member is zero-force when a joint has two non-collinear members and no load.

Seen in Practice exam 3483.

$F_j$ are the real member forces, $f_j$ the forces from the unit virtual load.

Unit 4Internal Forces

A vertical strip has its own centroid at half its height.

Differentiate the moment to get shear, again to get the distributed load.

Each step raises the order of the curve: a constant load gives linear shear and a parabolic moment. See the shear and moment diagrams note.

Seen in Practice exam 3483.

The second moment of area; measures resistance to bending.

Shift an inertia from the centroidal axis to a parallel axis a distance $d$ away.

Always add when moving away from the centroid.

For a rectangle, $J_O = \tfrac{bh(h^2 + b^2)}{12}$ .

Seen in Practice exam 3483.

Zero when either axis is an axis of symmetry.

The centroidal value is smaller because of the parallel axis shift.

Centroidal value about an axis through the centroid, parallel to the base.

Stack disks along the axis of revolution and integrate.

Tap a card to unfold the notes.

Unit 1Basic Definitions and Vectors

Angle between two vectorscos⁡α=V⃗1⋅V⃗2V1V2\cos\alpha = \frac{\vec{V}_1 \cdot \vec{V}_2}{V_1 V_2}cosα=V1​V2​V1​⋅V2​​

Cross productV⃗3=V⃗1×V⃗2=∣i⃗j⃗k⃗V1xV1yV1zV2xV2yV2z∣\vec{V}_3 = \vec{V}_1 \times \vec{V}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ V_{1x} & V_{1y} & V_{1z} \\ V_{2x} & V_{2y} & V_{2z} \end{vmatrix}V3​=V1​×V2​=​iV1x​V2x​​j​V1y​V2y​​kV1z​V2z​​​

Dot productV⃗1⋅V⃗2=V1V2cos⁡θ=V1xV2x+V1yV2y+V1zV2z\vec{V}_1 \cdot \vec{V}_2 = V_1 V_2 \cos\theta = V_{1x} V_{2x} + V_{1y} V_{2y} + V_{1z} V_{2z}V1​⋅V2​=V1​V2​cosθ=V1x​V2x​+V1y​V2y​+V1z​V2z​

Newton's second lawF⃗=m⋅a⃗\vec{F} = m \cdot \vec{a}F=m⋅a

Vector magnitude∣V⃗∣=Vx2+Vy2+Vz2|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}∣V∣=Vx2​+Vy2​+Vz2​​

WeightW⃗=m⋅g⃗\vec{W} = m \cdot \vec{g}W=m⋅g​

Unit 2Geometry and Trigonometry

Direction cosinesFx=Fcos⁡α,Fy=Fcos⁡β,Fz=Fcos⁡γF_x = F\cos\alpha, \quad F_y = F\cos\beta, \quad F_z = F\cos\gammaFx​=Fcosα,Fy​=Fcosβ,Fz​=Fcosγ

Law of cosinesa=b2+c2−2bccos⁡αa = \sqrt{b^2 + c^2 - 2 b c \cos\alpha}a=b2+c2−2bccosα​

Law of sinesasin⁡α=bsin⁡β=csin⁡γ\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}sinαa​=sinβb​=sinγc​

SOHCAHTOAsin⁡α=opphyp,cos⁡α=adjhyp,tan⁡α=oppadj\sin\alpha = \frac{\text{opp}}{\text{hyp}}, \quad \cos\alpha = \frac{\text{adj}}{\text{hyp}}, \quad \tan\alpha = \frac{\text{opp}}{\text{adj}}sinα=hypopp​,cosα=hypadj​,tanα=adjopp​

Spherical force componentsFx=Fcos⁡θsin⁡ϕ,Fy=Fcos⁡ϕ,Fz=Fsin⁡θsin⁡ϕF_x = F\cos\theta\sin\phi, \quad F_y = F\cos\phi, \quad F_z = F\sin\theta\sin\phiFx​=Fcosθsinϕ,Fy​=Fcosϕ,Fz​=Fsinθsinϕ

Unit 3Forces, Moments and Centroids

Centroid by integrationxˉ=∫x dA∫dA,yˉ=∫y dA∫dA\bar{x} = \frac{\int x \, dA}{\int dA}, \qquad \bar{y} = \frac{\int y \, dA}{\int dA}xˉ=∫dA∫xdA​,yˉ​=∫dA∫ydA​

Circle: areaA=πr2=πd24A = \pi r^2 = \frac{\pi d^2}{4}A=πr2=4πd2​

Composite centroidXˉ=∑xˉi Ai∑Ai\bar{X} = \frac{\sum \bar{x}_i\, A_i}{\sum A_i}Xˉ=∑Ai​∑xˉi​Ai​​

Cone: volume and centroidV=13πr2h,yˉ=h4V = \tfrac{1}{3}\pi r^2 h, \quad \bar{y} = \frac{h}{4}V=31​πr2h,yˉ​=4h​

Cylinder: volume and centroidV=πr2h,yˉ=h2V = \pi r^2 h, \quad \bar{y} = \frac{h}{2}V=πr2h,yˉ​=2h​

Ellipse: areaA=πabA = \pi a bA=πab

Force along a directionu⃗=AB⃗∣AB⃗∣,F⃗=F⋅u⃗\vec{u} = \frac{\vec{AB}}{|\vec{AB}|}, \qquad \vec{F} = F \cdot \vec{u}u=∣AB∣AB​,F=F⋅u

Hemisphere: volume and centroidV=23πr3,yˉ=38rV = \tfrac{2}{3}\pi r^3, \quad \bar{y} = \frac{3}{8} rV=32​πr3,yˉ​=83​r

Member elongationΔL=Fk,k=EAL0\Delta L = \frac{F}{k}, \qquad k = \frac{E A}{L_0}ΔL=kF​,k=L0​EA​

Moment componentsMx=yFz−zFy,My=zFx−xFz,Mz=xFy−yFxM_x = y F_z - z F_y, \quad M_y = z F_x - x F_z, \quad M_z = x F_y - y F_xMx​=yFz​−zFy​,My​=zFx​−xFz​,Mz​=xFy​−yFx​

Moment of a force (scalar)(shows up on exams)M=F⋅d=F⋅rcos⁡θM = F \cdot d = F \cdot r \cos\thetaM=F⋅d=F⋅rcosθ

Moment of a force (vector)M⃗=r⃗×F⃗\vec{M} = \vec{r} \times \vec{F}M=r×F

Principle of virtual workP′⋅Δ=∑jpj δj(P′=1)P' \cdot \Delta = \sum_j p_j\, \delta_j \qquad (P' = 1)P′⋅Δ=j∑​pj​δj​(P′=1)

Pyramid: volume and centroidV=13 (base area) h,yˉ=h4V = \tfrac{1}{3}\,(\text{base area})\, h, \quad \bar{y} = \frac{h}{4}V=31​(base area)h,yˉ​=4h​

Quarter circle: area and centroidA=14πr2,xˉ=yˉ=4r3πA = \tfrac{1}{4}\pi r^2, \quad \bar{x} = \bar{y} = \frac{4r}{3\pi}A=41​πr2,xˉ=yˉ​=3π4r​

Rectangle: area and centroidA=bh,xˉ=b2,yˉ=h2A = b h, \quad \bar{x} = \frac{b}{2}, \quad \bar{y} = \frac{h}{2}A=bh,xˉ=2b​,yˉ​=2h​

Right triangle: area and centroidA=12bh,xˉ=b3,yˉ=h3A = \tfrac{1}{2} b h, \quad \bar{x} = \frac{b}{3}, \quad \bar{y} = \frac{h}{3}A=21​bh,xˉ=3b​,yˉ​=3h​

Semicircle: area and centroidA=12πr2,yˉ=4r3πA = \tfrac{1}{2}\pi r^2, \quad \bar{y} = \frac{4r}{3\pi}A=21​πr2,yˉ​=3π4r​

Simple truss condition(shows up on exams)b=2n−3b = 2n - 3b=2n−3

Sphere: volumeV=43πr3V = \tfrac{4}{3}\pi r^3V=34​πr3

Truss joint displacementΔ=∑j=1bfj Fj L0jEj Aj\Delta = \sum_{j=1}^{b} \frac{f_j\, F_j\, L_{0j}}{E_j\, A_j}Δ=j=1∑b​Ej​Aj​fj​Fj​L0j​​

Unit 4Internal Forces

Arc length elementdL=1+(y′)2 dx=R dθdL = \sqrt{1 + (y')^2}\, dx = R\, d\thetadL=1+(y′)2​dx=Rdθ

Centroid of an area by stripsdA=y dx,x~=x,y~=y2dA = y\, dx, \quad \tilde{x} = x, \quad \tilde{y} = \frac{y}{2}dA=ydx,x~=x,y~​=2y​

Circle: moment of inertiaIˉx=Iˉy=πr44=πd464,JO=πr42\bar{I}_x = \bar{I}_y = \frac{\pi r^4}{4} = \frac{\pi d^4}{64}, \quad J_O = \frac{\pi r^4}{2}Iˉx​=Iˉy​=4πr4​=64πd4​,JO​=2πr4​

Ellipse: moment of inertiaIˉx=πab34,Iˉy=πba34\bar{I}_x = \frac{\pi a b^3}{4}, \quad \bar{I}_y = \frac{\pi b a^3}{4}Iˉx​=4πab3​,Iˉy​=4πba3​

Load, shear, and moment relationships(shows up on exams)V=dMdx,q=dVdx=d2Mdx2V = \frac{dM}{dx}, \qquad q = \frac{dV}{dx} = \frac{d^2 M}{dx^2}V=dxdM​,q=dxdV​=dx2d2M​

Moment of inertia (area)Ix=∫y2 dA,Iy=∫x2 dAI_x = \int y^2 \, dA, \qquad I_y = \int x^2 \, dAIx​=∫y2dA,Iy​=∫x2dA

Parallel axis theoremI=Iˉ+Ad2I = \bar{I} + A d^2I=Iˉ+Ad2

Polar moment of inertia(shows up on exams)JO=Ix+Iy=∫r2 dAJ_O = I_x + I_y = \int r^2 \, dAJO​=Ix​+Iy​=∫r2dA

Product of inertiaIxy=∫xy dAI_{xy} = \int x y \, dAIxy​=∫xydA

Quarter circle: moment of inertiaIx,base=Iy,base=πr416,Iˉx≈0.0549 r4I_{x,\text{base}} = I_{y,\text{base}} = \frac{\pi r^4}{16}, \quad \bar{I}_x \approx 0.0549\, r^4Ix,base​=Iy,base​=16πr4​,Iˉx​≈0.0549r4

Radius of gyrationkx=IxA,ky=IyA,kO=JOAk_x = \sqrt{\frac{I_x}{A}}, \quad k_y = \sqrt{\frac{I_y}{A}}, \quad k_O = \sqrt{\frac{J_O}{A}}kx​=AIx​​​,ky​=AIy​​​,kO​=AJO​​​

Rectangle: moment of inertiaIˉx=bh312,Iˉy=hb312,Ix,base=bh33\bar{I}_x = \frac{b h^3}{12}, \quad \bar{I}_y = \frac{h b^3}{12}, \quad I_{x,\text{base}} = \frac{b h^3}{3}Iˉx​=12bh3​,Iˉy​=12hb3​,Ix,base​=3bh3​

Semicircle: moment of inertiaIx,base=πr48,Iˉx≈0.1098 r4I_{x,\text{base}} = \frac{\pi r^4}{8}, \quad \bar{I}_x \approx 0.1098\, r^4Ix,base​=8πr4​,Iˉx​≈0.1098r4

Triangle: moment of inertiaIˉx=bh336,Ix,base=bh312\bar{I}_x = \frac{b h^3}{36}, \qquad I_{x,\text{base}} = \frac{b h^3}{12}Iˉx​=36bh3​,Ix,base​=12bh3​

Volume of revolution (disk)dV=πr2 dydV = \pi r^2 \, dydV=πr2dy