Mechanics of Material Cheat Sheet

= Complete Mechanics of Material Formula Sheet 𝐿𝑓 − 𝐿0 (100) 𝐿0 𝐴0 − 𝐴 𝑓 (100) 𝐴0 ∑𝐹𝑥 = 0; ∑𝐹𝑦 = 0; ∑𝐹𝑧 = 0 ∑𝑀𝑥 = 0; ∑𝑀𝑦 = 0; ∑𝑀𝑧 = 0 ∑𝐹𝑥 = 0; ∑𝐹𝑦 = 0; ∑𝑀𝐴 = 0 Two-Dimensional Problems ∑𝐹𝑥 = 0; ∑𝑀𝐴 = 0; ∑𝑀𝐵 = 0 Alternative Sets: or ∑𝑀𝐴 = 0; ∑𝑀𝐵 = 0; ∑𝑀𝐶 = 0 Hooke’s law = 𝜎 = 𝐸𝜖 𝑃 𝜎= 𝐴 Bearing stress Shearing stress 𝜏𝑎𝑣𝑔 𝑉 = 𝐴 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑠𝑡𝑒𝑛𝑔𝑡ℎ Factor of safety 𝑛𝑠 = 𝑃 𝜎𝑏 = 𝐴𝑝 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑃 𝜎𝑎𝑙𝑙 Required cross-sectional area 𝐴= STRAIN AND MATERIAL PROPERTIES Normal strain Shear strain 𝜖= 𝛿 𝐿 𝜋 𝛾 = − 𝜃′ 2 Δ𝑢 Δ𝑣 𝜖𝑥 = 𝜖 = Δ𝑥 𝑦 Δ𝑦 Δ𝑢 Δ𝑢 + 𝛾𝑥𝑦 = Δ𝑦 Δ𝑥 Components of strain Percent elongation 𝜏 = 𝐺𝛾 𝑒 = (1 − 2𝜈)𝜖𝑥 = ( 1 − 2𝜈 Unit volume change (axial stress) 𝜎𝑦 𝜎𝑥 −𝜈 𝜖𝑥 = 𝐸 𝐸 𝜎𝑦 𝜎𝑥 𝜖𝑦 = −𝜈 𝐸 𝐸 𝜏𝑥𝑦 𝛾𝑥𝑦 = 𝐺 𝐸 ) 𝜎𝑠 1 [𝜎 − 𝜈(𝜎𝑦 + 𝜎𝑧 )] 𝐸 𝑥 1 𝜖𝑦 = [𝜎𝑦 − 𝜈(𝜎𝑥 + 𝜎𝑧 )] 𝐸 1 𝜖𝑧 = [𝜎𝑧 − 𝜈(𝜎𝑥 + 𝜎𝑦 )] 𝐸 𝜏𝑥𝑦 𝛾𝑥𝑦 = 𝐺 𝜏𝑦𝑧 𝛾𝑦𝑧 = 𝐺 𝜏𝑥𝑧 𝛾𝑥𝑧 = 𝐺 𝑈=∫ 𝛿= 𝑛 Multiple prismatic bar 𝐸 3(1 − 2𝜈) Relation between elastic moduli 𝐺= 𝑛 Non-prismatic bar 𝛿=∫ Unit volume change (three-dimensional stress) 𝜖 = 𝜖𝑥 + 𝜖 𝑦 + 𝜖𝑧 1 − 2𝜈 (𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧 ) = 𝐸 𝐸 𝑘= 3(1 − 2𝜈) ____________________________________ Bulk modulus of elasticity Strain energy density (for normal stress) 𝐿𝑃 𝑥 0 𝐴𝑖𝐸𝑖 𝜖𝑥 = 𝛼Δ𝑇 Thermal strain 𝜙=∫ Power 𝑃𝑖 𝐿𝑖 𝐴𝑥 𝐸 Torque 𝑃= 𝑇= 𝑇= 𝑑𝑥 𝜎𝑥′ = 𝜎𝑥 cos2 𝜃 𝜏𝑥′ 𝑦′ = 𝜎𝑥 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 Shear stress 𝛾= Torsion formula 𝜌𝜙 𝜌 = 𝛾𝑚𝑎𝑥 𝑐 𝐿 𝜏= 𝑃 𝐴 𝜌 𝜏 𝑐 𝑚𝑎𝑥 𝑇𝑐 𝐽 𝑇𝜌 𝜏= 𝐽 𝜏𝑚𝑎𝑥 = 𝜎𝑥′ = 𝜏𝑠𝑖𝑛2𝜃 𝑇𝑥 𝑑𝑥 𝐺𝐽𝑥 𝐿 159𝑃 9550𝑃 , 𝑇 = 𝑁𝑚 = 𝑛 𝑓 1050𝑃 63000𝑃 = , 𝑇 = 𝑙𝑏. 𝑖𝑛 𝑓 𝑛 𝐽 𝑇 = 𝑐 𝜏𝑎𝑙𝑙 Required shaft parameters 𝛿𝑡 = 𝛼(Δ𝑇)𝐿 TORSION Shear strain 𝐺𝑖 𝐽𝑖 2𝜋𝑛𝑇 , (𝑛 = 𝑟𝑝𝑚) 60 Normal and shearing stress 𝜎𝑚𝑎𝑥 = 𝐾𝜎𝑛𝑜𝑚 = 𝐾 𝑇𝑖 𝐿𝑖 0 Thermal deformation Maximum normal stress 𝑇𝐿 𝐺𝐽 𝑖=1 Non-prismatic shafts 𝐴𝐸 𝑖=1 𝜙= 𝜙=∑ 𝑃𝐿 𝛿=∑ 𝜏𝑥′ 𝑦′ = 𝜏𝑐𝑜𝑠2𝜃 Angle of twist Prismatic shafts Multiple prismatic shaft AXIAL LOADED MEMBERS Deformation of prismatic bar Generalized Hooke’s law 𝜖𝑥 = 𝜏𝑥𝑦2 𝑑𝑉 2𝐺 Strain energy (for shear stress) Hooke’s law (two dimensional stress) CONCEPT OF STRESS Normal stress 𝑈=∫ 𝜖𝑦 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =− 𝜈=− 𝑎𝑥𝑖𝑎𝑙 𝑠𝑡𝑎𝑖𝑛 𝜖𝑥 Poisson’s ratio 𝜎𝑥2 𝑑𝑉 2𝐸 Strain energy (for normal stress) Percent reduction in area Equations of Equilibrium Three-Dimensional Problems 𝜎𝑥2 2𝐸 𝑈0 = Equations of transformation for stress Maximum shearing stress 𝜏𝑚𝑎𝑥 = 𝐾𝜏𝑛𝑜𝑚 = 𝐾 Yield torque 𝑇𝑦 = 𝑇𝑐 𝐽 𝜋𝑐 3 𝐽 𝜏 = 𝜏 2 𝑦 𝑐 𝑦 𝜋𝑐 3 𝜌03 (4 − 3 ) 𝜏𝑦 𝑐 6 4 1 𝜌03 ) 𝑇𝑦 = 𝑇𝑦 (1 − 3 4 𝑐3 Total torque 𝑇= Ultimate torque 𝑇𝑢 = 2 3 4 𝜋𝑐 𝜏𝑦 = 𝑇𝑦 3 3 Rectangular bars Maximum shear stress 𝜏𝑚𝑎𝑥 = 𝑇 𝛼𝑎𝑏 2 𝑀𝑦 𝑀𝑐 𝑀 𝜎𝑚𝑎𝑥 = = 𝑆 𝐼 𝐼 Maximum bending stress at a notch 𝑀𝑐 ) 𝜎𝑚𝑎𝑥 = 𝐾𝜎𝑛𝑜𝑚 = 𝐾 ( 𝐼 SHEAR AND MOMENT IN BEAMS Shear formula Shear force and load relation 𝑉𝑄 𝑑𝑉 𝜏𝑥𝑦 = =𝑤 𝐼𝑏 𝑑𝑥 𝐵 First moment of (shaded) cross𝑉𝐵 − 𝑉𝐴 = ∫ 𝑤𝑑𝑥 sectional area 𝐴 𝑄 = 𝐴∗ 𝑦 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑙𝑜𝑎𝑑 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴&𝐵 Shear flow Moment and shear force relation 𝑞 = 𝑉𝑄/𝐼 𝑑𝑀 =𝑉 Required section modulus 𝑑𝑥 𝐵 𝑀𝑚𝑎𝑥 𝑆= 𝑀𝐵 − 𝑀𝐴 = ∫ 𝑉𝑑𝑥 𝜎𝑎𝑙𝑙 𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑒𝑎𝑟 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴&𝐵 Combined normal stress 𝑀𝑧 𝑦 𝑀𝑦 𝑧 Macauley function + 𝜎𝑥 = − 0 𝑤ℎ𝑒𝑛 𝑥 < 𝑎 𝐼𝑦 𝐼𝑧 〈𝑥 − 𝑎 〉𝑛 = { 𝑛 (𝑥 − 𝑎) 𝑤ℎ𝑒𝑛 𝑥 ≥ 𝑎 Equation of the neutral axis Step function 𝑀𝑦 𝐼𝑧 0 𝑤ℎ𝑒𝑛 𝑥 < 𝑎 𝑦= 〈𝑥 − 𝑎 〉0 = { 𝑀𝑧 𝐼𝑦 1 𝑤ℎ𝑒𝑛 𝑥 ≥ 𝑎 0 𝑤ℎ𝑒𝑛 𝑥 < 𝑎 Angle defining orientation of the 〈𝑥 − 𝑎〉1 = { 𝑥 − 𝑎 𝑤ℎ𝑒𝑛 𝑥 ≥ 𝑎 neutral axis Rule of integration 𝐼𝑧 𝑛+1 𝑡𝑎𝑛𝜙 = 𝑡𝑎𝑛𝛼 〈𝑥 − 𝑎〉 𝐼𝑦 ∫ 〈𝑥 − 𝑎〉𝑛 𝑑𝑥 = 𝑓𝑜𝑟 𝑛 ≥ 0 Angle of twist 𝑇 𝛽𝛼3 𝐺 𝜙= 𝑛+1 Singularity functions Applied force 0 𝑤ℎ𝑒𝑛 𝑥 ≠ 𝑎 𝑤(𝑥) = 𝑃〈𝑥 − 𝑎〉1 = { 𝑃 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 0 𝑤ℎ𝑒𝑛 𝑥 ≠ 𝑎 𝑤(𝑥) = 𝑀0 〈𝑥 − 𝑎〉−2 = { 𝑀0 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 Rule of integration Applied moment ∫ 〈𝑥 − 𝑎〉𝑛 𝑑𝑥 = 〈𝑥 − 𝑎〉𝑛 𝑓𝑜𝑟 𝑛 ≤ 0 (𝑛 = −1, −2, … . ) STRESS IN BEAMS 𝜖𝑥 = − Normal strain Flexure formula 𝑦 = −𝐾𝑦 𝜌 𝜎𝑥 = − Inelastic bending Yield moment 𝐼 1 𝑀𝑦 = 𝜎𝑦 = 𝑆𝜎𝑦 = 𝑏ℎ2 𝜎𝑦 𝐶 6 Elastic-plastic moment 2 1 𝑦 3 0 ) ) 𝑀 = 𝑀𝑦 (1 − ( ⁄ 2 3 ℎ2 Plastic moment 3 1 𝑀𝑝 = 𝑀𝑢 = 𝑀𝑦 = 𝑏ℎ2 𝜎𝑦 4 2 Shape factor 𝐴 𝑅 = 𝑑𝐴 ∫𝐴 𝑟 Distance between centroid and neutral axis 𝑒 = 𝑟 − 𝑅 𝑀(𝑅 − 𝑟) 𝐴𝑒𝑟 Combined normal stress 𝑃 𝑀(𝑅 − 𝑟) 𝜎= − 𝐴 𝐴𝑒𝑟 Normal stress 𝜎= − 𝜎𝑎 = Spherical vessels Tangential stress 𝑝𝑟 2𝑡 𝑝𝑟 2𝑡 Equation for transformation for strain 𝜎= 𝛾𝑥𝑦 𝜖 𝑥 + 𝜖𝑦 𝜖 𝑥 − 𝜖𝑦 + cos 2𝜃 + sin 2𝜃 2 2 2 𝛾𝑥𝑦 𝜖 𝑥 + 𝜖𝑦 𝜖 𝑥 − 𝜖𝑦 − cos 2𝜃 − 𝜖𝑦 ′ = sin 2𝜃 2 2 2 𝛾𝑥′ 𝑦′ = −(𝜖𝑥 − 𝜖𝑦 ) sin 2𝜃 + 𝛾𝑥𝑦 𝑐𝑜𝑠2𝜃 𝜖𝑥 ′ = TRANSFORMATION OF STRESS AND STRAIN Equations of transformation for stress Principal strains 𝜎𝑥 ′ = 1 1 (𝜎 + 𝜎𝑦 ) + (𝜎𝑥 − 𝜎𝑦 ) cos 2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃 2 𝑥 2 𝜏𝑥 ′ 𝑦 ′ 𝜎𝑦′ = 1 = − (𝜎𝑥 𝜎 ) sin 2𝜃 + 𝜏𝑥𝑦 cos 2𝜃 𝑦 2 1 1 (𝜎 + 𝜎𝑦 ) − (𝜎𝑥 − 𝜎𝑦 ) cos 2𝜃 − 𝜏𝑥𝑦 sin 2𝜃 2 𝑥 2 Principal stresses 𝜎𝑚𝑎𝑥,min = 𝜎1,2 = 𝜎𝑥 + 𝜎𝑦 2 ± √( 2 𝜎𝑥 − 𝜎𝑦 ) + 𝜏2𝑥𝑦 2 Planes of principal stresses 2𝜏𝑥𝑦 tan 2𝜃𝑝 = 𝜎𝑥 − 𝜎𝑦 Average normal stress 𝜎𝑥 + 𝜎𝑦 𝜎𝑎𝑣𝑔 = 𝜎 ′ = 2 Maximum shear stresses 𝜏max = ±√( 𝜎𝑥 −𝜎𝑦 2 2 2 = ) + 𝜏𝑥𝑦 𝜎1 −𝜎2 2 Planes of maximum shear stresses 𝜎𝑥 − 𝜎𝑦 tan 2𝜃𝑠 = − 2𝜏𝑥𝑦 Radius of Mohr’s circle for stress 𝑅 = √( 𝜎𝑥 − 𝜎𝑦 2 2 ) + 𝜏𝑥𝑦 2 𝜖1,2 = 𝜖 𝑥 − 𝜖𝑦 2 𝛾𝑥𝑦 2 𝜖 𝑥 + 𝜖𝑦 ) = 𝜖1 − 𝜖2 ± √( ) +( 2 2 2 Planes of principal strains 𝛾𝑥𝑦 tan 2𝜃𝑝 = 𝜖𝑥− 𝜖𝑦 Maximum shear strains 𝛾max = ±2√( 𝜖 𝑥 − 𝜖𝑦 2 𝛾𝑥𝑦 2 ) ) +( 2 2 Absolute maximum shear strain (𝛾max)𝑎 = 𝜖1 − 𝜖3 Radius of Mohr’s circle of strain 𝑅 = √( 𝜖𝑥 − 𝜖 𝑦 2 𝛾𝑥𝑦 2 ) ) +( 2 2 ____________________________________________________ COMBINNED LOADINGS AND FAILURE CRITERIAS Normal stress due to axial loading Torsion formula 𝜎𝑥 = 𝜏= 𝑃 𝐴 𝑇𝑟 𝐽 True or absolute maximum shear stress Flexure formula 1 𝑀𝑐 (𝜏max)𝑎 = (𝜏13)𝑎 = (𝜎1 − 𝜎3 ) 𝜎𝑥 = 2 𝐼 Cylindrical vessels Shear formula 𝑉𝑄 Tangential stress 𝜏= 𝑝𝑟 𝐼𝑏 𝜎𝑡 = 𝑡 Axial stress Tangential stress in cylindrical vessels 𝑝𝑟 𝜎𝑡 = 𝑡 Axial stress in cylindrical vessels 𝑝𝑟 𝜎𝑎 = 2𝑡 Tangential stress in spherical vessels 𝑝𝑟 𝜎= 2𝑡 Combined normal and shear stress 𝜎1,2 = 𝜎𝑥 𝜎𝑥 2 2 ± √( ) + 𝜏𝑥𝑦 2 2 𝜏𝑚𝑎𝑥 = √( 𝜎𝑥 2 2 ) + 𝜏𝑥𝑦 2 Planes of principal stresses 2𝜏𝑥𝑦 𝑡𝑎𝑛2𝜃𝑃 = 𝜎𝑥 Maximum shear stress in helical spring 16𝑃𝑅 𝑑 ) 𝜏𝑚𝑎𝑥 = (1 + 4𝑅 𝜋𝑑3 Total deflection of a helical spring 𝑃 𝑀𝑧 𝑦 𝑀𝑦 𝑧 𝛿= − + 𝐼𝑧 𝐼𝑦 𝐴 𝑃 𝑀𝑦 𝜎𝑥 = − 𝐼 𝐴 Equation of neutral axis 𝑃 𝑀𝑦 − =0 𝐴 𝐼 Maximum shear stress theory |𝜎1 − 𝜎2 | = 𝜎𝑦 Maximum energy of distortion theory 2 𝜎1 − 𝜎1 𝜎2 + 𝜎22 = 𝜎𝑦2 Maximum normal stress theory |𝜎1 | = 𝜎𝑢 𝑜𝑟 |𝜎2 | = 𝜎𝑢 DEFLECTION IN BEAMS Equation of the deflection curve 𝑑2𝑣 𝑀 = 𝑑𝑥 2 𝐸𝐼 Basic equation for the elastic curve 𝑑2𝑣 =𝑀 𝐸𝐼 𝑑𝑥 2 𝑑3𝑣 𝐸𝐼 3 = 𝑉 𝑑𝑥 𝑑4𝑣 =𝑤 𝐸𝐼 𝑑𝑥 4 Boundary conditions Fixed or clamped support 1 𝜎𝑦 𝐿𝑒 2 ) 𝜎𝑐𝑟 = 𝜎𝑦 = ( 𝑣(𝑎) = 0 𝐸 2𝜋 𝑟 ′ 𝜃(𝑎) = 𝑣 (𝑎) = 0 Secant formula Simple support 𝑃 𝑒𝑐 𝜋 𝑃 𝑣(𝑎) = 0 𝜎𝑚𝑎𝑥 = [1 + 2 sec ( √ )] 𝑀(𝑎) = 𝐸𝐼𝑣 ′′ (𝑎) = 0 𝑟 𝐴 2 𝑃𝑐𝑟 Free end 𝑀(𝑎) = 𝐸𝐼𝑣 ′′ (𝑎) = 0 𝑉(𝑎) = 𝐸𝐼𝑣 ′′′ (𝑎) = 0 ENERGY METHODS AND IMPACT Guided support Strain energy in bars 𝑣 ′ (𝑎) = 0 Bars with variable cross section and 𝑉(𝑎) = 𝐸𝐼𝑣 ′′′ (𝑎) = 0 varying axial loads First moment-area theorem 𝜃𝐵 𝐴⁄ 𝐿2 𝑃 𝑥 𝑑𝑥 𝑀 𝑈=∫ 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴 𝑎𝑛𝑑 𝐵 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝐸𝐼 0 2𝐴𝑥 𝐸 Second moment-area theorem 𝑡𝐵⁄𝐴 Prismatic bars subjected to end 𝑀 = (𝑎𝑟𝑒𝑎 𝑜𝑓 Loads 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴 𝑎𝑛𝑑 𝐵) 𝑥1 𝐸𝐼 2 BUCKLING OF COLUMNS Critical load Pin-ended columns 𝜋 2 𝐸𝐼 𝐿2 𝜋 2 𝐸𝐴 𝑃𝑐𝑟 = (𝐿 𝑟⁄ )2 𝑃𝑐𝑟 = Critical load Columns with other end conditions 𝜋 2 𝐸𝐼 𝑃𝑐𝑟 = 2 𝐿𝑒 𝜋 2 𝐸𝐴 𝑃𝑐𝑟 = (𝐿𝑒⁄ 𝑟)2 Critical stress Long columns 𝜋 2𝐸 𝑃𝑐𝑟 = 𝜎𝑐𝑟 = (𝐿𝑒 ⁄𝑟 )2 𝐴 Short columns 𝑃 𝜎𝑚𝑎𝑥 = 𝐴 Intermediate columns Tangent modulus formula 𝑃𝑡 𝜋 2 𝐸𝑡 𝜎𝑐𝑟 = = 𝐴 (𝐿𝑒 ⁄𝑟 )2 Johnson’s formula 𝑃 𝐿 2𝐴𝐸 Strain energy in shafts Shafts with variable cross section and varying toques 𝑇𝐿𝑥2 𝑑𝑥 𝑈=∫ 0 2𝐺𝐽𝑥 Prismatic shafts subjected to end torques 𝑇2𝐿 𝑈= 2𝐺𝐽 Strain energy in bending and shear Pure bending 𝐿 2 𝑑𝑥 𝑀 𝑈=∫ 0 2𝐸𝐼 Pure shear 𝐿 2 𝑑𝑥 𝑇 𝑈=∫ 0 2𝐴𝐺 Conservation of energy Linearly elastic member 1 𝑈 = 𝑊 = Σ𝑝𝑘 𝛿𝑘 2 Displacement by the work-energy method Axially loaded bars 𝑈= Torsion bars 𝛿= 𝜙= 𝑃𝐿 𝐴𝐸 𝑇𝐿 𝐺𝐽 Beams in pure bending 𝑀𝐿 𝜃= 𝐸𝐼 Castigliano’s Theorem Displacement 𝜕𝑈 𝛿𝑖 = 𝜕𝑃𝑖 Slope 𝜕𝑈 𝜃𝑖 = 𝜕𝑀𝑖 Angle of twist 𝜕𝑈 𝜙𝑖 = 𝜕𝑇𝑖 Impact factor 𝐾 = 1 + √1 + Maximum dynamic Load 𝑃𝑚𝑎𝑥 = 𝐾𝑊 Stress 𝜎𝑚𝑎𝑥 = 𝐾𝜎𝑠𝑡 Deflection 𝛿𝑚𝑎𝑥 = 𝐾𝛿𝑠𝑡 2ℎ 𝛿𝑠𝑡 FINITE ELEMENT ANALYSIS Global force-displacement relations for an element {𝐹}𝑒 = [𝑘]𝑒 {𝛿 }𝑒 Force matrix 𝐹1𝑥 𝐹 {𝐹 }𝑒 = 1𝑦 𝐹2𝑥 { 𝐹2𝑦 } Stiffness matrix for an element 𝑐2 𝑐𝑠 −𝑐 2 −𝑐𝑠 2 −𝑐𝑠 −𝑠2 ] [𝑘]𝑒 = [ 𝑐𝑠2 𝑠 −𝑐 −𝑐𝑠 𝑐 2 𝑐𝑠 2 −𝑐𝑠 −𝑠 𝑐𝑠 𝑠 2 2 𝑐 𝑐𝑠 −𝑐2 −𝑐𝑠 𝐴𝐸 𝑠 2 −𝑐𝑠 −𝑠2 ] = 𝐿 [ 𝑐 2 𝑐𝑠 𝑠2 Axial force in the bar element 𝑢𝑗 − 𝑢𝑖 𝐴𝐸 𝐹𝑖𝑗 = ( ) [𝑐 𝑠]𝑖𝑗 { 𝑣 − 𝑣 } 𝑗 𝑖 𝐿 𝑖𝑗 𝐴𝐸 𝐿 System equations {𝐹} = [𝐾 ]{𝛿 } Global nodal force matric 𝑛 {𝐹} = ∑{𝐹 }𝑒 1 𝑛 Global stiffness matrix [𝑘] = ∑[𝑘 ]𝑒 1