Mechanics of Material Cheat Sheet (Autosaved) PDF

Preview

Summary

Mechanics of Material Cheat Sheet (Autosaved).pdf...

Description

=

Complete Mechanics of Material Formula Sheet (By KIM KINAL)

𝐿𝑓 − 𝐿0 𝐿0

(100)

𝐴0 − 𝐴𝑓 (100) = 𝐴0

Percent reduction in area Equations of Equilibrium Three-Dimensional Problems

∑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

𝜎𝑏 =

𝑃 𝐴𝑝

𝜏𝑎𝑣𝑔 =

𝑉 𝐴

𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑠𝑡𝑒𝑛𝑔𝑡ℎ 𝑛𝑠 = 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

Factor of safety

𝑃 𝜎𝑎𝑙𝑙

Required cross-sectional area

𝐴=

STRAIN AND MATERIAL PROPERTIES Normal strain

Shear strain

𝜖=

𝛾=

𝛿 𝐿

𝜋 − 𝜃′ 2

Δ𝑢 Δ𝑣 𝜖𝑥 = 𝜖 = Δ𝑥 𝑦 Δ𝑦 Δ𝑢 Δ𝑢 𝛾𝑥𝑦 = + Δ𝑦 Δ𝑥

Components of strain

Percent elongation

𝜖𝑦 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =− 𝑎𝑥𝑖𝑎𝑙 𝑠𝑡𝑎𝑖𝑛 𝜖𝑥

Poisson’s ratio

𝜈=−

1 − 2𝜈 ) 𝜎𝑠 𝑒 = (1 − 2𝜈)𝜖𝑥 = ( 𝐸

Unit volume change (axial stress)

𝜎𝑦 𝜎𝑥 −𝜈 𝐸 𝐸 𝜎𝑦 𝜎𝑥 𝜖𝑦 = −𝜈 𝐸 𝐸 𝜏𝑥𝑦 𝛾𝑥𝑦 = 𝐺

𝜖𝑥 =

Hooke’s law (two dimensional stress)

CONCEPT OF STRESS Normal stress

𝜏 = 𝐺𝛾

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

𝑈0 =

𝐸

𝑈=∫

𝜏𝑥𝑦2 𝑈=∫ 𝑑𝑉 2𝐺

𝑃𝑖 𝐿𝑖 𝛿=∑ 𝐴𝑖𝐸𝑖

𝑘=

𝐸

3(1 − 2𝜈) ____________________________________

Strain energy density (for normal stress)

Torque

𝐿

𝑃=

𝑇=

𝑖=1

0

𝑃𝑥 𝑑𝑥 𝐴𝑥 𝐸

𝑇=

𝜖𝑥 = 𝛼Δ𝑇

Thermal strain

𝜎𝑥 ′ = 𝜎𝑥 cos2 𝜃 𝜏𝑥 ′ 𝑦 ′ = 𝜎𝑥 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃

𝜌𝜙 𝜌 = 𝛾𝑚𝑎𝑥 𝛾= 𝑐 𝐿

Torsion formula

𝜏=

𝑃 𝐴

𝜌 𝜏 𝑐 𝑚𝑎𝑥

𝑇𝑐 𝐽 𝑇𝜌 𝜏= 𝐽

𝜏𝑚𝑎𝑥 =

𝜎𝑥 ′ = 𝜏𝑠𝑖𝑛2𝜃

𝑇𝑥 𝑑𝑥 𝐺𝐽𝑥

𝐿

2𝜋𝑛𝑇 , (𝑛 = 𝑟𝑝𝑚) 60

159𝑃 9550𝑃 = , 𝑇 = 𝑁𝑚 𝑓 𝑛

1050𝑃 63000𝑃 , 𝑇 = 𝑙𝑏. 𝑖𝑛 = 𝑛 𝑓

𝑇 𝐽 = 𝑐 𝜏𝑎𝑙𝑙

Normal and shearing stress

𝜎𝑚𝑎𝑥 = 𝐾𝜎𝑛𝑜𝑚 = 𝐾

𝑇𝑖 𝐿𝑖 𝐺𝑖 𝐽𝑖

Required shaft parameters

𝛿𝑡 = 𝛼(Δ𝑇)𝐿

Maximum normal stress

𝑖=1

0

Thermal deformation

Shear stress

Bulk modulus of elasticity

𝜙=∫

Power

𝑛

Unit volume change (three-dimensional stress)

𝜖 = 𝜖𝑥 + 𝜖𝑦 + 𝜖𝑧 1 − 2𝜈 (𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧 ) = 𝐸

Non-prismatic shafts

Multiple prismatic bar

𝛿=∫

𝑇𝐿 𝐺𝐽

𝜙=∑

𝑃𝐿 𝛿= 𝐴𝐸

Non-prismatic bar

𝜙=

Multiple prismatic shaft

AXIAL LOADED MEMBERS Deformation of prismatic bar

TORSION Shear strain

3(1 − 2𝜈)

𝜏𝑥 ′ 𝑦 ′ = 𝜏𝑐𝑜𝑠2𝜃

Angle of twist Prismatic shafts

𝑛

Strain energy (for shear stress)

Relation between elastic moduli

𝐺=

𝜎𝑥2 𝑑𝑉 2𝐸

Strain energy (for normal stress)

Generalized Hooke’s law

𝜖𝑥 =

𝜎𝑥2 2𝐸

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 )𝑇 = 𝑇𝑦 (1 − 4 𝑐3 𝑦 3

Total torque

𝑇=

Ultimate torque

𝑇𝑢 =

2 3 4 𝜋𝑐 𝜏𝑦 = 𝑇𝑦 3 3

Rectangular bars Maximum shear stress

𝜏𝑚𝑎𝑥 =

𝑇 𝛼𝑎𝑏 2

𝑇

𝑀𝑦 𝑆 𝑀 𝑀𝑐a notch Maximum 𝜎𝑥 = −bending stress at = 𝜎𝑚𝑎𝑥 = 𝐼 𝐼 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 𝐼𝑧 𝐼𝑦 𝑛 〈𝑥 − 𝑎〉 = { (𝑥 − 𝑎)𝑛 𝑤ℎ𝑒𝑛 𝑥 ≥ 𝑎 Equation of the neutral axis Step function 𝑀𝑦 𝐼𝑧 𝑦= 𝑤ℎ𝑒𝑛 𝑥 < 𝑎 〈𝑥 − 𝑎〉0 = {0 𝑀𝑧 𝐼𝑦 1 𝑤ℎ𝑒𝑛 𝑥 ≥ 𝑎 0 𝑤ℎ𝑒𝑛 𝑥 < 𝑎 Angle defining orientation of the 1 〈𝑥 − 𝑎〉 = { 𝑥 − 𝑎 𝑤ℎ𝑒𝑛 𝑥 ≥ 𝑎 neutral axis Rule of integration 𝐼𝑧 𝑡𝑎𝑛𝜙 = 𝑡𝑎𝑛𝛼 〈𝑥 − 𝑎〉𝑛+1 𝐼𝑦 ∫ 〈𝑥 − 𝑎〉𝑛 𝑑𝑥 = 𝑓𝑜𝑟 𝑛 ≥ 0 𝑛+1 Inelastic bending Singularity functions Yield moment Applied force 1 𝐼 𝑤(𝑥) = 𝑃〈𝑥 − 𝑎〉1 = { 0 𝑤ℎ𝑒𝑛 𝑥 ≠ 𝑎 𝑀𝑦 = 𝜎𝑦 = 𝑆𝜎𝑦 = 𝑏ℎ2 𝜎𝑦 𝑃 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 6 𝐶 Applied moment Elastic-plastic moment 0 𝑤ℎ𝑒𝑛 𝑥 ≠ 𝑎 3 1 𝑦0 2 𝑤(𝑥) = 𝑀0 〈𝑥 − 𝑎〉−2 = { 𝑀0 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 𝑀 = 𝑀𝑦 (1 − ( ) ) 3 ℎ2 ⁄ 2 Rule of integration Plastic moment ∫ 〈𝑥 − 𝑎〉𝑛 𝑑𝑥 = 〈𝑥 − 𝑎〉𝑛 𝑓𝑜𝑟 𝑛 ≤ 0 1 3 (𝑛 = −1, −2, … . ) 𝑀𝑝 = 𝑀𝑢 = 𝑀𝑦 = 𝑏ℎ2 𝜎𝑦 2 4 Shape factor STRESS IN BEAMS 𝐴 𝑅= Normal strain 𝑑𝐴 𝑦 ∫ 𝐴 𝑟 𝜖𝑥 = − = −𝐾𝑦 𝜌 Distance between centroid and neutral Flexure formula axis Angle of twist

𝜙=

𝛽𝛼3 𝐺

Normal stress 𝑒 = 𝑟 − 𝑅

𝑝𝑟 Spherical vessels𝜎𝑎 = 2𝑡 Tangential stress

𝑀(𝑅 − 𝑟) 𝐴𝑒𝑟 𝜎= − Combined normal stress 𝑃 𝑀(𝑅 − 𝑟) 𝜎= 𝐴𝑒𝑟 − 𝐴

𝑝𝑟 𝜎 = 2𝑡 Equation for transformation for strain

TRANSFORMATION OF STRESS AND STRAIN Equations of transformation for stress 𝜎𝑥 ′ =

1 1 (𝜎 + 𝜎𝑦 ) + (𝜎𝑥 − 𝜎𝑦 ) cos 2𝜃 + 𝜏𝑥𝑦𝑠𝑖𝑛2𝜃 2 𝑥 2

𝜏𝑥 ′ 𝑦 ′

𝜎 𝑦′ =

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 =

2 𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2 ) + 𝜏 𝑥𝑦 ± √( 2 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

𝑅 = √(

Principal strains 𝜖1,2 =

𝜖𝑥 + 𝜖𝑦 𝜖𝑥 − 𝜖𝑦 2 𝛾𝑥𝑦 2 ± √( ) +( ) = 𝜖1 − 𝜖2 2 2 2

Planes of principal strains 𝛾𝑥𝑦 tan 2𝜃𝑝 = 𝜖𝑥− 𝜖𝑦 Maximum shear strains 𝛾max = ±2√(

𝜖𝑥 − 𝜖𝑦 2 𝛾𝑥𝑦 2 ) +( ) 2 2

Absolute maximum shear strain (𝛾max)𝑎 = 𝜖1 − 𝜖3 Radius of Mohr’s circle of strain

𝜎1−𝜎2 2

Planes of maximum shear stresses 𝜎𝑥 − 𝜎𝑦 tan 2𝜃𝑠 = − 2𝜏𝑥𝑦 Radius of Mohr’s circle for stress 𝜎𝑥 − 𝜎𝑦 2 ) + 𝜏𝑥𝑦 2 2

𝜖𝑥 + 𝜖𝑦 𝜖𝑥 − 𝜖𝑦 𝛾𝑥𝑦 sin 2𝜃 + cos 2𝜃 + 2 2 2 𝜖𝑥 + 𝜖𝑦 𝜖𝑥 − 𝜖𝑦 𝛾𝑥𝑦 𝜖𝑦 ′ = − cos 2𝜃 − sin 2𝜃 2 2 2 𝛾𝑥 ′𝑦′ = −(𝜖𝑥 − 𝜖𝑦 ) sin 2𝜃 + 𝛾𝑥𝑦 𝑐𝑜𝑠2𝜃 𝜖𝑥 ′ =

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 = 𝑡vessels Axial stress in cylindrical 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

𝜎𝑥 2 𝜏𝑚𝑎𝑥 = √( ) + 𝜏𝑥𝑦2 2

Planes of principal stresses 2𝜏𝑥𝑦 𝑡𝑎𝑛2𝜃𝑃 = 𝜎𝑥 Maximum shear stress in helical spring 16𝑃𝑅 𝑑 ) 𝜏𝑚𝑎𝑥 = (1 + 4𝑅 𝜋𝑑 3 Total deflection of a helical spring 𝑃 𝑀𝑧 𝑦 𝑀𝑦 𝑧 𝛿= − + 𝐼𝑦 𝐼𝑧 𝐴 𝑃 𝑀𝑦 𝜎𝑥 = − 𝐼 𝐴 Equation of neutral axis 𝑀𝑦 𝑃 − =0 𝐴 𝐼 Maximum shear stress theory |𝜎1 − 𝜎2 | = 𝜎𝑦 Maximum energy of distortion theory 𝜎12 − 𝜎1𝜎2 + 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 𝑑 3𝑣 𝐸𝐼 3 = 𝑉 𝑑𝑥 𝑑 4𝑣 𝐸𝐼 4 = 𝑤 𝑑𝑥 Boundary conditions

Fixed or clamped support

2 1 𝑣(𝑎) = 0 𝜎𝑦 𝐿𝑒 = 𝜎𝑦 = Secant𝜎𝑐𝑟formula 𝜃(𝑎) = 𝑣 ′ (𝑎) = 0 ) ( Simple support 𝐸 2𝜋 𝑟 𝑣(𝑎) = 0 𝑃 𝑀(𝑎) = 𝐸𝐼𝑣 ′′ (𝑎) = 0 𝜋 𝑃 )] 𝑒𝑐 𝜎𝑚𝑎𝑥 = [1 + 2 sec ( √𝑃𝑐𝑟 Free end 2 𝑟 𝐴 ′′ (𝑎) 𝑀(𝑎) = 𝐸𝐼𝑣 =0 ′′′ (𝑎) 𝑉(𝑎) = 𝐸𝐼𝑣 =0 ENERGY METHODS AND IMPACT Guided support Strain energy in bars ′ (𝑎) 𝑣 =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 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴 𝑎𝑛𝑑 𝐵) 𝑥1 Loads = (𝑎𝑟𝑒𝑎 𝑜𝑓 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 𝑃𝑐𝑟 𝜋 2𝐸 𝜎𝑐𝑟 = = (𝐿𝑒⁄ 𝑟)2 𝐴 Short columns 𝑃 𝜎𝑚𝑎𝑥 = 𝐴 Intermediate columns Tangent modulus formula 𝜋 2 𝐸𝑡 𝑃𝑡 = 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 𝐿2 𝑀 𝑑𝑥 𝑈=∫ 0 2𝐸𝐼 Pure shear 𝑇𝐿 2 𝑑𝑥 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=

𝑃𝐿 Torsion bars 𝛿 = 𝐴𝐸 𝑇𝐿 𝜙= 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

𝑐𝑠 𝑠2 −𝑐𝑠 −𝑠2 𝑐 2 2 𝑐𝑠 −𝑐 2 −𝑐𝑠 −𝑐 𝑐 2 𝑐𝑠 ] −𝑐𝑠 2 [ 𝐿 −𝑐𝑠 2−𝑠 𝐴𝐸 𝑠 −𝑐𝑠 𝑐𝑠 2 𝑠2 2 −𝑠 = 𝑐 2 𝑐𝑠 −𝑐 −𝑐𝑠 𝑐 2 𝑐𝑠 ] [ 𝐿 Axial force in the bar element 𝑠2 𝑢𝑗 − 𝑢𝑖 𝐴𝐸 𝐹𝑖𝑗 = ( ) [𝑐 𝑠]𝑖𝑗 { 𝑣 − 𝑣 } 𝑗 𝑖 𝐿 [𝑘]𝑒 =

𝐴𝐸

𝑖𝑗

System equations {𝐹} = [𝐾 ]{𝛿 } Global nodal force matric 𝑛

{𝐹} = ∑{𝐹 }𝑒 1

𝑛

Global stiffness matrix

[𝑘] = ∑[𝑘]𝑒 1...