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Geometric Properties of Cross Sections 1. Centroid: geometric center 𝑥" =

∑ % & '&

𝑦) =

∑ '&

∑ *& ' & ∑ '&

2. Second moment of area: an area which reflects how its points are distributed 𝐼%% = ∫' 𝑦 - 𝑑𝐴

𝐼** = ∫' 𝑥 - 𝑑𝐴

Polar moment of area: 𝐽1 = 𝐼%% + 𝐼**

3. Product second moment of area: measure of the symmetry of the shape (=0 if symmetry) 𝐼%* = ∫' 𝑥𝑦3𝑑𝐴 4. Parallel-axis of symmetry 𝐼%% = 𝐼% 4% 4 + (𝑦)- )𝐴

𝐼** = 𝐼* 4* 4 + ( 𝑥" -)𝐴

𝐼%* = 𝐼% 4* 4 + (𝑥" 𝑦))𝐴

5. Second moment area about an inclined axis 𝐼% 4 % 4 = 𝐼* 4 * 4 = 𝐼% 4 * 4 =

788 97:: -

788 97:: -

788 ;7:: -

+ −

788;7:: -

788;7:: -

cos 2𝜃 − 𝐼%* sin 2𝜃 cos 2𝜃 + 𝐼%* sin 2𝜃

sin 2𝜃 + 23𝐼%* cos 2𝜃

6. Mohr’s circle 7. Principal second moment of area: maximum/minimum Principal rotation: tan 2𝜃F = (7

;-78:

88;7:: )

𝐼%% + 𝐼** 𝐼%% − 𝐼** + cos 2𝜃FG − 𝐼%* sin 2𝜃FG 2 2 𝐼%% + 𝐼** 𝐼%% − 𝐼** − cos 2𝜃F- + 𝐼%* sin 2𝜃F333333𝐼- = 2 2 333333𝐼G =

Concept of Stress HI

1. Stress = H'

Describes the intensity of the internal force acting on a specific plane of area ⇒ Normal Stress (perpendicular) → volume change → 𝜎M = lim

HIP

H'→1 H'

Axial Force 𝑃 = ∫' 𝜎3𝑑𝐴

Axial force = resultant of normal/direct stress over the cross section 𝜎RST = 𝑃 Z𝐴33(Mpa =

⇒ Shear Stress (tangential) → shape change → 𝜎% = lim

HI8

H'→1 H'

Shear force 𝑉 = ∫' 𝜏3𝑑𝐴

𝜎* = lim

HI:

H'→1 H'

Shear force = resultant of shear stress over the cross section 𝜏RST = 𝑉⁄𝐴 2D Transformation of Stresses 1. Sign convention Positive normal stress acts outward from all faces Positive shear stress acts upward on the right-hand face of the element 2. Stress transformation

𝜎%% + 𝜎** 𝜎%% − 𝜎** + cos 2𝜃 + 𝜏%* sin 2𝜃 2 2 𝜎%% + 𝜎** 𝜎%% − 𝜎** 3333𝜎* 4* 4 = − cos 2𝜃 − 𝜏%* sin 2𝜃 2 2 𝜎** − 𝜎%% sin 2𝜃 + 𝜏%* cos 2𝜃 3333𝜏% 4* 4 = 2 3. Mohr’s circle of stress 3333𝜎% 4% 4 =

G

𝑎 = (𝜎%% + 𝜎** ) -

𝑟 - = (𝜎 − 𝑎)- + 𝜏 +ve angle is clockwise on Mohr’s circle 4. Principal stress Represents the maximum and minimum normal stress at the point Principal rotation tan 2𝜃 =

-3`8:

(a8;a: )

Principal in-plane normal stresses 𝜎G =

a89a: -

+

a8;a: -

cos 2𝜃 + 𝜏%* sin 2𝜃

𝜎% + 𝜎* 𝜎% − 𝜎* cos 2𝜃 − 𝜏%* sin 2𝜃 − 2 2 No shear stress acts on the principle planes

33333333333333333333333333333333333333333333333333333333333333333333333333𝜎- = 5. Maximum shear stress a8;a: ) -

33333𝜏XR% = b(

+ 𝜏%* - (the normal stress of max shear stress 𝜎XR% = a ;a

Mohr’s circle observation 𝜏XR% = c d - Y c

a89a: -

)

W ) XXY

Concept of Strains 1. Strain Defined as the ratio of the deformation to original size of the body (intensity of deformation) Objective description of the material behaviour ⇒ Normal strain: uniform through cross section f

Average axial strain 𝜀RST = g Axial strain at a point 𝜀 =

hf h%

g

⇒3Deformation3𝑒 = ∫1 𝜀3𝑑𝑥

⇒ Shear strain: measure of angle of distortion tan 𝛾 = 𝑑𝑒 ⁄𝑑ℎ ⇒ 𝛾 = 𝑑𝑒⁄ 𝑑ℎ 3(small angles) Transformation of Strains 1. Sign convention Positive normal strains 𝜀%% and 𝜀** causes elongation

Positive shear strain 𝛾%* causes small angle AOB

The angle 𝜃 is measured in counter clockwise direction

2. Strain transformation

𝜀%% + 𝜀** 𝜀%% − 𝜀** 𝛾%* 𝑠𝑖𝑛 2𝜃 𝑐𝑜𝑠 2𝜃 + + 2 2 2 𝜀%% + 𝜀** 𝜀%% − 𝜀** 𝛾%* 𝑠𝑖𝑛 2𝜃 3 3333𝜀* 4* 4 = 𝑐𝑜𝑠 2𝜃 − − 2 2 2 𝛾% 4* 4 𝜀** − 𝜀%% 𝛾%* cos 2𝜃 3333 sin 2𝜃 + = 2 2 2 3333𝜀% 4% 4 =

3. Principal rotation tan 2𝜃F =

q8:

r8;r:

4. Maximum shear strain qst8 -

r8;r: ) -

= b(

q8:

+(

-

)- (Average strain 𝜀RST =

r89r: -

)

Orientation of element representing maximum shear strain tan 2𝜃u = −

r8;r: q8:

Hooke’s Law 1. Normal Strain due to temperature: 𝜀 = 𝛼3∆𝑇 (𝛼: coefficient of thermal expansion)

Change in length(elongation) 3𝛿z = 𝜀3𝐿 = 𝛼3Δ𝑇3𝐿

Young’s Modulus: 𝜎 = 𝐸𝜀 (normal stress= young’s modulus* normal strain) r::

Poisson’s Ratio: 𝑣3𝑜𝑟3𝜇 = − €

88

2. Hooke’s law for shear stress G

𝛾%* = 𝜏%* •

𝛾*M =

Shear Modulus: 𝐺 =

G

𝜏 • *M ƒ

G

𝛾%M = 𝜏%M •

-(G9S)

3. Plane Stress and Plane Strain Plane Stress

Plane Strain

𝜎MM = 𝜏%M = 𝜏*M = 0

𝜀MM = 𝛾%M = 𝛾*M = 0

𝜀%% =

𝜀MM =

1 (𝜎 − 𝑣3𝜎** ) 𝐸 %%

1 (𝜎 − 𝑣3𝜎%% ) 𝐸 ** −𝑣 = (𝜎 + 𝜎** ) ≠ 0 𝐸 %%

𝜀** = 𝜀MM

𝛾%* =

1 3𝜏 𝐺 %*

𝛾%M = 𝛾*M = 0

1 [(𝜎 − 𝑣(𝜎%% + 𝜎** ) ≠ 0 𝐸 MM

𝜎MM = 𝑣3(𝜎%% + 𝜎** ) 𝜀%% = 𝜀** =

1 − 𝑣𝑣 [(𝜎%% − 𝜎** ) 𝐸 1−𝑣 𝑣 1 − 𝑣[(𝜎** − 𝜎%% ) 𝐸 1−𝑣

𝜏%M = 𝜏*M = 0

Axial Loading 1. Stress & Strain due to axial force ⇒ Stepped cross section If elastic, 𝑒 =

Wg ƒ'

W3XX

𝑚𝑚 = ‰FR3XX Y

If inelastic, 𝑒 = 𝜀𝐿

3333⇒ Varying area gG W

𝑒 = ∫g-

𝑑𝑥 ( 𝐴 = ƒ'

ŠhY ‹

)

2. Compatibility: certain restriction on deformation in regard to geometry and boundary ⇒ Rigid

fY;fd gd

=

fŒ;fd gY

Wg

⇒ Constrained bar 𝛼𝐿∆𝑇 + ƒ' = 0 Indeterminant Problems 1. Wires supporting rigid bar 2. Turning bolt problem 3. Fixed bar

Elastic Beam Bending 1. Bending strain formula h•

h•

𝜀 = −𝑦3( h% ) [ ( h% ) bending curvature] 2. Bending stress formula 𝜎 = −𝑦 r

‰P

7PP

(𝑀𝑝𝑎 = −𝑚

‰

W∙XX )⇒ XX‘

𝜅 = = (𝑚𝑚;G = XX = ƒ37 *

Proof 𝜎 = 𝐸3𝜀 = −𝑦3(𝐸

h• ) h%

3𝑀 = −

W∙XX ) ‰FR3XX ‘3 ‰PP

ƒ37PP

a∙7 *

Note: unit 10“ ∙ 𝑚𝑚;G = 𝑚;G

h•

= h%

3. Workshop: Maximum normal stress/ Maximum momentum 𝑀𝑜𝑑𝑢𝑙𝑖3𝑟𝑎𝑡𝑖𝑜, 𝑏𝑒𝑛𝑑𝑖𝑛𝑔3𝑚𝑜𝑚𝑒𝑛𝑡3𝑑𝑖𝑎𝑔𝑟𝑎𝑚 ⇒ 𝑚𝑎𝑥𝑖𝑚𝑢𝑚3𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚, 𝑠𝑒𝑐𝑜𝑛𝑑 3𝑚𝑜𝑚𝑒𝑛𝑡3𝑜𝑓3𝑎𝑟𝑒𝑎 𝜎 = −𝑦

‰ 7

(𝑀𝑝𝑎 = −𝑚

W∙XX 3) XX‘

⟺𝑀=−

a∙7 *

(Allowable max M is actually the minimum one)

Composite Beam Bending Stresses 1. Elastic section modulus 𝑍•žF/

ž••žX

7

= or 𝜎XR% = ¡ *

;‰

¢£¤¥/¦¤££¤s

¡

2. Transformed section ⇒ same centroid

' ⇒ same bending rigidity (EI)3⇒ 𝐸' 𝐼MM = 𝐸§ 𝐼'"MM

ƒ

⇒Moduli Ratio 𝑛 = ƒ ¨ ©

e.g. For a rectangular section (transformed width)

' = 𝐼MM

A

B

b

nb

𝜎

𝜎)𝑛

G 𝑏𝑑 “ G-

+ 𝑏𝑑 ( 𝑦')-

𝑦' distance to centroid

' " = 𝑛3𝐼MM 𝐼'MM

Inelastic & Plastic Beam Bending Stresses 1. First yield moment 𝑀*

3333𝑀* = 𝜎* ∙ 𝑍 = 𝑓* ∙ 𝑍 (yield stress: 𝜎* or 𝑓* ) 2. Strain- plane sections remain plain No matter what material properties, strain distribution will be linear

3. Plastic section modulus 𝑍F 𝑀F = 𝑓* ∙ 𝑍F

Shape factor 𝛼 =

‰¥

‰:

=

¢¥ ¢

(how long after yield moment is reached the fully plastic moment)

4. Fully plastic moment 𝑀F 𝑀F = 𝐶𝐿 = 𝑇𝐿 '

𝐶 = 𝑇 = 𝑓* -

5. Neutral axis

Elastic ⇒ Centroid

Fully plastic ⇒3Half across the section Partial plastic ⇒ 𝐶 = 𝑇

suppose neutral axis by 𝑑« and y and x; separate area according to shape and width;

𝐹 = 𝑃 ∙ 𝐴 = 𝑃𝑏ℎ = 𝐴𝑟𝑒𝑎3𝑜𝑛3𝑠𝑡𝑟𝑒𝑠𝑠3𝑑𝑖𝑎𝑔𝑟𝑎𝑚 ∙ 𝑤𝑖𝑑𝑡ℎ3 Beam Deflection 1. Moment curvature hYS

h•

𝑀M = 𝐸3𝐼MM h% ≈ 𝐸3𝐼MM h% Y ⇒ 𝑣(𝑥) = ∬ ‰

h•

𝜅 = ƒ37 P = h% PP

2. Boundary condition 𝑣=0

33𝑣 = 0

𝑣 = 0,

3. Step function < 𝑥 − 𝑎 >« =

hS h%

=0

03333333333333333333𝑥 < 𝑎 (𝑥 − 𝑎)« 33333𝑥 ≥ 𝑎

4. Bending moment diagram

‰(%) ƒ7

𝑑𝑥𝑑𝑥 + 𝐶G 𝑥 + 𝐶-

𝑀=𝐹∙𝐿

Shear Stresses 1. Shear Force Shear Force exists only when bending moments varies Shear force = slope of bending moment of diagram 𝑉 =

h‰ h%

2. Shear stress: longitudinal stress arises from resistance 𝜏=

³´ WXXŒ (XX‘XX 7

33333𝑉:shear force

= 𝑀𝑝𝑎)

33333𝑄:3first moment of area

·Œ

33333𝐼:second moment of area (horizontal: 𝐼 =

“

33333𝑏:width at the cut

, vertical: 𝐼 = ¸G

The percentage of force carried: 𝑉 = ∫' 𝜏 𝑑𝐴 = ∫¸1 𝜏 3𝑏3𝑑𝑙

Shear flow 1. Shear flow: a measure of force per unit 𝑞=

³´ 7

»RFR»¼•*

I

Welds: 𝑠 = º (spacing= u·fR½3¾¸ž¿) (𝐹 ≥ 𝑞 ∙ 𝑠)

Shear center: 𝑒 =

IÀ · ³

·Œ

G-

)

Torsion 1. Torsion Formula: 𝜏 =

zÁ Â

33333𝜏:shear stress at any radial distance 𝜌 33333𝑇:internal torque in a shaft

33333𝜌:radial distance from centre to cross section 33333𝐽:polar moment of inertia Š

Š

(Solid circular shaft: J= 𝑟 ‹ Hollow shaft: J= (𝑟1 ‹ − 𝑟¼ ‹)) Š

2. First yield: 𝑇* = 𝜏* 𝑐 “ (𝜌* = 𝑐) Elastic-plastic torque: 3𝑇 = Plastic torque: 𝑇F =

Š 𝜏 (4𝑐“ Ä *

-Š 𝜏 3𝑟“ “ *

‹

− 𝜌* “ )

= “ 𝑇*

3. Angle of twist: Uniform/ constant cross-section torque and shear modulus: ∅ = z 3g

Stepped shafts: ∅ = ∑ •& 3Â &

zg

•Â

& &

Note: If Elastic-Plastics diagram: Hooke law: 𝜏 = 𝐺𝛾 → zg

∅ = •Â =

q3g Á

/𝛾=

q Á

∅3Á g

z

q

= • → Á 3𝑠𝑎𝑚𝑒 (use this formula to determine the condition for cross section)...