Week 1-3 slides

Mechanics of Solids

Internal Resultant Loadings

Example Determine the resultant internal loadings at D, E and F.

4 kN

4 kN

4 kN

D

E

F

B

A

1.2 m 1.2 m

1.8 m

1.8 m

D:

Free-body diagram D

ND

MB_D

VD

x

y

Equations of equilibrium

VD = 0

ND = 0

MB_D = 0

C

D

Internal resultant loadings

F:

Free-body diagram

Equations of equilibrium 4 kN

4 kN

4 kN

D

E

F

B

A

1.2 m 1.2 m

1.8 m

1.8 m

NF

MB_F

VF

4 kN

4 kN

4 kN

F

B

A

x

y

VF = 0

FFy – 4 – 4 – 4 = 0 NF =12 kN

MB_F – 4 x 1.2 – 4 x 1.2 + 4 x 1.2 = 0 MB_F = 4.8 kN m

C


F


E:

Free-body diagram


4 kN

4 kN

D

E

F

B

A

1.2 m 1.2 m

1.8 m

1.8 m

NE

ME

VE

E

A

x

y

FAy

FAx (Support reactions)

C

4 kN

4 kN 4 kN

B

A

C

A:

F’Ay

F’Ax

FCy

FCx

x

y

F’Ax + FCx = 0

F’Ay + Fcy – 12 = 0

F’Ax = - FCx

MAC – 4 x 1.2 – 4 x 1.2 + 4 x 1.2 = 0 MAC = Fax x 1.8

MAC = 4.8 kN m F’Ax = - 2.67 kN


C


D

E:

Free-body diagram


4 kN

4 kN

D

E

F

B

A

1.2 m 1.2 m

1.8 m

1.8 m

NE

MB_E

VE

E

A

x

y

FAy


C

4 kN

4 kN

B

A

A:

F’Ay

F’Ax FBy

FBx x

y

F’Ay x 1.2 – (-2.67 x 0.9) – 4 x 2.4 = 0

F’Ay = 6 kN

F’Ax = - 2.67 kN

FAy = - F’Ay = - 6 kN

FAx = - F’Ax = 2.67 kN

F’Ax = - 2.67 kN


B



E:

Free-body diagram


4 kN

4 kN

D

E

F

B

A

1.2 m 1.2 m

1.8 m

1.8 m

NE

MB_E

VE

E

A

x

y

FAy


C

FAy = - F’Ay = - 6 kN

FAx = - F’Ax = 2.67 kN

VE + FAx = 0

NE + FAy = 0 NF = 6 kN

MB_E – 2.67 x 0.9 = 0 MB_E = 2.4 kN m

VE = - 2.67 kN


a

a


Fx : Shear Force (V)

3-D

Mx : Bending Moment

FZ : Normal Force (N)

Fy : Shear Force (V)

My: Bending Moment

Mz : Torsional Moment or torque (T)

Mechanics of materials: external loads and intensity of internal forces F1

F2

a

a

z

y

x

o

Mz

My

Fx

Fy

Mx

Fz

F1

F2

Mechanics of Solids

Stress

Stress

Stress: the intensity of the internal force on a specific plane passing through a point

The material is continuous (no voids) and cohesive (no cracks, breaks and defects)

Shear Stress, τ,

Normal Stress, σ Tensile Stress:

Compressive Stress:

[MPa]

Stress


The material is continuous (no voids) and cohesive (no cracks, breaks and defects)

Average Normal Stress: (Uniform Uniaxial Stress) a

a

a-a

A

The beam is prismatic bar.

σ

F

Average Shear Stress:

Stress

When the stress (intensity of force) of an element exceeds some level, the structure will fail. For convenience, we usually adopt allowable force or allowable stress to measure the threshold of safety in engineering.


An 80 kg lamp is supported by a single electrical copper cable. if the maximum allowable stress for copper is σCu,allow=50MPa, please determine the minimum size of the wire/cable from the material strength point of view.

σ ≤ σallow

Stress

Factor of Safety, F.S., a ratio of failure load Ffail (found from experimental testing) divided by the allowable one Fallow

Uniform uniaxial stress,

An 80 kg lamp is supported by a single electrical copper cable. if the maximum allowable stress for copper is σCu,fail=50MPa, and the factor of safety of 2 it be achieved, please determine the minimum size of the wire/cable from the material strength point of view.

σCu,allow= σCu,fail/F.S. = 25 MPa

Stress Analysis Engineering Design

Stress

1-71 w = 2.5 kN/m. Member AB has a square cross section of 12 mm. Determine the average normal stress and average shear stress acting at sections a-a and b-b. (p. 45)

w

C

A

b

b

1.2 m B 60o

Week 1-3 slides

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