# Institute of Engineering

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```					Institute of Engineering   Mechanics of Materials
410211
Dr.Tanongsak Bisarnsin
Email: tanong@ccs.sut.ac.th
Phone: 224227
Office Hours:
Monday      13.30-15.30

2
Mechanics I
• Structural analysis

ΣF = 0;

P                  ΣMA = 0;
C
• Internal forces
M
A   Ax        B                    V
N
Ay        By                                 B

By
• Section properties of member BC

A, I, x, y
3
Mechanics of Materials I

• Stresses             • Strain
M
P                   P
V                          L
N                                          ∆L
B

By                      ∆L
ε=
L
σn = N
A

τv = VQ
It

σb = My
I

4
• Stresses
M                          σn
V           N
N
B

By
V    τv

σbending
M

5
σ (MPa)
σu = 390 MPa
400
350
σfail = 295 MPa
300
250
200
σy = 225 MPa                             σpl = 205 MPa
150
100                    200
E=           = 200x103 MPa
50                    0.001
0                                          ε (mm/mm)
0.00     0.10    0.20      0.30   0.40      Upper scale
0.0000   0.0010 0.0020 0.0030 0.0040           Lower scale

6
•Beam

P

A                                        B
C        D

P
Ax   A                    NC
MC
C
Ay              VC

VC
D ND
MC                         MD
NC C
VD
P
Ax   A                          ND
MD
D
Ay                    VD
7
D ND
MD
VD

D          ND   MD

VD
Cross-section of D

8
P
Axial                                          σN =
A
1

N     NA            2

3

Cross-section of D
Normal Force              Normal stress

1          σN

2          σN

3          σN

State of stress at 1,2 and 3                                9
Bending Stress
(σb)top
My
1                   (σ b ) =
M                                   y               I
NA            2      O
3

(σb)bottom
Bending Moment            Bending stress

1         (σb)top

2

3         (σb)bottom

State of stress at 1,2 and 3                        10
Transverse Shear
1

NA                             (τV)max
2

3
V                                              VQ   A
(τ V ) =
Shear Force             Shear stress                     IT

1

2       τmax

3

State of stress at 1,2 and 3                   11
Combine Stress                         σV = 20 kPa
1

Normal stress      2
2

3

(σb)top = 50 kPa
1

Bending stress        2            O
3

(σb)bottom = 50 kPa
1

Shear stress                    (τV)max = 15 kPa
2

3

(σb)top = 30 kPa
1

Combine stress                   (τV)max = 15 kPa             (σb)NA = 20 kPa
2

3

(σb)bottom = 70 kPa   12
(σb)top = 30 kPa
1

N   M             (τV)max = 15 kPa           (σb)NA = 20 kPa
2

3
V
(σb)bottom = 70 kPa

1         σ = 30 kPa

2          σ = 20 kPa
τmax= 15 kPa

3         σ = 70 kPa

State of stress at 1,2 and 3                              13
• Shear diagram, bending moment diagram and deflected curve of a beam
Po
Mo                        wo

A
D       B                 C
L       L                 L
2       2                 2
RA                     RB

V                                      +
+
diagram
-

M M
0
diagram                 +
-

Deflected
Curve
Point of inflection      14
15 kN•m          20 kN                          2 kN/m
15 kN•m                     37.5 kN•m
A
D              B                   C
3m                 3m                  3m                       A            D-
RA = 7.5 kN                            RB = 24.5 kN      7.5 kN         3m       7.5 kN

7.5                              6
V (kN)                                                               37.5 kN•m 2 kN/m
+                                   +                                      18.5 kN

-12.5          -
D+           B-     9 kN•m
-18.5              12.5 kN        3m
37.5
M (kN•m) 15
+
-                9 kN•m          2 kN/m

-9
Deflected
B+           C
Curve
6 kN         3m
Point of inflection                                        15
• Shaft
Torsion Shear Stress
Tρ
τT =
T                        J          (τT)max

NA     O     ρ

(τT)max

1

2       3

State of stress at 1, 2 and 3         16
Combine Torsion and Shear Stress

1
T                                              (τT)max = 25 kPa

NA                 (τV)max = 15 kPa    O           ρ
2           V

(τT)max = 25 kPa
VQ                         Tρ
(τ V ) =                      τT =
IT                          J
3

1       25 kPa                        2     15 + 25 = 40 kPa            3    25 kPa

State of stress at 1, 2 and 3                                   17
•Connection
Simple Shear Stress
F1
F1

F1
F1

d

F1

F´1 = F1

F1   τaverage = F1 = F1
A   π d2
τaverage                        4
18
V   V
τaverage =     =
A   π d2
4

τaverage

19
Stress Transformation

σ

θ

τ

θ   σθ

τθ
20
Deflection of Beams

v(x) = f (M, EI)

Deflection curve

Column Buckling
Pcr = f (EI, A, L)

L

21

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