Statics Lecture Notes 2021 PDF

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Statics Lecture Notes 2021...

Description

Fy

FR q

Fx

x

+ F2

y 500 N

25

o

x

=

√

x

+ Fy2

453.22 + 211.32

y

12

o

x 200 N

y o

o

180 + 12

x 200 N

SIN

ALL

TAN

COS

150 N o

30

o

20 kN

35

o

60

o

15 78 kN

x

=

√

+ Fy2

502 + (−10)2

150 N

150 N o

30

15

o

78 N

Rx

+ F2

FRx

) = arctan(

54.81

150 N o

30

40 N

o

15

∑ ∑

Fy = 0 Fz = 0

F1y = 200 · cos 40 or F1y = 200 · sin 310 = −153.2N F2x = 300 · cos 15 = 289.8 F2y = 300 · sin 15 = 77.6N

y F2 F3 g

b x a

F1

F1x + F2x + F3x = 0 128.6 + 298.8 + F3 · cos(180 − γ) = 0

F1y + F2y + F3y = 0 −153.2 + 77.6 + F3 · sin(180 − γ) = 0

y

F2

F1

x

1.0 m

1.5 m

1.5 m 1.0 m

50 kg

50 kg

b

T1

T2

a W

β = 56.3◦

cos α

3

= T2 ·

2

sin α + T2 sin β − 490.5 = 0

2 T2 · · 0.5547 + T2 · 0.832 = 490.5 3 1.20 · T2 = 490.5

line of action of the force F

d o moment / lever arm

F

F

d

A

= 20 × 5 = −100 N · mm

F q

d

A

= 17.3N

= Lx · Fy − Ly · Fx

F q

Ly

A

d

Lx

Fy

= 86.60N = 100 · sin 30 = 50N

20 N

17

0

7.5 m

A

1m 3m q F A 2m

500 3m 2 kN

A 2m

F d F

10 kN 2m 10 kN 2.5 m A

= 10 · 2 = 20kN m

= −25kN m(clockwise)

= 45kN m(anti − clockwise)

= −25 + 45

∑

M=0

20kN

4m

6m

20kN

RAx

RAy

R By 6m

4m

RAy + 8 − 20 = 0

20kN

10m

20kN MA RAx

R Ay 10m

P1

P2

R By

R Ay a

b

c

Fully fixed

Pinned

Roller

M Rx

Rx Ry Ry Ry

P q

a

b

P

a

b

P

a

c

b

P1

a

P2

b

P W

rough surface

P W DFn

DNn

W

P h F

A

N x

F Fs Fk F=P

P

qs

W qs

y

Fs

x’

y’ N

x

W cos θs

= tan θs

y x

q

b q b b

q

y

, x

,

q

b q

b q b

P

W

h

rough surface

N =W

P h

W F

A

N x

N ·x= P ·h

P h

W F

A

N x=b/2

h

P

h

W

rough surface

P

a 30

o

c

b

2.0 m

3m P 1.5 m

4m

y

y’ W

P

5

3 P

W

4

N

F F

N

x

x’

5

=0

N = 44145 ·

4 5

= 35316N

5 = 15892N

− 0.3 · 35316

y’ 3

P

5

W

4 F N

x’

5

+ 0.3 · N − Pup0

5

+ 0.3 · 35316

2.0 m

3m P 1.5 m

4m

N1 = 28.2kN

1 2 q

= 16.9kN

N2 = 28.2 + 20 cos 20 = 47kN

F2 = 10.26 + 20 sin 20 = 17.1kN

y’

W1

F1

N1 x’

N1 y’

F1

W2

F2

N2 x’

= 11.75kN

2

frictionless pulley

1 2 a b

kN

1.0 0.5 2.0

1.5

1.0

2.0

2.0

m 0

1

2

3

4

5

6

7

8

1.0 0.5 2.0

kN

FL

1.5

m 0

1

3

2

4

FL = 5.0kN

FR

1.0

2.0

kN

2.0

m 4

5

6

7

8

FR = 5.0kN

FR

FL

FR

FL

Tension

Compression

joint

bar

(I)

(II)

(III)

B

4m

A C

10 kN

3m

FAB

4 5 3

A

F AC

10

B

RBx

4 5 3

FBC

= −

5 3

FA B

FAB

FBC

R Cx

FBC

FAC

C

R Cx

FAC

C R Cy

R Cy

3 × 10 + 4 × RBx = 0 RBx = −7.5kN

B RBx

4m

RCx

A

C

10 kN RCy 3m

E

D

4m

C B

A

75 N

6m

G

2.5 m

5 kN F

5 kN

8m E 5 kN

A

D B

4m

C

4m

4m

G

5 kN F

5 kN

8m E 5 kN A

D B

4m

C

4m

4m

E

D

A

B

4m

C

6m

75 N

2.5 m

10 kN H

J

I

L

K

N

M

2m A

G B

3m

D

C

3m

3m

E

3m

F

3m

3m

10 kN L

M

2m A

G E

3m

3m

3m

3m

F

3m

3m

L

M

10 kN

2m A R

E

F

RAy 3m

3m

3m

3m

3m

RGy · 3 − FEF · 2 = 0 FEF = 7.5 kN

√

13

= 0

FEM = 5 ·

√

13

√

13 = 9.01 kN 2

=0

3 FLM = −FEF − FEM · √ 13

L

M 2m G

E

F R Gy 3m

3m

3 · √ = −15 kN 2 13

15 − 30 + 5 ·

√

√

13

·2= 0

13 3 · √ · 2 = +15 − 30 + 15 = 0 correct 2 13

G 5 kN F

5 kN

8m E 5 kN A

D B

C

4m

4m

4m

1.0 kN

1.0 kN

1.0 kN 1.0 kN

1.0 kN 4m 0.5 kN

0.5 kN

1.5 m

1.5 m

2m

2m

1.5 m

1.5 m

h

h y-

y-x

xb

b

b 2 1 y¯ = h 2

b 3 1 y¯ = h 3

h y-c -x c b1

b2

h y- 1

-y 2 x-1

x- 2 b1

2

(h × b2 ) =

b2

1 (5 × 4) = 10 m2 2

Ac = A1 + A2 = 30 + 10 = 40 m2

1 b1 = 6 = 3 m 2 2 1 1 1 x¯2 = b1 + b2 = 6 + 4 = 7 m or 7.333 m 3 3 3

Ac

40 = 4.08 m

1

x¯n · An

∑n

h2

h1 h3

b1

b2

b3

h1 r h2

b1

b2

h1

1 2 b1

b2

b2 ) = WT ot · x¯ 2 2 78480 · 1 + 294300 · 4.5 = 372780 · x¯ x¯ = 3.76m + W2 · (b1 +

h2 = WT ot · y¯ 2 2 78480 · 2 + 294300 · 1 = 372780 · y¯ y¯ = 1.21m + W2 ·

h2

e h q a

b

c

300

40

40

120

40

y=ax

y dx

~y ~x

x

∫ ∫

x¯ = ∫

y dx

ax2 dx ax dx

[

]b

3

[

3

]

2 a x3 + c a b3 + c 0 ax dx 0 ] = [ 2 ∫b ]b = [ b 2 a2 +c 0 ax dx a x2 + c 0 2 = b 3

∫2

y¯ =

· y dx y dx

[

∫ a2 x2 ∫ 2

a2 x3

]b

2

0

dx 1 a 2 b3 6 = [ 2 ]b0 = 3 ab2 ax ax dx

1 = ab 3

3

h

y=x2

y

dx 1 x y=

y

∫

A

1 x · y dx = 0∫ 1 dA 0 y dx

∫

[

] x4 1

x3 dx 4 ∫01 = [ 3 ]10 = 2 x 0 x dx 3

∫

A

dA

2 x2 0 2 · x dx ∫1 2 0 x dx

=

0

∫1 y 0 2 ∫1 0

=

1 4 1 3

· y dx y dx

] x5 1 10 0 [ 3 ]1 x 3 0 [

= 0.75m

= 0.3m

x

y=x2

y

dy 1 x

x

y=x3

y

1

x

x

y 3

y=x3 1 x

x

w

A

L

B

L/2

W=wL

R By

R Ay L

A

B

2

2

=0

= (5 × 8) × RBy

8 = 160 kN.m 2 160 = = 20kN 8

w2

w1

L

A

B

2L/3 L/2 W 2

W1

R By

R Ay

A

L

1 1 W2 = (w2 − w1 ) × L = (10 − 4) × 6 = 18kN 2 2

RAy + RBy = W1 + W2 RAy + RBy = 42kN

B

2

− W2 ·

2L =0 3

2L L + W2 · 2 3 RBy × L = 24 × 3 + 18 × 4 RBy × L = 144kN.m

RBy × L = W1 ·

RBy = 24kN

1

2 0.5 m 1m

R By

R Ay 2m A

3m

7m B

2m

R By

R Ay 5m

5m

B

A

w2

w1

A

L

B

h

h

m2

)

(

)

m kg · g 2 · h(m) =ρ m3 s (

)

h F

p

2

(Ph + P0 ) =

1

p

1

1.5 m 1.5 m

2.0 m

F

p

2

p h

p

=

F

p 2

p

1

0

1

+

F2

p

2

p

1

=⇒ RB = 27.6kN

F1

p

1

A 1.0 m

RA

.75 m F1 F2

RB

B

A

h q B

h1 A h2 B

w

A

h q B

w h

FH

2

(PH − P0 ) · A

area A

volume V

h

b=

h = tan θ

2

h

10

·b·h·1=

1

R

a

b

c

2

(Pb − Pa) · A + Pa · A

a

R

Loading

Axial

Shear

Bending

Torsion

Deformation

2

F A

B

L/2

L/2

x A V

F/2

L/2

2

−V V

= 0 = +

F 2

F A V F/2

L/2

2

−F −V

= 0

V

= −

F 2

A

B

F/2

F/2

L/2

L/2

F/3

F/3

F/3

A

B

L/4

L/4

L/4

2

L/4

L/4

L/4

L/4

L/4

F/3 F/2 F/3 B

A F/3

F/2

w

B

A

L

2

2

− wx − V V

= 0 =

wL − wx 2

w

A

V

wL/2

x

wL 2 0 wL 2

L/2

L/2

F1

F2

A

B

L/2

L/4

L/4

W

L

W

W

L

F A

B C

L

∑

L/2

MA = 0 =⇒ RBy · L − F ·

3L 2

= 0

RRy =

RAy + RBy − F = 0

RAy +

3F −F = 0 2

1 RAy = − F 2

3F 2

(3/2)F -(1/2)F

-F

w

B

A

L

w

L/4 A

L/2

L/4 B

w

(average over dx)

V + dV

V

dx

V0

=−

∫ x

w dx

x0

w1 w2

2m A

∑

1m B

F

3m C

MB = 0 =⇒ (w1 · 2) ×

1m D

3m E

2 5 + REy × 5 = (w2 · 3) × + 10 × 8 2 2

F

5

= 17 kN

w1

A

V

2m A

Real

B

Free Body Diagram

A

= 0 − (5x) = −5xkN

w1 dx

w1

VB

A

B 2m

RBy 2m A

1m B

x FBD

C

Real

w dx x0

= −(5 · 2) + 9 = −1kN 2m ≤ x ≤ 3m

V

= 10kN

= 0

F

VE

V

E

R Ey

x

F

w

L/4

L/2

A

L/4 B

4 L ≤ x ≤ 3L4 4 3L ≤x≤L 4

10 kN/m

5 kN/m

12 m A

B

(5 · 12) × 4 = RBy × 12 2 RBy = 40 kN ∑

Fy = 0 =⇒ RAy + RBy = 60 + 30 RAy = 50 kN

10 kN/m

50 kN

x A

0

10 −

= 50 − 10x +

5 x dx 12

5 2 x +c 24

24

x2

12 kN/m

0 kN/m

6m A

B

W A

B

L/2

L/2

2

2

A Mx

W/2

x 2

2

2

≤x≤L

=

WL 4

+ve

deformed shape

W

A Mx

W/2 L/2

x 2

Mx =

( Wx L −W x− 2 2

2

+

Wx 2

)

B Mx’

W/2

x’

0

M

L

WL/4

F/3

F/3

F/3

A

B

L/4

4

4

≤x≤

L 2

2

≤x≤

3L 4

L/4

L/4

L/4

4

≤x≤L

w

B

A

L

2

w

A

Mx

wL/2

x

2

· x − FR ·

x 2

2

−

wx2 2

L w L · − 2 2 2 w L2 − · 4 2 4 wL2 wL2 = − 8 4 wL2 = 8 (

0

L

wL2 8

W

L

-ve

x

W

MA

RAy L

x A MA

Mx

F

F

B

Mx’ x’

-FL

0

+M

w

B

A

L

F A

B C

L

L/2

12 kN/m

0 kN/m

6m A

B

w

L/4 A

L/2

L/4 B

30 kN.m A

B

3m

3m

w

(average over dx)

V + dV

V

dx

2

− (V + dV ) · dx + M + dM = 0

dx

=0...