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Problem Set of Static of Rigid Bodies, Exercises of Differential Equations

University of Rizal SystemDifferential Equations

Problem Set of Static of Rigid Bodies

Typology: Exercises

2024/2025

Uploaded on 12/05/2024

angela-keizy-dela-vega
angela-keizy-dela-vega 🇵🇭

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bg1
BES
025
DELA VEGA, ANGELA KEIZY A.
BSCE2 - 02
SITUATION 1
1.1 & 1.2
SITUATION 2
VERTICAL LINE
L
2
A3= 70
x3= 0
y3==70
2= 35
A1
A2
A3
= + +
LTL1L2L3
LT= 258.496
LT= 40π+20π+70
258.496x
= 40π(0) + 20π(25.465) + (70)(0)
258.496ȳ = 40π(135.465 ) + 20π(95.465) + (70)(35)
x
= 6.190 mm
ȳ= 98.536 mm
1.1
1.2
= + -
ATA1A2A3
AT= 6429.226
= 1012.5π + 4500 - 400π
AT
2.1 & 2.2
A1
A2
A3
QUARTER ARC
A= = = 20π
y2=
x2=
=
2π
2rπ
2(40)
= 95.465
2r
π
2(40)
π
=
= 25.465
2r
π
2(40)
π
=
70+ 70 +
SEMI CIRCLE ARC
= L1πr
x1
y1
= 0
=
= = 40ππ(40)
2r
π= 135.465
2(40)
π
=
70+40+
SEMI CIRCLE
= A1πr
4=
2π(45)
4
2
= 1012.5π
x1
y1
90
2
=
=4(45)
3π
= 45
= 19.099
100r
CIRCLE
A3
x3= 45
y3= 100
πr2
= = 400π
TRIANGLE
A=
bh
2=
90(100)
2= 4500
y2=
x2=
2
2
3(90) = 60
2
3(100)= 66.667
6429.226x
= 1012.5π(45) + 4500(60) - (400π)(45)
6429.226ȳ = 1012.5π(19.099) + 4500(66.667) - (400π)(100)
x
= 55.507 mm
ȳ= 86.108 mm
lx1= 0.11r4
lx1= 0.11(45)4
lx1= 451068.75
lx2=
lx2=
lx2=
bh
36
3
90(100)
36
3
6
2.5 x10
ly1=
ly1=
ly1= 1610311.672
πr
8
4
π(45)
8
4
ly2=
ly2=
ly2=
hb
36
3
100(90)
36
6
2.025 x10
lx3=
lx3=
lx3=
πr
4
4
π(20)
4
4
4000π
ly3=
ly3=
ly3=
πr
4
π(20)
4
4
4000π
4
lxc= ∑l + A (dy)
x2
lxc= 451068.75 + 1012.5π(86.108 - 119.099) + + 4500(86.108 - 66.667) - 4000π - 400π(86.108-100)
22 2
6
2.5 x10
lxc
6
= 7.746 x10 mm
4
lyc= ∑l + A (dx)
y2
lyc= 1610311.672 + 1012.5π(55.507 - 45) + + 4500(55.507 - 60) - 4000π - 400π(55.507 - 45)
26
2.5 x10
lyc
6
= 3.813 x10 mm 4
2 2
2.1
2.2
3
pf2

Partial preview of the text

Download Problem Set of Static of Rigid Bodies and more Exercises Differential Equations in PDF only on Docsity!

BES

DELA VEGA, ANGELA KEIZY A.

BSCE2 - 02

SITUATION 1

SITUATION 2

VERTICAL LINE
L
A

x 3 = 0

y 3 = =

A

1

A

2

A

3

L = + +

T

L

1

L 2 L

3

L
T = 258.

L T = 40π+20π+

258.496x̄ = 40π(0) + 20π(25.465) + (70)(0)

258.496ȳ = 40π(135.465 ) + 20π(95.465) + (70)(35)

x̄ = 6.190 mm

ȳ= 98.536 mm

A = + -

T

A

1

A 2 A

3

A T = 6429.

A = 1012.5π + 4500 - 400π T

A

1

A

2

A

3

QUARTER ARC

A = = = 20π

y 2

x 2 =

π

r

π

2r

π

π

2r

π

π

SEMI CIRCLE ARC
L =

1

πr

x 1

y 1

= π(40)= 40π

2r

π

π

SEMI CIRCLE
A 1 =

πr

2

π(45)

2

= 1012.5π

x 1

y 1

3π

100r

CIRCLE
A

3

x 3

y 3

πr

2

= = 400π

TRIANGLE
A

bh

y 2 =

x 2

2

6429.226x̄ = 1012.5π(45) + 4500(60) - (400π)(45)

6429.226ȳ = 1012.5π(19.099) + 4500(66.667) - (400π)(100)

x̄ = 55.507 mm

ȳ= 86.108 mm

l x

1

= 0.11r

4

l x

1

4

l x 1

l x

2

l x

2

l x

2

bh

3

3

6

2.5 x

l y

1

l y

1

l y

1

πr

4

π(45)

4

l y 2

l y

2

l y

2

hb

3

6

2.025 x

l x 3

l x

3

l x 3

πr

4

π(20)

4

4000π

l y

3

l y

3

l y 3

πr

π(20)

4

4000π

4

l x

c = ∑l + A (dy)

x

2

l x c = 451068.75 + 1012.5π(86.108 - 119.099) + + 4500(86.108 - 66.667) - 4000π - 400π(86.108-100)

2 2 2 6

2.5 x

l x c

6

= 7.746 x10 mm

4

l y c = ∑l + A (dx)

y

2

l y

c

= 1610311.672 + 1012.5π(55.507 - 45) + + 4500(55.507 - 60) - 4000π - 400π(55.507 - 45)

2 6

2.5 x

l y

c

6

= 3.813 x10 mm

4

2 2

3

( ) (25m) = 6250 N

∑M = 0

T

2

T A
T = 14443.150 N

A

= 13020.833 N
T

o

T o

L

Wx

2

) dx

a

b

T

o

L

= 51.859 m

L

2

) dx

25

250x

T

o

3.2 length

T = 14443.150 N
AD ( ) - 900N = 0
250N

m

250N

m

SITUATION 3

SITUATION 3

w =

Ty - 6250N = 0

Ty = 6250 N

Tax = FH

∑F x = 0

∑F = 0

y

2

2 2

T
T

x

Ty

2

2

T

3.1 maximum

∑F = 0

z

-AD ( ) - AC ( ) = 0

∑F y = 0

AB - AC ( ) = 0
∑F = 0

x

AD = 1125 N
-1225 ( ) - AC ( ) = 0
AC = 843.75 N
AB - 843.75 ( ) = 0
AB = 506.3 N