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PROBLEM-1.

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PROBLEM-1

The hanger assembly is used to support a distributed loading of w=16kN/m. Determine the average stress in the 12-mm-diameter bolt at A and the average tensile stress in rod AB, which has a diameter of 15 mm. If the yield shear stress for the bolt is ty = 180 MPa, and the yield tensile stress for the rod is sy = 275 MPa, determine factor of safety with respect to yielding in each case.

B

1 m

C

A

w

4/3 m

2/3 m

FAB

+ MC = 0

Cx

a

A

C

Cy

4/3 m

2/3 m

We get

FAB = 40 kN

1 m

1 m

W

Shear force:

V = FAB/2 = 20 kN

V

V

Shear stress:

Bolt A

FAB=40 kN

PROBLEM-1Equation of equilibrium

(FABsina)(4/3) – W(1) = 0

a = tan-1(3/4) = 36.87o

For bolt A

t = 176.8 N/mm2

FAB

C

A

PROBLEM-1Factor of safety for bolt A:

For rod AB

= 226.4 N/mm2= 226.4 MPa

Factor of safety for rod AB:

PROBLEM-2

Two bars are used to support a

load P. When unloaded, AB is

125 mm long and AC is 200 mm

long, and the ring at A

has coordinates (0,0). If a load is

applied to the ring at A, so that it

moves it to the coordinate

position (6.25 mm, -18.25 mm),

determine the normal strain in

each bar.

B

C

60o

200 mm

125 mm

A

P

B

C

60o

125 mm

200 mm

A

230.67 mm

C

B

y

108.25 mm

x

A’

18.25 mm

62.5 mm

6.25 mm

PROBLEM-2Initial Geometry

BD = ABcos60o = 62.5 mm

AD = ABsin60o = 108.25 mm

= 168.17 mm

Final Geometry

A’B = ??

= 143.975 mm

= 205.47 mm

B

143.975 – 125

125

A

=

A’

A’B – AB

AB

A’C – AC

AC

eAB =

eAC =

205.47 – 200

200

=

PROBLEM-2Average normal strain

= 0.1518 m/m

= 0.0274 m/m

PROBLEM-3

The bar DBA is rigid and is originally

held in the horizontal position. When the

weight W is supported from D, it causes

the end D to displace downward 0.5 mm.

The wires are made of A-36 steel and have

a cross-sectional area of 0.8 mm2, and

Est= 210 GPa.

Determine:

- The normal strain & stress in wire BE.
- The reaction force FBE and weight W.
- The normal stress & strain in wire CD.

1.5 m

D

A

B

B’

D’

dB = 0.3 mm

s = (210x109)(2x10-4)

= 42x106 N/m2

PROBLEM-31) Normal strain & stress for wire BE:

Displacement:

DD’ = dD = 0.5 mm

BB’ = dB

Normal strain for wire BE:

Applying Hook’s law for wire BE:

s = E e

FBE

1 m

1.5 m

D

A

B

Dy

Ay

FBE = (42x106)(0.8x10-6)

= 33.6 N

Thus,

Dy = 0.6FBE = 20.16 N

PROBLEM-32) Reaction force FBE & weight W

- Free-body diagram for the bar

- Equations of equilibrium

MA = 0;

Applying uni-axial normal stress,

where FBE is a tensile force:

Dy

C

W

PROBLEM-3The weight W

FBD of point C:

Equilibrium at point C:

Fy = 0;

W = Dy = 20.16 N

3) Normal stress and strain for wire CD

Wire CD undergoes tensile force

PROBLEM-4

B

The rigid pipe is supported by a pin

at C and an A-36 steel guy wire AB.

If the wire has a diameter of 5 mm,

determine how much it stretches

when a load of P = 1500 N acts on

the pipe. The material remains elastic.

E = 210 GPa.

2 m

60o

A

C

“how much it stretches”

What does it mean??

stretch = change in length =dAB

P

FBA)X

P

B

FBA)Y

FBA

Cx

Cx

C

C

Cy

Cy

PROBLEM-4B

2 m

60o

A

C

Free-body diagram

FBA)X = FBAcos 60o

Component of FBA:

FBA)Y = FBAsin 60o

FBA)X (2) – P(2) = 0

PROBLEM-4Equations of equilibrium

B

FBA)X

P

MC = 0;

FBA)Y

FBA)X = P = 1500 N

Cx

C

Applying uni-axial normal stress,

Cy

FBA

PROBLEM-4Normal strain for wire BA:

Change in length of wire BA:

dAB = eBALBA = (7.27x10-4)(2000/sin60o) mm

= 1.67 mm

PROBLEM-5

d

A shear spring is made from two blocks

of rubber, each having a height h, width

b, and thickness a. The blocks are

bonded to the three plates as shown. If

the plates are rigid and the shear

modulus of the rubber is G, determine

the displacement of plate A if a

vertical load P is applied to this plate.

Assume that the displacement is small

so that d = a tan g ag

h

a

a

P/2

g

d

V

a

b

Rubber cross section

PROBLEM-5d

h

a

a

Free-body diagram after deformation

Shear force: V = P/2

Displacement:

Shear stress:

d = a tan g a g

Shear strain:

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