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Aleks [24]
3 years ago
15

Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diam

eter of 12 mm and an allowable normal stress of 110 MPa. Pipe (2) has an outside diameter of 48 mm, a wall thickness of 5 mm, and an allowable normal stress of 65 MPa. Assume x1Determine the maximum load Pmax that can be supported by the structure without exceeding either allowable normal stress.
Engineering
1 answer:
Galina-37 [17]3 years ago
8 0

Answer:

P_max = 25204 N

Explanation:

Given:

- Rod 1 : Diameter D = 12 mm , stress_1 = 110 MPa

- Rod 2: OD = 48 mm , thickness t = 5 mm , stress_2 = 65 MPa

- x_1 = 3.5 mm ; x_2 = 2.1 m ; y_1 = 3.7 m

Find:

- Maximum Force P_max that this structure can support.

Solution:

- We will investigate the maximum load that each Rod can bear by computing the normal stress due to applied force and the geometry of the structure.

- The two components of force P normal to rods are:

               Rod 1 : P*cos(Q)  

               Rod 2: - P*sin(Q)

where Q: angle subtended between x_1 and Rod 1 @ A. Hence,

               Q = arctan ( y_1 / x_1)

               Q = arctan (3.7 / 2.1 ) = 60.422 degrees.

- The normal stress in each Rod due to normal force P are:

               Rod 1 : stress_1 = P*cos(Q)  / A_1

               Rod 2: stress_2 = - P*sin(Q)  / A_2

- The cross sectional Area of both rods are A_1 and A_2:

               A_1 = pi*D^2 / 4

               A_2 = pi*(OD^2 - ID^2) / 4

- The maximum force for the given allowable stresses are:

               Rod 1: P_max =  stress_1 * A_1 / cos(Q)

                          P_max = (110*10^6)*pi*0.012^2 / 4*cos(60.422)

                          P_max = 25203.61848 N

               Rod 2: P_max =  stress_2 * A_2 / sin(Q)

                          P_max = (65*10^6)*pi*(0.048^2 - 0.038^2) / 4*sin(60.422)

                          P_max = 50483.4 N

- The maximum force that the structure can with-stand is governed by the member of the structure that fails first. In our case Rod 1 with P_max = 25204 N.

             

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Explanation:

From  the given question:

Using the distortion energy theory to determine the  factors of safety  FOS can be expressed  by the relation:

\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma  y^2-\sigma x \sigma y+3 \tau_{xy^2}}

where; syt = strength in tension and compression = 350 MPa

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\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000

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FOS=\dfrac{350}{105.3565}

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USing the maximum-shear stress theory;

\dfrac{350}{2 FOS}  =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2

\dfrac{350}{2 FOS}  ={ (\dfrac{110-100}{2})^2

\dfrac{350}{2 FOS}  =25

FOS = 350/2×25

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Using distortion- energy theory formula:

\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}

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\dfrac{350}{FOS}= 88.88

FOS = 350/88.88

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USing the maximum-shear stress theory;

\dfrac{350}{2 FOS}  =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2

\dfrac{350}{2 FOS}  =\sqrt{ (35)^2+ (-20)^2

\dfrac{350}{2 FOS}  =\sqrt{ 1225+ 400

\dfrac{350}{2 FOS}  =40.31

FOS}  =\dfrac{350}{2*40.31}

FOS = 4.341

7 0
3 years ago
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