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kolbaska11 [484]
3 years ago
6

A column has a 4.8 cm by 8.7 cm rectangular cross section and a height 4 mm . The column is fixed at both ends and has a lateral

support at mid-height for the weak axis. The material has E = 200 GPa and σY = 250 MPa.a. The factor of safety for the column yielding is FS 1.5. What is the maximum load that can be applied based on the column yielding due to the normal stress? Assume the load is applied at the centroid of the section. Express your answer with appropriate units to three significant figures. b. The design is to have a factor of safety for buckling FS 2.5. What is the maximum load based on buckling in the strong axis of the column? Express your answer with appropriate units to three significant figures.
Engineering
1 answer:
Elden [556K]3 years ago
3 0
I’m literally so sorry
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A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.
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Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is 8.621 \times 10^{-4} in/in

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

\frac{M}{I}=\frac{F}{y}

Here

  • M is the moment which is to be calculated
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                         I=\frac{bd^3}{12}

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  • b is the breath given as 0.75"
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                     I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4

  • y is given as

                     y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\

  • Force is 50 ksi

\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in

The yield moment is 400 k.in.

Part b:

The strain is given as

Strain=\frac{Stress}{Elastic Modulus}

The stress at the station 2" down from the top is estimated by ratio of triangles as

                        F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi

Now the steel has the elastic modulus of E=29000 ksi

Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in

So the strain is 8.621 \times 10^{-4} in/in

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For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi

The plastic moment is 600 ksi.

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