Answer:
Part a: The yield moment is 400 k.in.
Part b: The strain is [tex]8.621 \times 10^{-4} in/in[/tex]
Part c: The plastic moment is 600 ksi.
Explanation:
Part a:
As per bending equation
[tex]\frac{M}{I}=\frac{F}{y}[/tex]
Here
[tex]I=\frac{bd^3}{12}[/tex]
Here
[tex]I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4[/tex]
[tex]y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\[/tex]
[tex]\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[/tex]
The yield moment is 400 k.in.
Part b:
The strain is given as
[tex]Strain=\frac{Stress}{Elastic Modulus}[/tex]
The stress at the station 2" down from the top is estimated by ratio of triangles as
[tex]F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi[/tex]
Now the steel has the elastic modulus of E=29000 ksi
[tex]Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in[/tex]
So the strain is [tex]8.621 \times 10^{-4} in/in[/tex]
Part c:
For a rectangular shape the shape factor is given as 1.5.
Now the plastic moment is given as
[tex]shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi[/tex]
The plastic moment is 600 ksi.
