Question 4.SP.6: Determine the plastic moment Mp of a beam with the cross sec...

Neutral Axis.     When the deformation is fully plastic, the neutral axis divides the cross section into two portions of equal areas. Since the total area is

A = (100)(20) + (80)(20) + (60)(20) = 4800 mm²

the area located above the neutral axis must be 2400 mm². We write

(20)(100) + 20y = 2400      y = 20 mm

Note that the neutral axis does not pass through the centroid of the cross section

Plastic Moment.     The resultant \mathbf{R} _{i} of the elementary forces exerted on the partial area A_{i} is equal to

R_{i}=A_{i} \sigma_{Y}

and passes through the centroid of that area. We have

R_{1}=A_{1} \sigma_{Y}=[(0.100  m )(0.020  m )] 240  MPa =480  kN

R_{2}=A_{2} \sigma_{Y}=[(0.020  m )(0.020  m )] 240  MPa =96  kN

R_{3}=A_{3} \sigma_{Y}=[(0.020  m )(0.060  m )] 240  MPa =288  kN

R_{4}=A_{4} \sigma_{Y}=[(0.060  m )(0.020  m )] 240  MPa =288  kN

The plastic moment M_{p} is obtained by summing the moments of the forces about the z axis.

M_{p}=(0.030  m ) R_{1}+(0.010  m ) R_{2}+(0.030  m ) R_{3}+(0.070  m ) R_{4}

=(0.030  m )(480  kN )+(0.010  m )(96  kN )

+(0.030  m )(288  kN )+(0.070  m )(288  kN )

=  44.16  kN \cdot  m                           M_{p}=44.2  kN \cdot  m

Note: Since the cross section is not symmetric about the z axis, the sum of the moments of \mathbf{R} _{1} and \mathbf{R} _{2} is not equal to the sum of the moments of \mathbf{R} _{3} and \mathbf{R} _{4}.