Determine moment of this normal force - rectangular beam:
A rectangular beam contains a width of 100 mm and a depth of 200 mm. This is utilized as a simply supported beam & the maximum bending stress is restricted to 10 N/mm2. Then determine the following :
(a) Entire normal force on the left bottom corner area of size 40 mm × 60 mm, and
(b) moment of this normal force around the neutral axis.
Solution
Maximum bending stress, σ_{max} = 10 N/mm^{2}
Normal force on the shaded area, left bottom corner is following
= (10/100) × (40 × 60) × (40 + 30)
Here, A = Shaded area = 40 × 60 mm^{2}
= Centroid of shaded area through the neutral axis
= (100 - 60) + ( 60/2) = (40 + 30) = 70 mm
Figure
∴ Normal force on the partial beam section is following
= (10/100) × 2400 × 70 = 16800 N = 16.8 N
The partial area is subjected into a tensile force
∴ Normal force on partial beam section = 16.8 kN (tension)
Moment of the force around the neutral axis
= (σ_{max} / y_{max}) × I_{s}
I_{s} = Moment of inertial of the shaded area around the neutral axis
= (1/12) × 40 × (60)^{3} + (40 × 60) × (70)^{2} = 1248 × 10^{4} mm^{4}
∴ Moment of normal force around the neutral axis
= ( 10 /100) × 1248 × 10^{4} = 1248 × 10^{3} Nm = 1248 m