Question

Q1 (30 pts): Three forces are applied on the column shown in the figure. Neglect the self-weight of the material and find the stress at Points A, B, C, D 30 kN 16 kN 20 kN Y 30 kN 16 kN --> x 60 mm 20 kN 90 mm Top View Side View

Answer 1

we can transform all the eccentric loading with a single load F and with moment M_X and M_Y at the center.

Loads(Compression) in KN	Moment about x-axis in KN-mm	Moment about y-axis in KN-mm
16	0	16x(-45)=-720
20	20x30=600	0
30	30x(-30)=-900	30x45=1,350
F=66KN	Mx=-300 KN-mm	My=630 KN-mm

now, area moment of inertia about neutral axis,

$I_{YY}=\frac{60\times 90^{3}}{12}=3.645\times 10^{6}mm^{4}$

area of cross section, A = 90x60 = 5400 mm²

from bending theory,

$\frac{M_{x}}{I_{xx}}=\frac{\sigma}{y}\Rightarrow \sigma_{1}=\frac{M_{x}}{I_{xx}}y=\frac{-300\times 10^{3}}{1.62\times 10^{6}}\times 30=-5.56N/mm^{2}$

( $\sigma_{1}$ will be tensile towards downward from X-X and compressive towards upward from X-X)

$\frac{M_{y}}{I_{yy}}=\frac{\sigma}{x}\Rightarrow \sigma_{2}=\frac{M_{y}}{I_{yy}}x=\frac{630\times 10^{3}}{3.645\times 10^{6}}\times (45)=7.78N/mm^{2}$

( $\sigma_{2}$ will be tensile towards leftward from Y-Y and compressive towards rightward from Y-Y)

and stress due to axial load

$\sigma_{3}=\frac{F}{A}=\frac{-66\times 10^{3}}{5400}=-12.22N/mm^{2}$   

( $\sigma_{3}$ will be compressive all over the section)

Stress at A:   $\sigma_{A}=-5.56+7.78-12.22 = -10N/mm^{2} (Compressive)$

Stress at B:   $\sigma_{B}=5.56+7.78-12.22 = 1.12N/mm^{2} (Tensile)$

Stress at C:   $\sigma_{C}=5.56-7.78-12.22 = -14.44N/mm^{2} (Compressive)$

Stress at D:   $\sigma_{D}=-5.56-7.78-12.22 = -25.56N/mm^{2} (Compressive)$