Question 5 of 8: Eccentric Inclined Load on a T-Shaped Column

Question 5: Eccentric Inclined Load on a T-Shaped Column (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A T-shaped element fixed at its 600×400 mm base, loaded by a 6000 kN force at 53.13° below horizontal applied at the peak of the trapezoidal top, 300 mm right of the stem centreline and 1500 mm above the base.

Find. Normal and shear stress distributions at the fixed base.

Approach. Resolve the inclined load into horizontal and vertical components, transfer both to the base centroid to get the axial force, shear, and bending moment, then superpose $\sigma=N/A\pm Mc/I$ for the normal stress and $\tau=1.5V/A$ for the shear.

1. Resolve the load. With P=6000 kN at 53.13° below horizontal, pointing down-and-left into the apex ($\cos53.13^\circ=0.6$, $\sin53.13^\circ=0.8$): $$F_x=-6000(0.6)=\boxed{-3600\text{ kN}},\qquad F_y=-6000(0.8)=\boxed{-4800\text{ kN}}$$
2. Locate the load point. The 1600 mm-wide top spans 600+600+400 mm; the apex sits at 1200 mm from the left edge, while the 600 mm-wide stem (Section A-A) spans 600–1200 mm (centreline at 900 mm). So the apex is $e=1200-900=300$ mm to the right of the stem centreline, and $h=300+400+800=1500$ mm above the base.
3. Internal forces at the base. The vertical component gives the compressive axial force, the horizontal component gives the shear, and BOTH the eccentric vertical component and the horizontal component (acting at height h) contribute to the bending moment about the base centroid: $$N=F_y=\boxed{-4800\text{ kN}}\ (\text{compression}),\qquad V=F_x=\boxed{-3600\text{ kN}}$$ $$M_z=eF_y-hF_x=300(-4800)-1500(-3600)=\boxed{3960\text{ kN}\!\cdot\!\text{m}}$$ the two contributions PARTIALLY OFFSET (the eccentric vertical load adds compression on the loaded side, while the horizontal shear bends the column so that same side goes into tension) but the horizontal-shear term dominates.
4. Section properties. Base A=600×400 mm, bending about the axis through the 400 mm depth (so stress varies across the 600 mm width): $$A=240000\text{ mm}^2,\quad I=\frac{(400)(600)^3}{12}=7.2000e+09\text{ mm}^4,\quad c=300\text{ mm}$$
5. Combined normal stress, $\sigma=N/A\pm M_zc/I$. $$\sigma(+300\text{ mm})=\boxed{145.0\text{ MPa (tension)}},\qquad \sigma(-300\text{ mm})=\boxed{-185.0\text{ MPa (compression)}}$$ Both magnitudes are well under the 350 MPa yield.
6. Shear stress. For a solid rectangle, $\tau_{max}=1.5|V|/A$ at the neutral axis, tapering to zero at the extreme fibres: $$\tau_{max}=1.5\frac{3600{,}000}{240{,}000}=\boxed{22.5\text{ MPa}}$$ well under the 60 MPa allowable.