Question 6 of 8: Eccentric & Inclined Load on a Column
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-6: Mechanics of Materials — National Exams, December 2014
3 hours duration. Closed book (one hand-written aid sheet permitted). Any FIVE of the
eight questions constitute a complete paper; all eight are solved below as a complete
study resource.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed.; Beer, Johnston, DeWolf & Mazurek,
Mechanics of Materials, 7th ed.; Gere & Goodno, Mechanics of Materials, 8th ed.
Question 6: Eccentric & Inclined Load on a Column (20 marks)
600 kN, vertical, e=250 mm from the column centreline
Inclined load
80 kN on a collar, 3-4-5 slope (down and to the left), applied at the column’s left face, 300 mm above section a-a
Section a-a
200 mm (in-plane) × 50 mm rectangle
Material
σyield=400 MPa, E=200 GPa
Find. (a) the normal stress distribution at section a-a (max/min); (b) the shear stress distribution at section a-a (max/min). Ignore column buckling.
Column with the 600 kN eccentric load, the 80 kN inclined (3-4-5) collar load, and section a-a 300 mm below the collar.
Approach. Reduce every force above section a-a to an equivalent axial force, bending moment, and shear force AT the section, then superpose σ=P/A±My/I and τ=1.5V/A (rectangular section).
Total axial (compressive) force at a-a: $$P_{total}=600+48=648\text{ kN}.$$
Total bending moment about the section centroid (moment increases compression on the +x/eccentricity side): from the 600 kN load, $M_1=600(250)=150{,}000$ kN·mm; from the 80 kN load's vertical component at x=−100 mm, $M_2=48(-100)/1000\cdot1000=-4800$ kN·mm... combining with the horizontal component acting at 300 mm above the section, $$M=e P_{600} + \left(z_{80}F_{x,80}-x_{80}F_{z,80}\right)=1.260e+08\text{ N}\cdot\text{mm} = 126.0\text{ kN}\cdot\text{m}.$$
Section properties (bending about the axis perpendicular to the 250 mm eccentricity, i.e. about the 50 mm-thick axis): $$A=10000\text{ mm}^2,\qquad I=\dfrac{(50)(200)^3}{12}=3.333e+07\text{ mm}^4,\qquad c=100\text{ mm}.$$
Normal stress distribution $\sigma=P/A+Mx/I$ (compression positive): $$\boxed{\sigma_{max}=442.8\text{ MPa (compression, at the eccentric-load side)}}$$ $$\boxed{\sigma_{min}=-313.2\text{ MPa}}$$ (i.e. 313.2 MPa tension at the opposite edge — the 250 mm eccentricity lies far outside the section’s kern, so the far edge goes into net tension; the compressive peak exceeds the 400 MPa yield, worth flagging as a design concern even though the question only asks for the linear-elastic distribution).
Shear stress distribution. Only the 80 kN load’s horizontal component is a transverse (shear) force on this horizontal cut: $$V=64\text{ kN}.$$ For a rectangle, shear is parabolic, zero at the edges and maximum at the centroid: $$\boxed{\tau_{max}=\dfrac{1.5V}{A}=9.60\text{ MPa (at the centroidal axis)}},\qquad \tau=0\text{ at }x=\pm100\text{ mm.}$$
