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.
T-shape: bottom flange 150×30 mm, web 40×150 mm above it (total depth 180 mm)
Material
σallow=260 MPa, τallow=60 MPa, E=200 GPa
Point E
bottom-left corner of the flange (a free surface/corner)
Find. (a) the maximum (absolute) normal stress in the beam; (b) the maximum shear stress in the beam; (c) the shear stress at E, at a section 2 m from the left support.
T-beam elevation (30 kN load at 3 m) with the T cross-section and point E marked at the bottom-left corner of the flange.
Approach. Locate the centroid and I of the T-section, find the reactions and the moment/shear diagrams, then apply the flexure formula for (a) [at the section of maximum moment] and the shear formula τ=VQ/(Ib) for (b) and (c) [at x=2 m].
Centroid (measured up from the bottom): $$\bar{y}=\dfrac{A_f y_f + A_w y_w}{A_f+A_w}=66.43\text{ mm}$$ (Af=150×30=4500 mm² at yf=15 mm; Aw=40×150=6000 mm² at yw=105 mm).
Moment of inertia (parallel-axis theorem on each part): $$I=3.242e+07\text{ mm}^4,\qquad c_{top}=113.57\text{ mm},\qquad c_{bot}=66.43\text{ mm}.$$
Reactions and moment at the section of interest. With the 30 kN load at 3 m (4 m from the right support), $$R_A=17.1\text{ kN}.$$ The maximum moment occurs directly under the load: $$M_{max}=R_A(3\text{ m})=34.29\text{ kN}\cdot\text{m}.$$
Maximum normal stress (top of the web is farther from the low-lying centroid than the bottom, so it governs even though both fibres see the same moment): $$\sigma_{top}=\dfrac{M_{max}\,c_{top}}{I}=120.1\text{ MPa (compression)},\qquad \sigma_{bot}=\dfrac{M_{max}\,c_{bot}}{I}=70.3\text{ MPa (tension)}$$ $$\boxed{\sigma_{max}=120.1\text{ MPa}}$$ — under the 260 MPa allowable.
Shear at x=2 m (left of the load, so $V=R_A$): $$V=17.1\text{ kN}.$$ Maximum shear occurs at the neutral axis, using the web width b=40 mm there: $$Q_{NA}=257969\text{ mm}^3\qquad\Rightarrow\qquad \boxed{\tau_{max}=\dfrac{VQ_{NA}}{Ib}=3.41\text{ MPa}}$$
Shear at point E. E sits at the very bottom-left corner of the flange — a free surface in two directions at once. The first moment of area beyond that point is zero ($Q_E=0$), so $$\boxed{\tau_E=0}$$ This is the expected result at any free corner of an open section, and is the actual teaching point of this sub-part: a student who reports a non-zero τ at E has misapplied VQ/Ib without recognising E lies on a stress-free boundary.
