Question 8 of 8: W150x30 Overhang Beam: Tip Deflection by Integration
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Basic Studies, 04-BS-6 Mechanics of Materials, 2018-Dec. Closed book; one hand-written aid sheet permitted. The exam instructs "any FIVE of the eight questions constitute a complete paper," but every question is solved below as a full study resource.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial deformation & statically indeterminate assemblies ch.4; shear/moment diagrams ch.6; stress transformation & Mohr's circle ch.9; buckling of columns ch.13; torsion of circular shafts ch.5; transformed-section composite beams ch.6; shear flow/VQ/Ib ch.7; beam deflection by integration ch.9). Standard steel W-shape tables (Appendix C of the exam) per CISC Handbook of Steel Construction.
Question 8: W150x30 Overhang Beam: Tip Deflection by Integration (20 marks)
Given. Pin support at $A$ ($x=0$); roller support at $B$ ($x=4\text{ m}$); overhang tip $C$ ($x=6\text{ m}$), concentrated load $P=30\text{ kN}$ downward at $C$ only (no distributed load). Section $W150\times30$ (Appendix C table): $I_x=17.1\times10^6\text{ mm}^4$. $E=200\text{ GPa}$.
Find. The deflection and slope of the beam at the overhang tip C, by the Method of Integration.
Overhang beam A(pin)-B(roller)-C(tip); 30 kN load at the free overhang tip C only.
Approach. Find the (possibly negative) reactions from global equilibrium, write EIy''=M(x) separately for the main span (0 to 4 m) and the overhang (4 to 6 m), integrate each twice, and solve the four constants from y(0)=0, y(4)=0 (both supports) and continuity of slope/deflection at B, then evaluate the overhang-segment expressions at x=6 m.
Reactions. $\sum M_A=0$: $R_B(4)=P(6)\Rightarrow R_B=45\text{ kN}$. $\sum F_y=0$: $R_A=P-R_B=30-45=\boxed{-15\text{ kN}}$ — negative, meaning the pin at A must pull DOWN to hold the beam (the tip load's moment about B exceeds what the span alone can resist); this is possible because A is a PIN, not a roller.
Moment expressions. Main span ($0\le x\le4$): $M_1(x)=R_Ax=-15x$. Overhang ($4\le x\le6$): $M_2(x)=R_Ax+R_B(x-4)=-15x+45(x-4)=30x-180$ (check: $M_2(6)=0$ at the free tip, as required).
Boundary/continuity conditions. $y_1(0)=0\Rightarrow C_2=0$; $y_1(4)=0$ fixes $C_1$; slope continuity $y_1^{\prime}(4)=y_2^{\prime}(4)$ fixes $C_3$; deflection continuity $y_2(4)=y_1(4)=0$ fixes $C_4$. Solving the four equations gives the constants that make $EIy_2$ and $EIy_2^{\prime}$ fully explicit over the overhang.
Evaluate at C ($x=6\text{ m}=6{,}000\text{ mm}$, $EI=200{,}000\times17.1\times10^6=3.42\times10^{12}\text{ N}\cdot\text{mm}^2$). $y_C=\boxed{-70.18\text{ mm}}$ (downward), $y_C^{\prime}=\boxed{-0.0409\text{ rad}}$ ($2.35^{\circ}$). Both are cross-checked against the closed-form overhang-tip formulas $\delta_C=Pa^2(L_1+a)/(3EI)$ and $\theta_C=PaL_1/(3EI)+Pa^2/(2EI)$ (with $a=2\text{ m}$ the overhang length, $L_1=4\text{ m}$ the main span), which reproduce the same magnitudes exactly.
