Question 3 of 8: Beam Deflection by the Method of 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, 2019-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 of composite members 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; combined axial+bending+shear ch.8; beam deflection by integration ch.9). Standard W-shape table (exam Appendix C) is not needed by any question below.
Question 3: Beam Deflection by the Method of Integration (2+18 marks)
Given. Same beam as Q2 ($L=6\text{ m}$, $R_0=1\text{ kN}$, $R_6=5\text{ kN}$, UDL $3\text{ kN/m}$ over $4\le x\le6\text{ m}$). I-section $d=220\text{ mm}$, $b_f=100\text{ mm}$, all thicknesses $10\text{ mm}$: $I=\dfrac{100(220)^3}{12}-\dfrac{90(200)^3}{12}=28.73\times10^6\text{ mm}^4$. $E_{st}=200\text{ GPa}$.
Find. (1) Maximum deflection $v_{max}$ and its location, by double integration; (2) whether $v_{max}\le L/360$.
Same beam as Q2, with the (exaggerated) deflected shape sketched under the applied UDL.
Approach. Integrate $EI\,v^{\prime\prime}=M(x)$ twice using Macaulay's notation to keep one expression valid over the whole span, fix the two constants from $v(0)=v(L)=0$, then locate the zero-slope point.
Moment expression (Macaulay). $EI\,v^{\prime\prime}=M(x)=R_0x-\dfrac{w}{2}\langle x-4000\rangle^2$ (mm, N — $R_0=1000\text{ N}$, $w=3\text{ N/mm}$).
Boundary conditions. $v(0)=0\Rightarrow C_2=0$. $v(6000)=0$ gives $C_1=-5.667\times10^{9}\text{ N}\cdot\text{mm}^2$ (solved with the Macaulay term active, since $6000>4000$).
Locate zero slope. For $x<4000\text{ mm}$ (Macaulay term inactive): $EI\,v^{\prime}=\dfrac{R_0}{2}x^2+C_1=0\Rightarrow x=\sqrt{-2C_1/R_0}=\boxed{3366.5\text{ mm}\;(3.367\text{ m from the left support})}$.
Maximum deflection. Substituting into $EIv$ and dividing by $EI=200{,}000\times28.73\times10^6=5.747\times10^{12}\text{ N}\cdot\text{mm}^2$: $v_{max}=\boxed{2.21\text{ mm}}$ (downward), sketched as the single sagging hump above.
Part (2) — deflection check. $L/360=6000/360=\boxed{16.7\text{ mm}}$. Since $2.21\text{ mm}\ll16.7\text{ mm}$, the beam easily satisfies the deflection requirement (utilised only $\approx13\%$ of the limit).
