Question 8 of 8: Wood–Resin Composite; Borosilicate Glass Design; Sledgehammer Materials Selection

Question 8: Wood–Resin Composite; Borosilicate Glass Design; Sledgehammer Materials Selection (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) $V_0=20\ \text{cm}^3$ dry birchwood, $\rho_{dry}=0.56\ \text{g/cm}^3$, resin SG$=1.30$, impregnated true density $\rho_{true}=1.52\ \text{g/cm}^3$. (b) Pure SiO$_2$–B$_2$O$_3$ glass, O:Si $\le2.5$; atomic masses (page 1) B$=10.8$, Si$=28.1$, O$=16.0$ g/mol. (c) 8 lb sledgehammer head for driving steel fence posts.

Find. (a) Grams of resin and final bulk density. (b) Maximum wt% B$_2$O$_3$. (c) Recommended hammer-head material and justification.

Approach

Part (a) is a mass balance at (essentially) constant bulk volume — the resin is assumed to fill the wood's existing internal pore space rather than adding extra external volume, so the "true density" given for the finished composite already fixes both the required resin mass and the resulting bulk density. Part (b) converts the given oxygen-to-silicon mole-ratio constraint into a maximum mole fraction of B$_2$O$_3$ in the glass, then converts that mole fraction to a weight percent using the given atomic masses. Part (c) is a properties-to-service-conditions matching exercise: list the demands the service places on the material, then select a material/heat-treatment combination that satisfies all of them simultaneously.

1. (a) Resin mass balance. Dry birch mass: $m_{dry}=\rho_{dry}V_0=0.56\times20=11.2$ g. Since the resin is assumed to completely fill the existing pore volume without expanding the piece (bulk volume stays $20\ \text{cm}^3$), the final total mass is fixed by the given true density: $$m_{total}=\rho_{true}V_0=1.52\times20=30.4\ \text{g}\quad\Rightarrow\quad m_{resin}=m_{total}-m_{dry}=30.4-11.2=\boxed{19.2\ \text{g}}.$$
2. Final bulk density. Because the total volume is unchanged at $20\ \text{cm}^3$ (the resin displaced only pre-existing void space, not new bulk volume), $$\rho_{bulk,final}=\frac{m_{total}}{V_0}=\frac{30.4}{20}=\boxed{1.52\ \text{g/cm}^3}$$ — identical to the given "true density," confirming that full pore-filling at constant external volume is the self-consistent interpretation of the stated data.
3. (b) Set up the O:Si mole-ratio constraint. Mix $(1-x)$ mol SiO$_2$ with $x$ mol B$_2$O$_3$ (mole fraction basis). Each mol SiO$_2$ supplies 1 Si and 2 O; each mol B$_2$O$_3$ supplies 0 Si and 3 O (boron is not counted in this ratio — only silicon is, per the problem's explicit "O:Si" wording). Total Si $=(1-x)$; total O $=2(1-x)+3x=2+x$: $$\frac{O}{Si}=\frac{2+x}{1-x}\le2.5.$$
4. Solve for the maximum mole fraction. $$2+x\le2.5(1-x)=2.5-2.5x\ \Rightarrow\ 3.5x\le0.5\ \Rightarrow\ x\le\boxed{1/7=0.1429}\ (14.29\ \text{mol\%}).$$
5. Convert to weight percent. $M_{SiO_2}=28.1+2(16.0)=60.1$ g/mol, $M_{B_2O_3}=2(10.8)+3(16.0)=69.6$ g/mol. Per 1 mol total oxide mix at $x=1/7$: $$m_{SiO_2}=\tfrac67(60.1)=51.51\ \text{g},\qquad m_{B_2O_3}=\tfrac17(69.6)=9.94\ \text{g},$$ $$\text{wt\%}\,B_2O_3=\frac{9.94}{9.94+51.51}\times100=\boxed{16.2\%}.$$ Assumption: boron enters the glass network as trigonal BO$_3$ units and is not itself counted in the "Si" denominator of the stated O:Si design ratio (only the extra oxygen it contributes is counted); the mix is treated as a simple two-oxide blend with no additional network modifiers.
6. (c) Sledgehammer-head material selection. Service conditions for a striking tool driving steel fence posts: repeated high-energy impact loading (thousands of strikes over its life), a striking face that must resist wear/deformation (mushrooming) against a hard steel target, and a body that must not shatter under shock — i.e. the material needs simultaneously high hardness at the striking face and high fracture toughness/impact resistance through the body, a combination that rules out both very hard brittle materials (fully-hardened tool steel, cast iron, ceramics — would chip/shatter) and very soft/ductile ones (dead-soft mild steel, aluminum, most polymers — would deform/mushroom and fail to transfer impact energy efficiently). The standard, correct choice is a medium-carbon or low-alloy steel (e.g. AISI 1045, 4140, or a drop-forged tool steel), quenched and tempered to a moderate hardness (roughly 45–55 HRC at the striking faces, often via selective/induction hardening so the core stays softer and tougher) — this balances wear resistance at the impact surface against the toughness needed to survive repeated shock loading without brittle fracture, exactly the same design logic used for cold chisels, axe heads, and other forged striking tools.