1. POLY2 is ℎ = 1.5µμ𝑚 thick and the minimum feature size is 𝑤 = 2µμ𝑚 a. 1 pt. for each correct spring constant (4 pts. total) Assuming every loading case is a force at the tip and the beam is oriented with its length along the x-­‐axis, width along the y-­‐axis, and height along the z-­‐axis 𝐾!! =
=
150×10!
𝐸𝐴 𝐸𝑤ℎ
=
𝐿
𝐿
N
2µμm 1.5µμm
m!
100µμm
𝐾!! = 1.5 2 1.5 ×10!!!!!!!! N/m 𝑲𝒙𝒙 = 𝟒. 𝟓×𝟏𝟎𝟑 𝐍/𝐦 𝐾!!
=
=
𝐸ℎ𝑤 !
=
4𝐿!
N
1.5µμm 2µμm !
m!
4 100µμm !
150×10!
1.5 1.5 8
×10!!!!!!"!!" N/m 4
𝑲𝒚𝒚 = 𝟎. 𝟒𝟓𝐍/𝐦 𝐾!! =
=
=
𝐸𝑤ℎ!
4𝐿!
N
2µμm 1.5µμm !
m!
4 100µμm !
150×10!
1.5 2 3.4
×10!!!!!!"!!" N/m 4
𝑲𝒛𝒛 ≅ 𝟎. 𝟑𝐍/𝐦 𝐾!" =
2𝐸𝐼! 𝐸𝑤ℎ!
=
𝐿!
6𝐿!
=
=
N
2µμm 1.5µμm !
m!
6 100µμm !
150×10!
1.5 2 3.4
×10!!!!!!"!! N 6
𝑲𝜽𝒛 = 𝟐×𝟏𝟎!𝟓 𝐍 Note that the units of 𝐾!" are in Newtons/radian b. 1 pt. for the force 1 pt. for the moment The force at the tip of the beam is simply 𝐹 = 𝑚𝑔, where 𝑚 is the mass of the plate = 2300
kg
m!
𝐹 = 𝑚𝑔 = 𝜌𝑉𝑔 100µμm
!
1.5µμm (10m/s ! ) 𝑭 = 𝟑. 𝟓×𝟏𝟎!𝟏𝟎 𝐍 The moment is caused by the offset of the plate mass from the beam tip. The center of mass of the plate is 𝐿!" = 50µμm from the tip assuming the mass is uniform, so 𝑴 = 𝒎𝒈𝑳𝒄𝒎 = 𝟐×𝟏𝟎!𝟏𝟒 𝐍𝐦 c. 1 pt. for beam tip linear deflection 1 pt. for beam tip angular deflection 1 pt. for the deflection of the end of the plate 1 𝑀𝐿! 𝐹𝐿!
𝑧=
+
𝐸𝐼 2
3
=
12
150×10!
N
m!
2µμm 1.5µμm
!
2×10!" Nm 100µμm
2
𝒛 = 𝟐𝐧𝐦 𝑧 ! = 𝜃! =
1
𝐹𝐿!
𝑀𝐿 +
𝐸𝐼
2
𝜽𝒛 = 𝟒×𝟏𝟎!𝟓 (𝒓𝒂𝒅𝒊𝒂𝒏𝒔) !
+
3.5×10!!" N 100µμm
3
!
Since the angular deflection at the beam tip is small, the deflection of the end of the plate is given by simple trig approximations 𝒅𝒛𝒑𝒍𝒂𝒕𝒆!𝒕𝒊𝒑 = 𝜽𝒛 𝑳𝒑𝒍𝒂𝒕𝒆 = 𝟒𝒏𝒎 d. 1 pt. for the acceleration Assuming the fracture strain of silicon is about 2% 𝐹 = 𝑚𝑎 = 𝑘𝑥 =
𝑎=
𝑎=
𝐸𝐴
𝑥 𝐿
𝐸𝐴 𝑥 𝐸𝐴
=
𝜖 𝑚𝐿
𝑚
N
1.5µμm 2µμm
m!
(0.02) 3.5×10!!! kg
150×10!
𝒂 = 𝟑×𝟏𝟎𝟖
𝐦
𝐬𝟐
e. 1 pt. for the acceleration 1 pt. for noting where the beam will break The maximum strain in a bent cantilever is always at the base of the anchored end, furthest from the neutral axis. This is where the beam will break. At the top surface of the beam 𝑧
ℎ
ℎ
𝜖!"# =
=
→𝜌 𝑥 =
𝜌 𝑥
2𝜌 𝑥
2𝜖!"#
At the base of the beam 𝑀 0 =
𝐸𝐼!
= 𝑀! + 𝐹! 𝐿 = 𝑚𝑎! (𝐿!" + 𝐿) 𝜌 0
𝑎! =
𝐸𝐼!
𝜌 0 (𝑚(𝐿!" + 𝐿))
Combining these equations, we get 𝒂𝒚 =
𝟐𝝐𝑬𝑰𝒚
𝐦
= 𝟖×𝟏𝟎𝟓 𝟐 𝒉(𝒎(𝑳𝒄𝒎 + 𝑳))
𝐬
2. 1 pt. for each scaling relation (11 pts. total) a. In part 1a. 𝐾!!
𝐴 𝑆!
∝ ∝
→ 𝑆 𝐿
𝑆
𝐾!! ∝
ℎ𝑤 !
∝ 𝑆 𝐿!
𝐾!! ∝
𝑤ℎ!
∝ 𝑆 𝐿!
𝐾!"
𝑤ℎ!
∝ ! ∝ 𝑆 ! 𝐿
In part 1b. 𝐹 ∝ 𝑚 ∝ 𝑉 ∝ 𝑆 ! 𝑀 ∝ 𝑚𝐿! ∝ 𝑆 ! In part 1c. 𝐹𝐿! 𝑆 !
𝑧∝
∝ ! → 𝑆 ! 𝐼
𝑆
𝜃! ∝
𝐹𝐿!
∝ 𝑆 𝐼
𝑑𝑧!"#$% ∝ 𝜃! 𝐿 ∝ 𝑆 ! In part 1d. 𝑎!"# ∝
𝐴 𝑆!
∝
→ 𝑆 !! 𝑚 𝑆!
In part 1e. 𝑎!"# ∝
𝐼
𝑆!
∝ ! → 𝑆 !! ℎ𝑚𝐿 𝑆
b. 1 pt. for each scaling relation (2 pts. total) The acceleration scales as 𝑆 !! In x 𝑆 !! =
𝑆=
𝑎!"#!
𝑎!"#!
𝑎!"#! 4×10! 𝑔
=
𝑎!"#!
𝑔
𝑺 = 𝟒×𝟏𝟎𝟕 In y 𝑺=
𝒂𝒎𝒂𝒙𝟏
= 𝟒×𝟏𝟎𝟒 𝒂𝒎𝒂𝒙𝟐
3. 1 pt. for each cross sectional area (4 pts.) 1 pt. for commenting on safety 𝐹 = 𝑘𝑥 =
𝐸𝐴
𝑥 = 𝐸𝐴𝜖 𝐿
𝐴=
𝐹
𝐸𝜖
Assuming 𝜖 = 0.02 m
50kg 10 !
𝑚𝑔
s
𝐴=
=
N
𝐸𝜖
!
150×10 ! 0.02
m
𝑨𝑺𝒊 = 𝟐×𝟏𝟎!𝟕 𝐦𝟐 The ultimate tensile strength for aluminum is about 300MPa. For glass it is 33MPa. For Steel, it varies a lot, but stainless steel has a max tensile strength of about 800MPa. All values from https://en.wikipedia.org/wiki/Ultimate_tensile_strength 𝑨𝑨𝒍 =
𝑭
𝝈𝒎𝒂𝒙
=
𝒎𝒈
= 𝟐×𝟏𝟎!𝟔 𝒎𝟐 𝝈𝒎𝒂𝒙
𝑨𝒈𝒍𝒂𝒔𝒔 = 𝟐×𝟏𝟎!𝟓 𝒎𝟐 𝑨𝒔𝒕𝒆𝒆𝒍 = 𝟔×𝟏𝟎!𝟕 𝐦𝟐 You would probably not feel safe hanging from the brittle materials (silicon, glass) since they are prone to snapping. 4. 1 pt. for an approximately correct final cross section 1 pt. for the dimple under poly1 1 pt. for putting metal on poly2 1 pt. for having poly2 contact poly1 and poly0 1 pt. for no metal on poly1 (it gets taken away during the oxide etch) Nitride
Poly0
Substrate
Poly1
Poly2
5. 1 pt. for each part correct (5 pts. total) -­‐1 pt. total if you forgot the nitride a. Substrate, nitride, poly 0, 1st oxide, poly 1, 2nd oxide, poly 2, metal b. Substrate, nitride, poly 0, poly 1, poly 2 c. Substrate, nitride, poly2, metal d. Substrate, nitride, 1st oxide, 2nd oxide, metal e. Substrate, nitride, 1st oxide, 2nd oxide 6. 5 pts. distributed as you see fit