Unformatted text preview: EML 4321 – HW #7 – Due 4 March, 2011 1. A cm square bar(430 steel), 30 mm wide, radius of the bend is 1.5 cm. a. F= wYh2 10 mm ⋅ 430 N / mm2 ⋅ (10 mm)2
=
= 14333 N = 1.46tonnes
L
30 mm b. 30 tan 45° = 15
→ x = 15mm x X c. 3 ⎛ 430 N / mm 2 ⋅1.5 mm ⎞
⎛ 430 N / mm 2 ⋅1.5 mm ⎞
ρ
⎛Yρ ⎞
⎛Yρ ⎞
= 1+ 4⎜
− 3⎜
= 1+ 4 ⎜
− 3⎜
⎟
⎟ = 0.999
⎟
⎟
2
2
ρf
⎝ Eh ⎠
⎝ Eh ⎠
⎝ 200000 N / mm ⋅10 mm ⎠
⎝ 200000 N / mm ⋅10 mm ⎠
3 αf
ρ αf
=
→ 0.999 =
∴α f = 89.91°
ρf α
90° Interior angle =180° − 89.91° = 90.09° 2. A sheet of 240 steel, thickness=1.2 mm, 300 mm wide, substantial friction and strain hardening a. Assume, C = 1.5, F = CY0 hw = 1.5 ⋅ 240 N / mm2 ⋅1.2 mm ⋅ 300 mm = 129600 N = 13.22tonnes b. l0t0 w0 = l1t1w1 (Volume )
e240 Steel (= 0.25) = l1 − l0
→ 1.25l0 = l1 , w0
l0 w1 ∴ t0 = 1.25t1 →1.2 = 1.25t1 ⇒ t1 = 0.96mm c. In case of 430 Steel ,
F = CY0 hw = 1.5 ⋅ 430 N / mm2 ⋅1.2 mm ⋅ 300 mm = 232200 N = 23.68 tonnes
e430 Steel (= 0.19) = l1 − l0
→ 1.19l0 = l1 , w0
l0 ∴ t0 = 1.19t1 →1.2 = 1.19t1 ⇒ t1 = 1.01mm w1 3. 4. A circular blank of 2.7 mm 240 Steel, drawn in to a cup (dia.: 10 cm, deep: 8 cm) a. F = Y π dh = 240 N / mm 2 ⋅ π ⋅100 mm ⋅ 2.7 mm = 203575 N = 20.76 tonnes b. Volume of steel in cup = π 4 ⋅ (10 cm) 2 ⋅ 8 cm = π 4 ⋅ d 2 ⋅ 2.7 cm ∴ d = 17.21cm 5. a. A = bh = 5 mm ⋅ 20 mm = 100 mm 2 b. Q = MRR = bhv = 0.5 cm ⋅ 2 cm ⋅ 7000 cm / min = 7000 cm 3 / min c. Ft = K s A = 2300 N / mm 2 ⋅100 mm 2 = 230000 N = 23.45 tonnes cf . K s = 2300 N / mm 2 ( for normalized 1035) inTable 7.1 d. P = K s Q = 2300 N / mm 2 ⋅ 7000cm 3 / min⋅ min / 60 sec ⋅ = 268 kW 6. 1045 5140
302 1020 4140 1035
Ti‐6Al‐4V 7075‐T6 ...
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This note was uploaded on 09/06/2011 for the course EML 4321 taught by Professor Staff during the Spring '08 term at University of Florida.
 Spring '08
 Staff

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