2. Two lines, each 10mm in length, are drawn on a thin-walled cylindrical pressure vessel as shown. The tank is then subjected to internal pressure p = 1MPa. This pressure causes the line oriented in the axial direction to become 10.0026mm long. The line in the circumferential direction becomes 10.0088mm. What are the normal strains in the axial and circumferential directions? What are Young's modulus and Poisson T r=100mm 1 It=1mm + SECTION OF CYLINDRICAL PRESSURE VESSEL Hooke's Law ①월 2 - 1 2 = @2a = 2 - V 얕 0 20-Yields 0= 0 ₁ - 2t • • 9 IMPa (100mm) (Imm) Uxial IMPa (100mm) 2 (1mm) 0.0026mm 10mm 0.0088mm why no ? /ircum. 10mm 100MPa immmm -→ BOMPA min 10.0026mm-10mm 10mm 10.0088mm -10mm 10mm 0.0026mm Tomm 0.0088mm 10mm 2(E(0.001008)= 100MPa - J (ROMPa)) ⇒E(0.00176) = 1200MPa - X (100 Mpa) -(E(0.001026)=50MPR + 7 (100MPa)) 0.00026 mmmmm * Put known values into Ⓒ & 2 & bring E to L.H.S * DE(Ec)=0-√(a) E(0.00806088) = 100 MPA – ✓ (BOMPA) - @E(E)= a√ (c) E(0.001026)=50MPR-7 (100MPa) How do you Solve? 0.00088 munun 1 = E(0.00100) = 150MPa 0.00150 0.00150 →E=100G Pa| Young's Modulus √=0.24 Poisson ratio
2. Two lines, each 10mm in length, are drawn on a thin-walled cylindrical pressure vessel as shown. The tank is then subjected to internal pressure p = 1MPa. This pressure causes the line oriented in the axial direction to become 10.0026mm long. The line in the circumferential direction becomes 10.0088mm. What are the normal strains in the axial and circumferential directions? What are Young's modulus and Poisson T r=100mm 1 It=1mm + SECTION OF CYLINDRICAL PRESSURE VESSEL Hooke's Law ①월 2 - 1 2 = @2a = 2 - V 얕 0 20-Yields 0= 0 ₁ - 2t • • 9 IMPa (100mm) (Imm) Uxial IMPa (100mm) 2 (1mm) 0.0026mm 10mm 0.0088mm why no ? /ircum. 10mm 100MPa immmm -→ BOMPA min 10.0026mm-10mm 10mm 10.0088mm -10mm 10mm 0.0026mm Tomm 0.0088mm 10mm 2(E(0.001008)= 100MPa - J (ROMPa)) ⇒E(0.00176) = 1200MPa - X (100 Mpa) -(E(0.001026)=50MPR + 7 (100MPa)) 0.00026 mmmmm * Put known values into Ⓒ & 2 & bring E to L.H.S * DE(Ec)=0-√(a) E(0.00806088) = 100 MPA – ✓ (BOMPA) - @E(E)= a√ (c) E(0.001026)=50MPR-7 (100MPa) How do you Solve? 0.00088 munun 1 = E(0.00100) = 150MPa 0.00150 0.00150 →E=100G Pa| Young's Modulus √=0.24 Poisson ratio
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Can someone please explain the parts marked on the photo. Like why is there no two included in the other equation and the steps taken to find Poisson equation. Please actually show step by step including the calculations to find the final answer. Thanks
Transcribed Image Text:2. Two lines, each 10mm in length, are drawn on a thin-walled cylindrical pressure vessel as shown. The
tank is then subjected to internal pressure p 1MPa. This pressure causes the line oriented in the axial
direction to become 10.0026mm long. The line in the circumferential direction becomes 10.0088mm. What
are the normal strains in the axial and circumferential directions? What are Young's modulus and Poisson
• Ea=
→0.00026
↑
r=100mm
L.
t=1mm
QE²
+
SECTION OF CYLINDRICAL
PRESSURE VESSEL
Hooke's Law
Oa
0²₁= €²-√²/²0²-
0&c=
Oa
E
2x1-2 Yields
✓
Oc
Pr
Qt
•
IMPa (100mm)
(Imm)
ircum.
IMPa (100mm)
2 (1mm)
0.0026mm
10mm
why no 2 ?
100MPa
0.0000mm
10mm
→ бомра
10.0026mm-10mm
10mm
10.0088mm -10mm
10mm
0.0026mm
10mm
0.0088mm
10mm
* Put known values into Ⓒ && bring E to L.H.S*
(DE(Ec)= 0c-V (0a) ⇒E(0.000008) = 100MPA - J (ROMPA)
-
@E(E₂) = 0₁-√(0₁) ➜ E(0.001026)=50MPa - J (100MPa)
2€ (0.001008) = 100 MPa - J (BOMPa) ⇒ E(0.001760) = 1200MPa - J (100 MPa)
-(E(0.001026)=50MPa + 7 (100MPa)))
0.00088
How do you
solve?
= E(0.00190) = 150MPa
0.00150
0.00150
|E=100G Pa| Young's Modulus
√= 0.24| Poisson ratio
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