The ply stress σα was calculated for AS1/3501-6 Graphite/Epoxy laminates [0/+-45]s under load with N1=10,000 N/m and N2=N6=0. When the strength of this material is as follows, determine the stability of the laminate according to the maximum stress theo
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The ply stress σα was calculated for AS1/3501-6 Graphite/Epoxy laminates [0/+-45]s under load with N1=10,000 N/m and N2=N6=0. When the strength of this material is as follows, determine the stability of the laminate according to the maximum stress theory.
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- Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in ) that can be stored in each or the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit. DATA FOR PROBLEM 2.7-5 Material Weight Density (lb/in3) Modulus of Elasticity (ksi) Proportional Limit (psi) Mild sleel 0.284 30,000 36,000 Tool steel 0.284 30,000 75,000 Aluminum 0.0984 10,500 60,000 Rubber (soft) 0.0405 0.300 300The fluctuating stresses listed in the table are found at a critical location of a component made of steel with Se = 40 ksi, Sy = 80 ksi, Sut = 110 ksi and f= 0.87. These stresses are applied on the part within 10 s. What is the accumulative damage of this part? What is the life of the part in hours if this stress pattern continues to repeat for the remained of the part's life? Use Goodman criterion and Miner's rule in your solution. Loading order |Omin o max |Number of cycles -20 30 -10 50 1 3. -30 30 1Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…
- Example: Convert the change in length data in Table 3-2 to engineering stress and strain and plot a stress-strain curve Homework- help Table 3-2 The results of a tensile test of a 0.505 in. diameter aluminum alloy test bar, initial length (1o) = 2 in. Calculated LTO Load (Ib) Change in Length (in.) Stress (psi) Strain (in./in.) 0.000 1000 0.001 0.0005 4,993 14,978 24,963 34,948 37,445 39,442 39,941 39,691 37,944 3000 0.003 0.0015 5000 0.005 0.0025 7000 0.007 0.0035 7500 0.030 0.0150 7900 0.080 0.0400 8000 (maximum load) 0.120 0.0600 7950 0.160 0.0800 7600 (fracture) 0.205 0.1025An ASTM cast iron has minimum ultimate strengths of 32 kpsi in tension and 95 kpsi in compression. Find the factors of safety using Coulomb Fragile Mohr (CMF) and Modified Mohr (MM) theories for the following stress state. σx = −3 kpsi, σy = −9 kpsi, τx y = −4 kpsiGiven: A part is made of ductile material and subjected to principal stresses o1, 02, 03. It is statically loaded. Find: Determine the factor of safety for the part. Use the following parameters in your analysis: ơ1 40 kpsi, Sut = 90 kpsi, Suc = 20 kpsi, 02 = 5 kpsi, oz = -10 kpsi, Sy = %3D = –130 kpsi
- Find the equivalent stresses at Points 1 and 2 of the element with given geometry and loading conditions according to the Maximum shear stress hypothesis and the Maximum strain energy hypothesis. Shaft diameter: 20 mm, Shaft Length 120 mm, F1 = 750N, F2 = 3000N, Mb = 2400 N.mm. Steel if St37 and Safety coefficient is 2 If taken, will this stick work safely under these operating conditions? (Yield of given steel Strength 225 Mpa, Tensile strength 370 Mpa)Part A The stress-strain diagram for a steel alloy having an original diameter of 0.40 in. and a gage length of 9 in. is shown in the figure below. (Figure 1) Determine the modulus of elasticity for the material. Express your answer to three significant figures and include appropriate units. HA ? Eapprox = Value Units Submit Request Answer Figure < 1 of 1 Part B a (ksi) 80 Determine the load on the specimen that causes yielding. 70 Express your answer to three significant figures and include appropriate units. 60 50 40 HA 30 20 Py = Value Units 10 € (in./in.) 0.04 0,08 0.12 0,16 0,20 0,24 0,28 O 0.0005 0.001 0.0015 0.002 0.0025 0.0030.0035 Request Answer SubmitUsing the Maximum Strain Criterion, determine the uniaxial failure stress, Ox, for off-axis loading of the unidirectional laminar in Fig. if the material is AS/3501 graphite/epoxy and the angle 0=30 .
- 4. The maximum stress a human tendon can withstand is estimated to be 1200 MPa. If you were to test it for rupture what load cell you should use (25ON or 5kN) what problems may occur if you select an incorrect load cell. 5. The figure below is a J-shaped (or concave upward) stress-strain curve. What does a J-shaped stress stain curve indicate about a material's response to stress and tendency to yield? Name two materials that have J-shaped stress strain curves.A material has the following properties, ultimate Sul = σul = 350mpa, the strain hardening exponent n=0.20. Determine the value for the strength coefficient K.See the set of bars for made of a material that has a yield strength of 350 MPa, determine the minimum h1 and h2 dimensions required. Apply a safety factor F.S. = 1.5 without flow. Each bar is 12 mm thick