20,000 in lb 1,000 lb 15,000 in lb 6,000 lb 100 lb/in 25" *5"* 10" 10"> Cross-Section: Solid Circular with 2" Diameter Material Properties (Aluminum): E=10x10 psi E v = 0.3 G= 2(1+v) 1.1 For the beam shown below, (a) Draw the V-Diagram (b) Draw the M-Diagram (c) Find the largest value of σ and give the (x,y,z) coordinates of its location. x (d) Find the largest value of xy, and give the (x,y,z) coordinates of its location. ху
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- REINFORCEB=(335mm)H=(435mm)All steel Fy- 50 ksi, Fu = 65 ksi E = 29,000 ksi 1) What is the purpose of "U" in equation Ae=U An ?(40 How A Rismatic steel bar of a Square cro ss section (looxlo0)mm is loaoled by Compressive force P= 1300 K, the bar has length 2.5 m, Eg=200GPay V=0.3,determine the shortening 8, the increase in the dimensions of cross Section, the change in Volume DV of the bar? loomm loomm
- The compound bar containing steel bronze and aluminum segments carries the axial loads shown in the figure. The properties of the segments are listed in the table. Determine the maximum allowable value of P in pounds if the change in length of the entire bar is limited to +0.07in. Steel Bronze Aluminm (A) 22,250 B) 26,250 25,250 D) 27,250 A (in²) 0.75 1 0.5 2P Steel 2 ft Bronze 4 ft 5P E (psi) 30000000 12000000 10000000 4P+ Aluminum 3 ft P L (FT) 4 3Consider a 1 m-long bar of 2024-T81 aluminum alloy with a 20mm-diameter and the following property data: E=70GPa Y.S.=410MPa T.S.=480MPa %Elongation at failure=8% What will be the length of the bar under a structural load of 1.2×10^5N in tension? Your answer should be in units of meters (m).c). Interpret the following diagram? The effect of carbon and heat treatment on the properties of plain carbon steels. 160 140 Tensile 120 strength 100 Impact energy | 100 Annealed 80 80 Normalized 60 60 Annealed 40 `Yield strength 20-% Elongation Annealed 20 Normalized- 0.2 0.4 0.6 0.8 1.0 Weight percent carbon Figure-4 Yield and tensile strength (ksi) Normalized % Elongation or impact energy (ft • Ib)
- Q5 A specimen of magnesium having a rectangular cross section of dimensions (3.2 mm X 19.1 mm) and gage length 63.5 mm is deformed in tension. Using the load-elongation data tabulated as follows load (N) and Extension (mm), Find the standard mechanical properties. load 1380 2780 5630 7430 8140 9870 12850 | 14100 14340 13830 12500 Ext. 0.03 0.06 0.12 0.2 0.25 0.64 1.91 3.18 4.45 5.72 6.99 a) Draw the stress strain curve. b) Find the young modulus. c) Find the proof stress d) Find the ultimate tensile strength. e) Find the fracture stress f) Find the maximum strain.A force of 100,000 N is applied to an iron bar with a cross-sectional area of 10 mm × 20mm and having a yield strength of 400 MPa and a tensile strength of 480 MPa,Determinea. whether the bar will plastically deform; andb. whether the bar will experience neckingV:EV I+ %9r NO O "l zain IQ 10/10 FIND STRESS IN Aluminum. IF E st= 200000 MPa, Ealu.=70000 .MPa 100 kN RALU=30 MM Aluminum Steel no one 31 MPa 69 MPa 19 MPa 55 MPa RST= 20 MM
- 1-13. The density (m ass/volume) of aluminum is 5.26 slug/ft. Determine its density in SI units. Use an appropriate prefix.} 8) A tensile load of 3000 N is applied to a steel bar of cross-sectional area 500 mm?. If a tensile load was applied to an aluminum bar to achieve the same lateral (transvers) strain with the former (steel bar), what would be the load? Note: Original dimensions of the two bars are the same (Est = 2.1 x 105 MPa, Eat = 0.703 × 105 MPa, vg = 0.3, vat = 0.33) E = º , v = - Saterat Eaxial Fatuminum™ 3000 N A-500 mm² A=500 mm² Fatuminum=? 3000 N Steel Aluminum6. Give the properties table below. A992 Steel 7178 T76 Aluminum Yield Strength 50 ksi 73 ksi Ultimate strength 75 ksi 85 ksi Modulus of Elasticity 29,000 ksi 10,500 ksi Assume you are designing a rectangular member with a specific allowed height. For a design where deflection criterion controls which material would require a larger cross section? For a design where tension stress controls, which material requires the larger cross section?