An aluminum [E = 10,000 ksi] control rod with a circular cross section must not stretch more than 0.22 in. when the tension in the rod is 2100 lb. If the diameter of th
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An aluminum [E = 10,000 ksi] control rod with a circular cross section must not stretch more than 0.22 in. when the tension in the rod is 2100 lb. If the diameter of the rod is 0.372 in., determine the maximum length of the rod.
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- This arm is attached to a wall on the left. The right end is free hanging. Each section is 2.0m wide and 3.0m high. If the force at the end of the arm is F = 200.0N, what is the force in member 9? Use the Method of Sections. Use a positive number if it is in tension and a negative number if it is in compression.7. A 20-mm-wide block is firmly bonded to rigid plates at its top and bottom. When the force P is applied the block deforms into the shape shown by the dashed line. Determine the magnitude of P. The block's material has a modulus of rigidity of G = 26 GPa. Assume that the material does not yield and use small angle analysis. 150 mm - 0.5 mm 150 mm [Ans: P= 260 kN2. Two 1-mm diameter wires support the rigid bar as shown. Initially, the 300 mm long rigid bar is horizontal. Determine the value of dimensions a and b so that when the load is applied in the bar, the bar will remain horizontal. After the load is applied using the computed values of a and b, the temperature increases by 15 degrees. What will be the angle of the bar with respect to the horizontal after the temperature change? Use E = 200 GPa. 100 cm 75 cm a b Rigid bar 100 N
- Question 4 4. A string has a diameter of 1 cm and the original length of 2 m. The string is pulled by a force of 200 N. Determine the change in length of the string! Young's modulus of the string = 5 x 109 N/m2.The two rods (A-B and B-C) shown in the left panel are connected by pins at A, B, and C. The cross sections of A-B and B-C are 10mm×13mm and 10mm×10mm, respectively. The bilinear stress-strain relation shown in the right panel is for the material used to make the two rods, where Ei is the slope of the stress-strain curve for stresses between 0 and 80 MPa and E2 is the slope for stresses larger than 80 MPa. Note that Hooke's law is only valid for stresses up to 80 MPa. A 1 1 B 0.900 m 3 P = 30 kN 2.44 m C σ MPa 80 E₂ = 40 GPa E₁ = 80 GPa E a) Given the conditions above, find the axial elongation for each rod when subjected to P=30KN. b) Find the final position of Point B due to the applied load P at that location. Hint: Assume point B moves by unknown amounts Ax and Ay from its initial position, and then use right triangles to write the known final lengths of the cables in terms of Ax and Ay. The equations can be simplified by assuming terms Ax² and Ay² are so small they can be…Figure 3: Given the figure, what is the maximum tension of that member? (use 2 decimal point, indicate the answer in kN. Ex: 3.87kN just encode 3.78)
- A 1∕2-in.-diameter round steel rod 40 ft long supports a loadof 4 kips. How much will it elongate? e= ΔL /L f = P/A E = (P∕A)/ (ΔL∕L) = (PL) / (A(ΔL))A 129 ft tower is located on the side of a mountain that is inclined 19* to the horizontal. A guy wire is to be attached to the top of the tower and anchored at a point 42 ft downhill from the base of the tower. Find the shortest length of wire needed. 129 ft 42 ft 19° NOTE: The picture is NOT drawn to scale. length of guy-wire = ftA 16-mm-diameter steel [E= 200 GPa] rod (2) is connected to a 50-mm-wide by 8-mm-thick rectangular aluminum [E= 70 GPa] bar (1). Determine the force Prequired to stretch the assembly 3.00 mm.
- Compound axial member ABC has a uniform diameter of d-2.4 in. Segment (1) is an aluminum [E₁ - 10,000 ksi] alloy and segment (2) is a copper [E₂ - 17,000 ksi] alloy. The lengths of segments (1) and (2) are L₁ - 78 in. and L₂ - 150 in., respectively. Determine the force P required to stretch compound member ABC by a total of 0.35 in. L₁ A Part 1 Aluminum Answer: A- B Save for Later Calculate the cross-sectional area of the compound member. L2 (2) Copper in.² C P Attempts: 0 of 1 used Submit Answer Part 2 The parts of this question must be completed in order. This part will be available when you complete the part above. Part 3 The parts of this question must be completed in order. This part will be available when you complete the part above.Q1: Determine the force in members EG, HG, and HJ in the given frame as shown in figure 1 using Method of Sections. [10] 1 kN 2 kN 2 kN 2 kN J 1 kN Н 0.46 m F D 2.62 m I A C| E G| 2.4 m 2.4 m 2.4 m 2.4 m Figure 1The bolt AB in (Figure 1) has a diameter of 18 mm and passes through a sleeve that has an inner diameter of 42 mm and an outer diameter of 52 mm. The bolt and sleeve are made of A-36 steel and are secured to the rigid brackets as shown. If the bolt length is 220 mm and the sleeve length is 200 mm, determine the tension in the bolt when a force of 50 kN is applied to the brackets.