Element 1Ук**Answer in MPa.2 cmElement 2Assume the reaction forces at the origin weresolved to be Rf = [325, 100, 0] N and=Rm [10, -2, 5] Nm. We will consider ahypothetical cut at the connection of the dolly tothe axle. Z is positive out of the page.What is the value of normal stress in thex-direction Ox for element 2?Answer options:a) Ox < -300b) -300 ≤ Ox < -100c) -100 ≤ Ox < -200d) 0 ≤ Ox < 100e) 100 ≤ Ox < 300f) 300 ≤ Ox

To determine the normal stress in the x-direction Ox for element 2, we need to calculate the force…

Answered: Assume the reaction forces at the…

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.6.7P: A wine of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The...

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A wine of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by =18,0001+30000.03(=ksi) in which is nondimensional and has units of kips per square inch (ksi). (a) Construct a stress-strain diagram for the material. (bj Determine the elongation, of the wire due to the Forces P. (c) IF the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit?
Two steel wines support a moveable overhead camera weighing W = 28 lb (see figure part a) used For close-up to viewing of field action at sporting, events. At some instant, wire I is at an angle a = 22° to the horizontal and wire 2 is at angle fi = 40°. Wires I and 2 have diameters of 30and 35 mils, respectively. (Wire diameters are often expressed in mils; one mil equals 0.001 in.) (a) Determine the tensile stresses s and s2 in the two wires. (b) If the stresses in wires 1 and 2 must be the same, what is the required diameter of wire 1 ? (c) To stabilize the camera for windy outdoor conditions, a third wire is added (see figure part b). Assume the three wires meet at a common point coordinates (0, 0. 0) above the camera at the instant shown in figure part b. Wire I is attached to a support at coordinates (75 ft, 48 ft, 70 Ft). Wire 2 is supported at (-70 ft. 55 ft, 80 Ft). Wire 3 is supported at (-10 ft. -85 Ft, 75 ft). Assume that all three wires have a diameter of 30 mils. Find the tensile stresses in all three wires
The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension, use a characteristic tensile stress obtained from a stress-strain curve as a measure of strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus, the strength-to-weight ratio RS/Wfor a material in tension is defined as Rs/w= in which a is the characteristic stress and 7 is the weight density. Note that the ratio has units of length. Using the ultimate stress Uas the strength parameter, calculate the strength-to-weight ratio (in units of meters) for each of the following materials: aluminum alloy 606I-T6, Douglas fir (in bending}, nylon. structural steel ASTM-A57.2, and a titanium alloy. Obtain the material properties from Tables [-1 and 1-3 of Appendix I. When a range of values is given in a table, use the average value.
Three round, copper alloy bars having the same length L but different shapes are shown, in the figure. The first bar has a diameter d over its entire length, the second has a diameter d over one-fifth of its length, and the third has a diameter d over one-fifteenth of its length. Elsewhere, the second and third bars have a diameter Id. All three bars are subjected to the same axial load P. Use the following numerical data: P = 1400 kN, L = 5m,d= 80 mm, E= 110 GPa. and v = 0.33. (a) Find the change in length of each bar. (b) Find the change in volume of each bar.
A high-strength steel bar used in a large crane has a diameter d = 2.00 in. (sec figure). The steel has a modulus of elasticity E = 29 × 10 psi and Poisson’s ratio is v = 0.29. Because of clearance requirements, the diameter of the bar is limited to 2.001 in. when it is compressed by axial forces. What is the largest compressive load Pmaxthat is permitted?
The 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 1
Find the stress in members DF, FI, and HI if the cross- sectional area of each member is 800 mm?. Indicate Tension or Compression. E 3m -2m- 2m- 3m 30KN 10KN 3m K Find the following DF: Force P Stress FI: Force P Stress HI: Force P Stress C.
A simple rod acted by fluctuating axial force produced a fluctuating stresses as „Sigma_min= - 500MPA, Sigma_max using mod-Goodman were applicable, answer Q6- Q7 100MPA (Sy = 400MPA, Sut = 600MPA) The equivalent fully reversed stress is
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 0.1025
Figure 1 shows a composite shaft ABCD acted upon horizontal forces. The moduli of elasticity and the cross-section areas of the segments of the shaft are also provided. Eal = 70 GPa AAB = 58 mm? Ecu = 126 GPa Est 200 GPa %3D %3D Авс 3D 77 mm? AcD = 39 mm2 АВ Сopper 20kN 8 kN Aluminum Steel 7 kN A B 20kN 8 kN 450 mm -300 mm- 400 mm Figure 1: Axially loaded composite shaft 2.1. Construct the diagram of internal normal forces for each cross-section of the shaft 2.2. Calculate the displacement of the point C with respect to the fixed-point A. 2.3. If the shaft must have one uniform diameter, determine the required minimum diameter of the shaft if the normal yield stress in the composite shaft must not exceed 450 MPa with a factor of safety equal to 1.8.
Figure 1 shows a system of three bars each made of different materials, and connected together and the temperature is initially kept at 12°C. Thereafter, the temperature of the system was raised to 50°C. Steel Brass Сopper Es = 200 GPa as = 12(10-0)/°C apr = = Epr = 100 GPa 21(10-6)/°C acu = 120 GPa Ecu Ebr %3D %3D 17(10-6)/°C %3D Ast abr Acu = 515 mm² = 450 mm² %3D Ag 200 mm² Apr st 300mm 200 mm 100mm Figure 1: An assembly of three bars 1.1. Determine the thermal stress raising in each bar if the assembly is placed between two rigid walls. 1.2. Determine the force exerted on the right rigid support if an allowance of 0.05 mm is provided between the left end of the steel bar and the left wall.
Figure 1 shows a system of three bars each made of different materials, and connected together and the temperature is initially kept at 12°C. Thereafter, the temperature of the system was raised to 50°C. Steelbno abm Est = 200 GPa Epr = 100 GPa ag = 12(10-)/°C abr = Brass Сopper Ecu 120 GPa %3D 21(10-6)/°C acu = 17(10-6)/°C %3D %3D Acu = 515 mm? Ast = 200 mm? Abr = 450 mm² - 300 mm- -200 mm 100 mm Figure 1: An assembly of three bars 1.1. Determine the thermal stress raising in each bar if the assembly is placed between two rigid walls. 1.2. Determine the force exerted on the right rigid support if an allowance of 0.05 mm is provided between the left end of the steel bar and the left wall.

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Assume the reaction forces at the origin were solved to be Rf = [325, 100, 0] N and Rm [10, -2, 5] Nm. We will consider a hypothetical cut at the connection of the dolly to the axle. Z is positive out of the page. What is the value of normal stress in the x-direction Ox for element 2?