Chapter 6 hw edit

Compound axial member ABC has a uniform diameter of d = 1.45 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy rod with a length of L₁ = 85 in. Segment (2) is a copper [E₂ = 17,000 ksi] alloy rod with a length of L₂ = 121 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 1800 μin./in. in the longitudinal direction. What is the total elongation of member ABC? SABC = i eTextbook and Media GO Tutorial Save for Later L Aluminum in. B L₂2 Copper CP Attempts: 0 of 5 used Submit Answer

● Example 3.2 A 60mm diameter solid shaft has a strain gauge mounted at 65° to the axis of the shaft. In service a torque is applied to the shaft and the strain gauge reads 200 x 10-6. Calculate the value of the torque if the shaft is made from steel with E = 207 GPa and v = 0.3. (BC&A question 11.18) Strain Transformations Mechanical Engineering Mechanics of Materials 2 Solution 3.2 Applied torque alone will produce only shear stress/strain in the directions parallel and perpendicular to the torque. The strain gauge is mounted at 65° to the axis of the shaft. Strain Transformations Mechanical Engineering Mechanics of Materials 2 Now Strain gauge From the strain circle 65⁰ Strain Transformations Note that for circular shaft 4 D J = ² (2) ²* 33 Dr PJ Martin Lecture Notes 34 Dr PJ Martin Lecture Notes 35

Problem F2 stress (ksi) 130 120 110 100 90 80 70 60 50 40 30 20 10 0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 strain (in/in) Above you will find the experimental stress-strain diagram of 1045 steel. Calculate the permanent set if a cylindrical specimen with a diameter of 2 in is loaded to the ultimate stress and then unloaded. Provide your answer in units of in/in with 3 significant figures after the decimal. The elastic modulus of 1045 steel is 29,000 ksi.

A rectangular steel block is 300mm in the x direction, 200mm in the y direction and 150mm in the z direction. The block is subjected to a triaxial loading consisting of three uniformly distributed forces as follows: 250kN tension in the x direction, 320kN compression in the y direction and 180kN tension in the z direction. For steel, v=0.30 and E=200,000MPa. Determine the total strain in the x direction due to the forces applied in the three directions.

Example 1: (i) Find the compatibility condition for the strain tensor e; if e1 , e22 , e12 are independent of x3 and e31 = e32 = e33 = 0. (ii) Find the condition under which the following are possible strain components. eii = k(x,? – x2), e12 = k' x1 x2 , e22 = k x1 X2 , %3D e31 = e32 = e33 = 0, k & k' are constants (iii) when eij given above are possible strain components corresponding displacements , given that u; = 0. find the

uniform diameter d= 1.75 in. Segment (1) is an aluminum Ę₁ = 10000 ksi L₁ = 90 in. Segment (2) is a copper Ę₂ = 17000 ksi L₂ = 126 in. When Pis applied, a strain gage attached to copper segment (2) measures a normal strain of &2 = 1700 uin./in. in the longitudinal direction. What is the elongation of segment (1)? Answer in inch rounded-off to 3 decimal places L₁ A (1) Aluminum B L2 (2) Copper

On the free surface of a component, a strain rosette was used to obtain the following normal strain data: €a = 300µɛ, ɛp = 400µs, and Ec = 200µe. Calculate the normal and shear strains in the x-y plane.

Tensile test specimens are extracted from the "X" and "y" directions of a rolled sheet of metal. "x" is the rolling direction, "y" is transverse to the rolling direction, and "z" is in the thickness direction. Both specimens were pulled to a longitudinal strain = 0.15 strain. For the sample in the x-direction, the width strain was measured to be ew= -0.0923 at that instant. For the sample in the y-direction, the width strain was measured to be gw=-0.1000 at that instant. The yield strength of the x-direction specimen was 50 kpsi and the yield strength of the y-direction specimen was 52 kpsi. Determine the strain ratio for the x direction tensile test specimen. Determine the strain ratio for the y-direction tensile test specimen. Determine the expected yield strength in the z-direction. Give your answer in units of kpsi (just the number). If the sheet is plastically deformed in equal biaxial tension (a, = 0, to the point where & = 0.15, calculate the strain, 6, that would be expected.

Problem 8. Re-write the strain compatibility condition €22,33 + €33,22 in the engineering notation. = 2 e23,23

Compound axial member ABC has a uniform diameter of d=1.45 in.  Segment (1) is an aluminum [E1=10,000 ksi] alloy rod with a length of L1=88 in.  Segment (2) is a copper [E2=17,000 ksi] alloy rod with a length of L2=128 in.  When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of ε2=2200 μin./in. in the longitudinal direction.  What is the total elongation of member ABC? answer is in inches

400 350 300 250 200 150 100 50 0.002 0.004 0.006 0.008 0.01 Strain Stress (MPa)

124,000s o(2) 0 s85 0.03 (o in MPa) 1+ 300ɛ (a) Find the axial normal strain in the cable and its elongation due to the load W = 6.8 kN. (b) If the forces are removed, what is the permanent set of the cable? Hint: Start with constructing the stress-strain dia- gram and determine the modulus of elasticity, E, and the 0.2% offset yield stress. D 2 m A 1.5 m B 1.5 m E m--1 m- 0.75 m 0.25 m W = 6.8 kN

A bar BD is held in a horizontal position when there is no load by the steel wire AB. When a weight Wis added at C, also hanging from a steel wire CE, then point C is displace down by 0.025in. The cross-sectional area of the wires is 0.002in². The wires are made of a steel alloy with E = 29000ksi and σy = 70ksi. Calculate the strain in both wires, and the weight W. B A N E Parameter ft L_1 4 L_2 1 L_3 2 L_4 4 The strain in AB is € AB = The strain in CE is ECE = The weight W = 30 W cc 080 BY NO SA 2021 Cathy Zupke Xlb D in./in. in./in.

An initially rectangular element of a material is deformed into the shape shown in the figure. Deformed Undeformed 16.6 0.1992 mm 15.7 0.2 mm Normal strain in the x- direction * -4.0 x 10^-3 4.0 x 10^-3 10 x 10^-3 -10 x 10^-3 Normal strain in the y- direction O -4.0 x 10^-3 4.0 x 10^-3 O 10 x 10^-3 O -10 x 10^-3 0.1515 mm 0.15 mm

In uniaxial loading the poisson's ratio is 0.30 and the strain in the x- direction is 250 x 10^-6. What is the strain in the z- direction? * A -8.5 x 10^-5 B -7.5 x 10^-5 C 7.5 x 10^-5 9.5 x 10^-5 D

Brinell hardness number = 0 BHN Rockwell hardness = . (a) The naval brass for which the stress-strain behavior is shown in Animated Figure 6.12. You might need to use Animated figure 6.19 Rockwell hardness Tensile strength = 0 MPA Tensile strength = 0 ksi 60 70 80 90 100 HRB Brinell hardness i HB 20 30 40 50 HRC Rockwell hardness i HRB 250 1500 200 Steels 150 1000 (b) The steel alloy for which the stress-strain behavior is shown in Animated Figure 6.22. You might need to use Animated 100 Figure 6.19. 500 Brass Cast iron (nodular) 50 Brinell hardness i HB Rockwell hardness i HRB 100 200 300 400 500 Brinell hardness number Tensile strength (MPa) Tensile strength (ksi)

Q. // (A) - An element of bone in plane strain where the state of strain are (Ex = 8 × 10¬4, €y = 5 × 10¬4, Yxy = 12 × 10-4). Find (a) the stresses at orientation a counterclockwise angle (40 deg.) from the x-axis, b) the strain energy and (c) the failure stress by Von Misses criteria. Known Modulus of elasticity = 2 × 10° N /m² and Poisson's ratio = 0.3

If an isotropic material has a shear modulus of 125 Gpa and a Poisson's ratio of 0.2, calculate its Young's modulus. Select one: O E= 80 Gpa O E= 450 Gpa O E= 60 Gpa %3D O E= 30 Gpa %3D O E= 300 Gpa %3D If a rubber material is deformed as shown in the following figure, determine the normal strain along diagonal BD.

Stress (MPa) 600 500- 400 300 200 100- 0 0.00 Stress (MPa) 0.04 500 400 300 200 100 0.000 1 0.08 0.002 0.004 Strain Strain 0.12 0.006 1 0.16 0.20 Figure 6.22 Tensile stress-strain behavior for a steel alloy. 6. A steel alloy with a rectangular cross section 13.8 mm x 6.4 mm and length 560 mm has the stress-strain behavior shown in Figure 6.22 above. This specimen is subjected to a tensile force of 42,000 N. (a) Determine the elastic and plastic strain in the specimen in the loaded condition. (b) Determine the final length of the specimen after the load is released.

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A bar of a uniform cross section is subjected to uniaxial tension and develops a strain in the direction of the force of 1/800. Calculate the change of volume per unit volume. assume v= 1/3

Calculate the strain (e) for a change in length (ō) and a length (L), where 8 = 0.015 in and L = 5 ft. 0.0025 0.00025 0.00215 0.00125

h.w: A compressive force was applied in the longitudinal direction of a cross-sectional bar ( A= 10 cm) and this resulted in a longitudinal strain of (0.0007) Calculate the amount of force? If you know that the elastic modulus (E= 206*10° m2

PROPAGATION OF ERRORS Determine the range of values to which the strain will be calculated.F = 150000 +/- 150 lbfSide length = 1.2 +/- 0.025 inModulus of Elasticity = 25000 +/- 11.5 ksiDetermine the relative error. Is it above or below the accepted 5% relative error?

Question 2 A torsion circular bar made of aluminum in Figure 1 is loaded by its weight and P kNm torsion (P = 10). Determine the strain energy stored in the bar. Given that the diameter of bar AB and BC is d₁= 0.5 m and d₂=0.3 m respectively. Take the specific weight of the aluminum, y as 27 kN/m³, Poisson's ratio, v= 0.3 and shear modulus, G as 25 GPa. 2.5 m 1m B C A Figure 1 d₁ d2 T=P kNm

A solid bar of length L = 4 m and diameter 200 mm is heated from 20 to 320 degrees celsius and restrained between two solid immovable walls. Young's modulus of the material is 90 GPa and the coefficient of thermal expansion is 22 x 10-6/°C. →Calculate the thermal strain, e, in micro-strain correct to two decimal places. micro-strain. Calculate the thermal stress, o, in megapascals (MPa) correct to two decimal places. E: σ: MPa. Hence calculate the force exerted by the bar in meganewtons (MN) correct to two decimal places: F MN. L