Week 4 Pre-Class

Can someone please help me to correctly solve all the following parts of this question. Thank you!

Can someone please help me to solve the following question using the FACTOR LABEL METHOD OF UNIT CONVERSION. Showing all work and formulas please and thank you!

Please solve last 2 subparts

Find the original cross sectional area and original length from the given information about a tensile test done on a specimen

The stainless steel specimen is shown in Figure 3 with tensile engineering stress-strain behaviour.   a)Calculate true stress at an engineering strain of 0.04.

We're measuring axial load in a rectangular bar with two strain gauges: one on the top in the axial direction and the other on the bottom in the transverse direction). An axial load, P, is applied and the ambient temperature is varying with time. P # Strain gages > Р (a) If one strain gauge is connected to leg 1 of the Wheatstone bridge, which leg (2, 3, or 4) would you connect the other strain gauge in order to achieve temperature compensation? Please explain your choice by writing out the equation. (b) Calculate the bridge constant of this arrangement, your (c) If some bending moment were inadvertently applied (in addition to the axial load), would choice in part (a) provide compensation for bending loading? Justify your answer by writing out the equation.

From the shear stress - shear strain diagram shown below, all the following are true except: Shear Stress (MPa) 500 450 400 350 300 250 200 150 100 50 0. 0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8 Shear Strain (rad) Shear Stress-Strain Curve for Brass Select one: O a. Shear stress and Shear strain are 430 MPa and 1.72 rad at fracture. O b. Shear stress and Shear strain are 250 MPa and 0.05 rad at Yielding. Oc. The elastic zone ends at a shear stress of 400 MPa and shear strain of 1 rad Od. The ultimate Shear stress and Shear strain are 430 MPa and 1.72 rad.

You need the shear modulus for a particular material in order to calculate shear strain for a known applied shear shear stress. You can't find shear modulus for the material, but you can find elastic modulus (200 GPa) and Poisson's ratio (0.3). Which of the following could you use as the shear modulus for your calculations? 200 GPa O 60 GPa O 77 GPa O 154 GPa

For the stress-strain curves below, which statements are correct? Stress (MPa) 60 50 40 30 20 10 0 0 A B C. 1 2 3 4 Strain 5 6 7 8 a. The mechanical behavior shown in C is that of an elastomer with low elastic modulus b. While the deformation in polymer A is fully reversible until breakage, polymer B and C only display reversible deformation until 0.1X and 5x elongations respectively The mechanical behavior shown in B is that of an elastomer with low elasticity d. Polymer B and C have an elastic behavior whereas A has a plastic behavior e. The polymer with the necking behavior has the highest strength Of. The polymer with brittle nature is the one that possesses the highest elastic modulus

The stainless steel specimen is shown in Figure 3 with tensile engineering stress-strain behaviour.   a)Calculate modulus of resilience.

we are to find the sheer strain, sheer stress and modulus of rigidity. table,specimen length,gauge length and diameter are given. specimen length= 140mm gauge length= 77.09mm diameter= 5mm

I need the solution please

The First half of this question has been answered. (First 3) Needing help with the last part. Answered: 1) Solid / Ductile Material 2) Spring Constant: K = 05 = 250 N/m 3) Time Threshold: 5 seconds Unanswered / Need: Viscoscity, Charasteristic Frequency & What real world model charasteristic does this describe?

The following data was obtained as a result of tensile testing of a standard 0.505 inch diameter test specimen of magnesium.  After fracture, the gage length is 2.245 inch and the diameter is 0.466 inch. a). Calculate the engineering stress and strain values to fill in the blank boxes and plot the data. Load(lb) Gage Length (in) Stress (kpsi) Strain 0 2     1000 2.00154     2000 2.00308     3000 2.00462     4000 2.00615     5000 2.00769     5500 2.014     6000 2.05     6200 (max) 2.13     6000 (fracture) 2.255     b). Calculate the modulus of elasticity c).  If another identical sample of the same material is pulled only to 6000 pounds and is unloaded from there, determine the gage length of the sample after unloading.

I need handwritten solution if u skip any part then i swear, it's simple do all otherwise skip

Please help. Do not just guess. Prove your answer. I will give a  good feedback. thank you :)

His can you find answers for these please and explain how ? Thank you

Given the following data for five different materials answer parts A B C D E & F with an explanation

Engineerin Engineerin Engineerin Engineering g g g strain Stress stress strain (in./in.) (in./in.) (ksi) (ksi) 0 0 76 0.08 30 0.001 75 0.10 55 0.002 73 0.12 60 0.005 69 0.14 68 0.01 65 0.16 72 0.02 56 0.18 74 0.04 51 (Fracture) 0.19 75 0.06 Plot the data as engineering stress (MPa) versus engineering strain (mm/mm) and determine the ultimate strength of the steel

Poisson's ratio is defined as (a) longitudinal strain/lateral strain (b) lateral strain/longitudinal strain (c) lateral strain x longitudinal strain 1 (d) (lateral strain) x (longitudinal strain) -

What do I do with the given Deformation (0.54 and 0.95) Even though I already have the Original Length (55mm) and Final Length (68.12mm) in finding the Strain(%)

Detailed steps

Calculate the young’s modulus from one of these two graphs?

Mechanics of materials I

Young Modulus of a material has characteristics of : (Choose TRUE or FALSE for each statement) Can be obtained by the gradient of stress versus strain graph. OTRUE OFALSE As a measurement for a material due to its resistance, subjected to the exerted forces. O TRUE O FALSE The two-dimensional plane of the material depends on Young Modulus. O TRUE OFALSE

4. We want to draw a 7.5 mm diameter copper wire having a yield strength of 140 MN.m²² into a 6 mm diameter wire. Will the drawn wire break during the drawing process? Strength (MN.m-2) 600 500 400 300 200 100 0 0 Tensile strength 20 Yield strength % Elongation 60 40 Percent cold work 80 100 90 80 70 60 50 40 30 20 10 100 % Elongation

. Data was collected from the tensile test demonstration for Aluminum 6061-T6. Below are the resulting stress-strain diagrams based on the tensile test we performed. o (MPa) Al 6061-T6 Stress-Strain Curve o (MPa) 350 320 300 250 200 150 100 50 0 Estimate 0 0.05 0.1 0.15 ε (mm/mm) 300 270 250 220 200 150 140 100 a) The modulus of Elasticity, E b) The yield stress, d, using the 0.2% offset method c) The strain at failure, & failure d) The modulus of resilience (approximately!) 50 0 0 Al6061-T6 Stress-Strain Curve, Magnified N 0.0032 0.002 0.004 0.006 0.008 € (mm/mm) The right diagram shows the magnified elastic region of stress-strain diagram to enhance the details.

Q3 : Rank the magnitudes of the moduli of elasticity for the following hypothetical X, Y, and Z materials from the greatest to the least.The appropriate A, B, and n parameters for these three materials are tabulated below; they yield EN in units of electron volts and r in nanometers: Material A В n 2.5 2.0 x 10-5 8. Y 2.3 8.0 x 10-6 10.5 Z 3.0 1.5 x 10-5 9.