Scenario. Three specimen of D12 rebar was subjected to a Tensile Test. Using a steel density of8000Kg/cu.m. With the following resultspecimen 1specimen 2M=0.895 kgM=0.89 kgL=98 cmL=99 cmLoad at yield strength= 5000 kgLoad at ultimate =8600 kgGauge length after test=21.5 cmLoad at yield strength= 5000 kgLoad at ultimate =8600 kgGauge length after test=21.6 cmspecimen 3M=0.897 kgL=100 cmLoad at yield strength= 5010 kgLoad at ultimate strength=8590 kgGauge length after test=21.6 cmspecimen Yield strengthYield strengthUltimate strength Ultimate strength STRAINbased on actual based on 12mmbased on actualbased on 12mmdiameter (MPa)diameter (MPa)diameter (MPa)diameter (MPa)1.2.average

Note: Dear Student! As per our guidelines, we are only allowed to answer a single question. Kindly…

specimen Yield strength based on actual Yield strength based on 12mm based on actual diameter (MPa) diameter (MPa) Ultimate strength Ultimate strength STRAIN based on 12mm diameter (MPa) diameter (MPa)

specimen Yield strength based on actual Yield strength based on 12mm based on actual diameter (MPa) diameter (MPa) Ultimate strength Ultimate strength STRAIN based on 12mm diameter (MPa) diameter (MPa)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

yields under a load of 8KN; it attains a max load of 15KN and breaks finally A tensile test is carried out on a bar of mild steel of diameter 20mm. The bar Еxample at a load of 7KN. Estimate yP (i) The tensile stress at the yield point (ii) The ultimate tensile test ii. Max loa (iii) The average stress at the breaking point if the diameter of neck is Omm.
A tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.
A sample test rod made of a ferrous material of 0.5 in. diameter was tested for tensile strength with the gauge length of 2.5 in. The following observations were recorded: Final length = 3.15 in; Final diameter = 0.275 in; Yield load = 7650 lb. and Ultimate load = 10375 lb. (a) Calculate: 1. yield stress, 2. ultimate tensile stress, 3. percentage reduction in area, and 4. percentage elongation. (b) What specific material, if any, is similar to this tested rod? Refer to AT7 and AT8. (c) This same material is then used as a tension member 55 inches long and is subjected to a maximum load of 6000 lb, repeated but not reversed. (d) Solve part (c) if the total elongation is not to exceed 0.025 in.? 1. %3D
The proof stress of a material under tensile loading can be considered as can be considered as equivalent to the ultimate tensile strength for aluminium alloys being only applied to engineering steels during a tensile test the intercept on the stress stain curve for a given strain where a line draw is parallel to the elastic response line. the stress at yield for materials have no defined yield stress
Force P and length change AL data are given in table below for the initial portion of a tension test on 7075-T651 Al alloy. The diameter before testing was 9.07 mm., and the gage length Linitial for t length change measurement was 50.8 mm. Determine the yield strength with offset limit, and %total elongation. P, kN AL mm 7.22 14.34 0.0839 0.1636 21.06 0.241 26.8 0.308 31.7 0.380 34.1 0.484 35.0 0.614 36.0 0.924 36.5 1.279 36.9 1.622 37.2 1.994
1. A steel of rotating-beam test specimen has an ultimate strength of 1600Mpa. Estimate the life of the specimen if it is tested at a completely reversed stress amplitude of 900MPA. Hint: You can either extrapolate the curve of f vs. Sut, or use the equation derived on P10 to estimate the value of f.
*O llAsiacell Strength of m. -> 02: A steel specimen 12mm diameter has gauge length 50mm. the steel specimen had tested via tensile test under maximum load 66KN with elongation 7.5mm, and the yield load of this specimen is 15KN with elongation 2.4mm. Calculate: 1- The engineering ultimate stress (ultimate tensile strength), and engineering strain at this point. 2- The engineering stress and strain at yield point. 3- The modulus of elasticity, and the modulus of resilience. 4- The final or fracture strain of a steel specimen, if you know that the final length of specimen after testing is 58.5mm. 5- The true stress and strain for ultimate point. Q3: The yoke-and-rod connection is subjected to a tensile force of 15 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members. Finally, find the shear strain in pin A. Take G teef 75GPA 40 mm 15 kN, 30 mm 30 mn 15 kN 04: In figure as shown below, assume that a 35mm diameter rivet joint the…
Q3: A tensile test on a cold worked brass gave the following results. The diameter of the test specimen d= 16 mm and the gauge lenth was 80 mm. After fracture the gauge length was 85 mm and the fracture point was was 15 mm diameter. The maxiumu load was 150 KN. The load and extension at the elastic limit were 70 KN and 0.5 mm, respectively. Calculate the modulus of elasticity. Calculate the % of elongation and % of reduced cross sectional area.
The brass has a shear modulus (G) of 40 GPa. In the torsion test calculated results. (G) was found to be 37 830 MPa. What is the percentage of errors in the experiment? Select one: O a. 0.05425% O b. 5.425% O c. 3.779% Od. 0.54%
Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…
A steel bar of 8 mm is heated from 10° C to 25° Cand the bar is free to expand. The bar will induce Tensile stress No stress Compressive stress Shear stress
TelkomSA E estion CALCULATIONS 4.1 Using the given pulley radius and the weight values, calculate the torque applied to the specimens. ← 4.2 Calculate the theoretical angle of twist. 4.3 Calculate the experimental angle of twist for the specimen by using the dial gauge readings and the lever distance to where the dial gauge makes contact. 4.4 Calculate the maximum torsional shear stress and compare the values to the shear stress limits of the material and discuss. 4.76% Rod - Experiment T mass. displacement cog 2009 300g 400g soog Torsion of a Diameter of pulley - 180, 4 mm Length of rad 304mm Diameter of rod at Centre Point - [6,7mm). Measure distance of lever - 19,52 mm Aluminum Solid rod. displacement mass 9,11 mm 9,23 mm 2,34 mm Q. 47 mm g s S8 mm 11:57 3 O