Procedure:1. Measure the dimensions of the specimen and the record them in the date sheet2. Turn the specimen on its side with respect to its position as molded and center in on life supportblocks.3. Center the loading system in relation to the applied force.4. Bring the load applying-block in contact with the surface of the specimen at the center andapply a load between 3 and 6% of the estimated load.5. Grind cap, or use leather shims on the specimen contact surface to eliminate any gap in excessof 0.004 inch (0.10 mm). Gaps in excess of 0.15 inch (0.38 mm) shall be eliminated by cappingor grinding.6. Apply the load on the specimen continuously and without shock. The load shall be applied atthe constant rate to the breaking. Apply the load at such a rate that constantly increases theextreme fiber stresses between 125 and 175 psi/min. (0.86 and 1.21 MPa/min) when calculateduntil rupture occurs.7. Take three measurements across each dimension (one at each edge and at the center) to thenearest0.05 in. (1 mm) to determine the average width and depth of the specimen at the pointof fracture. If the fracture occurs at a capped section, include the cap thickness in measurement. Calculation:For center Loading:MR = 3PL / 2bd?Where:MR- modulus of rupture, psi (MPa)P= maximum load applied as indicated by testing machine, in Ib (N)L= span length, in inches (mm)b= average width of specimen in inches (mm)d= average depth of specimen, at the fracture, in inches (mm)Note: The weight of the beam is not included in the above calculation.Tabulated Data and ResultSample No.L (mm)B (mm)D (mm)P (N)MR (MPa)145015015023000245015015021000345015015019000

Answered: Image /qna-images/answer/65d5585a-ed1c-4334-9e9c-11b0f52b2c9b.jpg

Calculation: For center Loading: MR = 3PL / 2bd? Where: MR- modulus of rupture, psi (MPa) P= maximum load applied as indicated by testing machine, in Ib (N) L= span length, in inches (mm) b= average width of specimen in inches (mm) d= average depth of specimen, at the fracture, in inches (mm) Note: The weight of the beam is not included in the above calculation. Tabulated Data and Result Sample No. L (mm) В (mm) D (mm) P (N) MR (MPa) 1 450 150 150 23000 450 150 150 21000 3 450 150 150 19000

Calculation: For center Loading: MR = 3PL / 2bd? Where: MR- modulus of rupture, psi (MPa) P= maximum load applied as indicated by testing machine, in Ib (N) L= span length, in inches (mm) b= average width of specimen in inches (mm) d= average depth of specimen, at the fracture, in inches (mm) Note: The weight of the beam is not included in the above calculation. Tabulated Data and Result Sample No. L (mm) В (mm) D (mm) P (N) MR (MPa) 1 450 150 150 23000 450 150 150 21000 3 450 150 150 19000

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: A uniaxial test was performed on a bone specimen having a central (gauge) region initially 6 mm long…

A: a uniaxial test results were given as followslength=6mmdia=2mm

Q: *3-20. The stress-strain diagram for a bone is shown and can be described by the equation € = 0.45…

Q: Can you please answer the following question in the picture and show all of your work in, please.

A: Detailed step by step solution is given below, still if you have any query please feel free to…

Q: Problem NO. 2. A student performed a tensile test in one of his samples, an aluminum specimen which…

Q: amplitude was 210 MPa. Compute the maximum and minimum stress. (Enter your answers to three…

Q: The following observations were made during a tensile test on a mild steel specimen 40 mm in…

A: Given data, Yield load =161 KN Maximum load = 242 KN Length of specimen at fracture = 249 mm…

Q: Problem 1.4. A tensile test was conducted on a mild steel bar. The following data was obtained from…

Q: A tensile test was carried out on a metal specimen. The gauge length of the specimen was 96 mm and…

A: Given data : gauge length=96 mmfinal length=107.4 mminital area =17.35 mm2final area=16.40 mm2

Q: Problem 1.4. A tensile test was conducted on a mild steel bar. The following data was tained from…

Q: 2. Given a strain gage with these properties: Rg=200 , Sg 2.5, r-2, Vs = 5 V, and Pg=0.015 W.…

A: To determine the sensitivity (Ss) of the strain gauge Wheatstone bridge, we can use the following…

Q: 9. (a) Sketch the Mode I, Mode II, and Mode III crack opening geometries, and clearly state which is…

A: Note : As per guidelines, we will only discuss the first one as it is not particularly mentioned…

Q: A concrete cylinder sample has a diameter of 6 inches and a height of 12 inches. The maximum load…

Q: The block is analysed in Cartesian system. The value of P is such that normal strain &=0. Modulus of…

Q: QUESTION 2 Test Engineers found that 22% of steel frames fracture within 15 years. The average…

A: Given: 30 Psi is the average stress Required: What refers 30 Psi Note: No shear and bending…

Q: Problem 1.4. A tensile test was conducted on a mild steel bar. The following data was Stained from…

Question

Procedure:
1. Measure the dimensions of the specimen and the record them in the date sheet
2. Turn the specimen on its side with respect to its position as molded and center in on life support
blocks.
3. Center the loading system in relation to the applied force.
4. Bring the load applying-block in contact with the surface of the specimen at the center and
apply a load between 3 and 6% of the estimated load.
5. Grind cap, or use leather shims on the specimen contact surface to eliminate any gap in excess
of 0.004 inch (0.10 mm). Gaps in excess of 0.15 inch (0.38 mm) shall be eliminated by capping
or grinding.
6. Apply the load on the specimen continuously and without shock. The load shall be applied at
the constant rate to the breaking. Apply the load at such a rate that constantly increases the
extreme fiber stresses between 125 and 175 psi/min. (0.86 and 1.21 MPa/min) when calculated
until rupture occurs.
7. Take three measurements across each dimension (one at each edge and at the center) to the
nearest0.05 in. (1 mm) to determine the average width and depth of the specimen at the point
of fracture. If the fracture occurs at a capped section, include the cap thickness in measurement.

Calculation:
For center Loading:
MR = 3PL / 2bd?
Where:
MR- modulus of rupture, psi (MPa)
P= maximum load applied as indicated by testing machine, in Ib (N)
L= span length, in inches (mm)
b= average width of specimen in inches (mm)
d= average depth of specimen, at the fracture, in inches (mm)
Note: The weight of the beam is not included in the above calculation.
Tabulated Data and Result
Sample No.
L (mm)
B (mm)
D (mm)
P (N)
MR (MPa)
1
450
150
150
23000
2
450
150
150
21000
3
450
150
150
19000

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

Problem 1.4. A tensile test was conducted on a mild steel bar. The following data was obtained from the test : (i) Diameter of the steel bar (ii) Gauge length of the bar (iii) Load at elastic limit (iv) Extension at a load of 150 kN. (v) Maximum load (vi) Total extension (vii) Diameter of the rod at the failure Determine : (a) the Young's modulus, = 3 cm = 20 cm = 250 kN = 0.21 mm = 380 kN = 60 mm - 2.25 cm. (c) the percentage elongation, and (b) the stress at lastic limit, (d) the percentage decrease in area.
A 42 meter solid shaft is under torsional loading 53 Nm. If in this application, the allowable shear stress for the material is 369 MPa. Calculate the required minimum diameter of the shaft: _____ mm ( input the value directly from your calculation before rounding it up to an engineering accepted diameter) Pay attention to units, and calculate your answer to one decimal place.
2 please include scketch
Solve it civil problem
A hot rolled steel tension test sample has a diameter of 6 mm and the original gage length of 30 mm. In a test to fracture, the stress and strain data below were obtained. Determine Generic Strain Value, %0.2 offset strain, and Calculated stress at %0.2 offset. Original diameter - 6mm Original Length - 30 mm Fracture diameter - 4.54 mm
You must have to draw FBD