Please correct any mistakes that I made. Thanks . The graph below represents a t

Question

Please correct any mistakes that I made. Thanks

. The graph below represents a typical stress-strain relaticashap for compressave test on O.15 X 0.30 mter coneree cylinder (w-0 67) at anage of 28 days Please include the units, and show all your work whenever neccssary for full credit. Using the Figure, a. Determine the ultimate stress in Mpa? (2 points 6b. Determine the 40% Ultimate stress in Mia? (2 peints) c. Determine the Secant Modulus in Gipa? (2 points) 9.5 locoto d. Compressive Strength (fc in Mpa? 4 points) 5e. Medulus of lasticity in Gpa? (4 points) Sf Modulus of from ACI Equation in Gpo (3 peints) Water/ Cement = 0.67 'er-17mea 25.0 20.0 15.0 10.0 5.0 21.0 0.0 13.5 750 15.5 900 17 100 19.5 1500 20001 200 3000 | 17.1) 4000 16.0 1000 2000 Strain, 10-6 3000 4000 Activate Windows

Explanation / Answer

5 a - Okay

5 b - Okay

5 c - Secant Modulus is generally determined by drawing straight line from origin to 85% of compressive strength,

that is 85% of 21 = 17.85 MPa. We may take closer value (17.7). Hence, =17.7/900 x10-6= 19.67 GPa.

(In your calculation, strain has to be 1100 x10-6 as per graph and not 1000x10-6).

5 d - generally, ultimate value is chosen as compressive strength. Hence, fc' = 21 MPa

5 e - SInce this is the slope of straight line portion. You can as well choose the data given Table.

15.5/750x10-6 = 20.67 GPa

5 f - Modulus (ACI) = 4700 SQRT(fc') = 21.54 GPa

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.