A marble column with a cross-sectional area of 45 cm^2 supports a load of 8.0 ti

Question

A marble column with a cross-sectional area of 45 cm^2 supports a load of 8.0 times 10^4 N. The marble has a Young's modulus of 6.0 times 10^10 Pa and a compressive strength of 2.0 times 10^8 pa. a. What is the stress in the column? b. What is the strain in the column? c. If the column is 2.0 m high, how much is the length changed by supporting the load? d. What is the maximum weight the column can support? Suppose air, with a density of 1.09 kg/m^3 is flowing into a venturimeter. The narrow section of the pipe at point A has a diameter that is 1/3^rd of the diameter of the larger section of the pipe at point B. The U-shaped tube is filled with water and the difference in height between the two sections of pipe is h = 1.45 cm. Answer the following: a. Using the value of h, calculate the pressure difference between A and B. Density of water = 1000 kg/m^3 b. Now write the continuity equation between points A and B. c. Now write Bernoulli equation between points A and B. d. Solve equation b and equation c to obtain the velocity of air at A and B.

Explanation / Answer

(a) Stress = force/area

= 8*104 /45*10-4 N/m2 = 1.7*107N/m2

(b) Strain = stress / young's modulus

= 2.8*107 /6*1010 = 2.963 * 10-4  

(c) L/L = strain

       so, L = L*strain =2*2.963*10-4 m = 5.925*10-4 m

(d) Maximum weight that can be supported = compressivestrength * area

= 2*108*45*10-4N

= 9*105 N.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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