1. A rod consists of three bars of unequal diameters with their diameters and lengths are shown in the Figure 1. Considerer the modulus of elasticity to be 200 kN/mm², find: 1.1. The stress in each bar 1.2. The elongation of the rod - 20 mm 20 kN- 1.5 m- 2.5 m 40 mm 90 KN- Figure 1 +++ - 30 mm 2.0 m- 60 KN

1. A rod consists of three bars of unequal diameters with their diameters and lengths are shown in the Figure 1. Considerer the modulus of elasticity to be 200 kN/mm², find: 1.1. The stress in each bar 1.2. The elongation of the rod - 20 mm 20 kN- 1.5 m- 2.5 m 40 mm 90 KN- Figure 1 +++ - 30 mm 2.0 m- 60 KN

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

Question 3 Determine the elongation of the square hollow bar when it is subjected to the axial force P = 100 kN, find the permanent elongation of the bar. The bar is made of a metal alloy having a stress-strain diagram which can be approximated as shown. o (MPa) 500 600 mm P 50 mm 250 5 mm e (mm/mm) 0.00125 0.05 50 mm ´5 mm a. Find the stress (MPa) of the bar. b. Find the permanent elongation (mm) of the bar.
F2 Q1) Axial displacement of point C in the system shown on the left is 0.01 cm. Find the maximum elongation of the bar and the maximum normal stress. A F1 A. GIVEN: F2 = 100 kN, (1 = 240 cm, €2= 160 cm, bi = 5 cm, bz=10 cm, h=5 cm, E=2.107 N/em², a=20 cm Note: neglect stress concentration. h bị b2 A-A section
2. The tires of an agricultural tractor shown below are assumed to be 25 cm wide for the front tires and 40 cm at the rear tires. Ground contact of each tire is 15 cm. Calculate the compressive stress between each tire and the ground surface. Weight = 2,500 kg 15 cm 1.3m Point A 2.2 m 2R 2F
3. The composite bar consists of a 20-mm diameter steel segment AB and 50-mm diameter brass end segments DA and CB. Find the stress on each segment due to the applied loads. E of steel is 200 GPa and E of brass is 101 GPa. 250mm 250mm 250mm 50mm 20mm 50mm 75KN 100KN brass steel brass D B C 75KN 100KN
2. A composite rod of overall length of 200 mm comprised of a steel rod and brass rod attached rigidly to the end. The diameter and length of the steel rod are given as 10 mm and 120 mm, respectively, the diameter and the length of the brass rod as 20 mm and 80 mm respectively. The rod is used as a tie in a link mechanism and the strain in the brass rod is limited to 0,53x 10-3. Given that the total extension of the composite rod must not exceed 0,162 4 mm and E for steel is 200 GPa, respectively. Calculate: d. Modulus of elasticity for the brass.
2. A composite rod of overall length of 200 mm comprised of a steel rod and brass rod attached rigidly to the end. The diameter and length of the steel rod are given as 10 mm and 120 mm, respectively, the diameter and the length of the brass rod as 20 mm and 80 mm respectively. The rod is used as a tie in a link mechanism and the strain in the brass rod is limited to 0,53x 10-3. Given that the total extension of the composite rod must not exceed 0,162 4 mm and E for steel is 200 GPa, respectively. Calculate: a. Strain in the steel rod
2. A composite rod of overall length of 200 mm comprised of a steel rod and brass rod attached rigidly to the end. The diameter and length of the steel rod are given as 10 mm and 120 mm, respectively, the diameter and the length of the brass rod as 20 mm and 80 mm respectively. The rod is used as a tie in a link mechanism and the strain in the brass rod is limited to 0,53x 10-3. Given that the total extension of the composite rod must not exceed 0,162 4 mm and E for steel is 200 GPa, respectively. Calculate: b. Load carried by the steel and brass rods
Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Take d₁ = 30 mm, d2 = 50 mm, and P = 50 kN. P 0.9 m 125 kN 125 KN B 1.2 m Find the average normal stress at the midsection of rod BC. (Input the answer with the appropriate sign.) The average normal stress at the midsection of rod BC is MPa.
2. A composite rod of overall length of 200 mm comprised of a steel rod and brass rod attached rigidly to the end. The diameter and length of the steel rod are given as 10 mm and 120 mm, respectively, the diameter and the length of the brass rod as 20 mm and 80 mm respectively. The rod is used as a tie in a link mechanism and the strain in the brass rod is limited to 0,53x 10-3. Given that the total extension of the composite rod must not exceed 0,162 4 mm and E for steel is 200 GPa, respectively. Calculate: a. Strain in the steel rod b. Load carried by the steel and brass rods d. Modulus of elasticity for the brass.
Calculate the combined stress (MPa) on the rod shown below using the principle stress method. Here, F1 = 0.35 kN, F2 = 0.12 kN, T = 428000 N-mm. Length and diameter of the rod are 75 cm and 75 mm, respectively. F2 F1 d
Let's consider a rod having a solid circular cross-section with diameter of 4 mm and it is made of a material having a Young's modulus E = 120 Gpa and a Poisson's ratio of 0.33. If a tensile force F is subjected to that rod cross-section, the diameter becomes 3.995 mm. determine the applied force F. Select one: O F= 2856 N OF = 2285 N OF = 7140N OF = 5712 N OF = 3427 N O F = 8568 N
60mm 2. H we measure a &=13µm decrease in the bolt diameter d. Bolt material properties are given as G=77 GPa and v=0.3. The bolt in the following figure when tightened, Determine the internal force at the bolt due to bolt tightening. i. 1 09 = p