Note: Please answer number 2, but the parameters and questions are based on number 1 as stated in the word problem. 1. The rectangular tube has the dimensions as shown. It is extruded from aluminumalloy 2014-T6. Determine the local compressive buckling stress. kh=5.2.04h2.0.04b 1.02. Same as Problem 1 but change the thickness of the h side to .072, but leave the bside .04 in thickness. kh= 4.3? 0.7 = 5300 n = 18.5

Here's a step-by-step approach to solve it:Step 1: Define Variables and Sign ConventionsBeam Length…

Answered: 1. The rectangular tube has the…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Determine the normal stress and change in length of the aluminum rod if the temperature was raised by 125°C. The 0.5-mm gap exists at 20°C. q = 5 t = 0
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Q4
2. A steel shaft (G = 80 GPa) of total length L = 3.0 m is encased for one-third of its length by a brass sleeve (G, = 40 GPa) that is securely bonded to the %3D %3D steel. The outer diameters of the shaft and sleeve are di = 75 mm and d2 = 90 mm, respectively. The maximum shear stress in brass is 70 MPa, maximum shear stress in steel is 110 MPa and the maximum allowable angle of twist is 5°. What is the maximum allowable torque that can be applied at the ends of the shaft? Brass Steel shaft di = 75 mm d2 = 90 mm sleeve T T A B 1.0 m- L = 2.0 m- 2 = 2.0 m d1 dy Brass- sleeve Steel shaft d2 d2
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The light rigid bar ABCD shown below is pinned at B and connected to two vertical rods. Assuming that the bar was initially horizontal and the rods stress-free, determine the stress in each rod after the load P = 20 kN is applied. steel L=1.5 m A=300 mm2 2 m- E= 200 GPa B D Aluminun L=2 m P = 20 kN - 1m A= 500 mm2 E= 70 GPa
A Sandwich block is comprised of brass in the centre with two aluminium plates attached to either side. The block is fastened by a top rigid end plate. An axial load (P = 443 kN) is applied to the centre of the block as shown below. Find the normal stress in the brass and the normal stress in the aluminium cover plates. Aluminum plates (E = 70 GPa) 300 mm Brass core (E = 105 GPa) 10 mm 40 mm 10 mm P Rigid end plate 60 mm OneDrive Screenshot saved The screenshot was added to your
a=2 b=5 80*a=160 90*b=450 100*b=500
Two steel tubes are joined at B by four pins (dp = 11 mm), as shown in the cross section a – a in the figure. The outer diameters of the tubes are dAB = 40 mm and dBC = 28 mm. The wall thicknesses are tAB = 6 mm and tBC = 7 mm. The yield stress in tension for the steel is (y = 200 MPa and the ultimate stress in tension is (U = 340 MPa. The corresponding yield and ultimate values in shear of the pine are 80 MPa and 140 MPa, respectively. Finally, the yield and ultimate values in bearing between the pins and the tubes are 260 MPa and 450 MPa, respectively. Assume that the factors of safety with respect to yield stress and ultimate stress are 4 and 5, respectively.Calculate the allowable tensile force Pallow considering tension in the tubes.Recompute the Pallow for shear in the pins.Finally, recompute Pallow for bearing between the pins and the tubes. What is the controlling value of P?
30 mm B T C 0.2 m PROBLEM 3.42 8 Two solid steel shafts, each of 30-mm diameter, are connected by the gears shown. Knowing that G = 77.2 GPa, determine the angle through which end A rotates when a torque of magnitude T = 200 N m is applied at A. . 0.4 m 0.2 m 60 mm D 0.1 m 90 mm 30 mm E 0.5 m
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