The compound bar containing steel bronze and aluminum segments carries the axial loads shown in the figure. The properties of the segments are listed in the table. Determine the maximum allowable value of P in pounds if the change in length of the entire bar is limited to +0.07in. A (in²) E (psi) Steel Bronze Aluminm 2P 0.75 1 0.5 Steel 2 ft Bronze 4 ft 5P 30000000 12000000 10000000 4P Aluminum 3 ft L (FT) 2 4 3 P : ...

The compound bar containing steel bronze and aluminum segments carries the axial loads shown in the figure. The properties of the segments are listed in the table. Determine the maximum allowable value of P in pounds if the change in length of the entire bar is limited to +0.07in. A (in²) E (psi) Steel Bronze Aluminm 2P 0.75 1 0.5 Steel 2 ft Bronze 4 ft 5P 30000000 12000000 10000000 4P Aluminum 3 ft L (FT) 2 4 3 P : ...

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter1: Introduction

Section: Chapter Questions

Problem 1.5.6P: The data in Table 1.5.3 were obtained from a tensile test of a metal specimen with a rectangular...

See similar textbooks

Similar questions

Q2\ A compound bar consisting of bronze, aluminum, and steel segments is loaded axially as shown in the figure. Determine the maximum allowable value of P if the change in length of the bar is limited to 2 mm and the working stresses prescribed in the table are not to be exceeded. A (mm') E (GPa) Ow (MPa) =TTT Aluminum Bronze 450 83 120 3P Bronze Steel 2P P AP- Aluminum 600 70 80 Steel -0.6 m→l- - 1.0 m- H-08 m- 300 200 140
A 6m long solid bar of circle section has 140mm diameter for aluminum for length of 4m, while it has 120mm diameter for a length of 2m for steel bar, as shown in figure. Find the elongation of the bar, when it is subjected to an axial tensile force of 300 kN. Take Est-200GPa, Eal- 70 GPa. (30 marks) 140 I *** 300 Kn Aluminum (30 marks) 4 m Steel 2 m 300 Kn 120
(1) Given A992 steel Ag = 10 in %3D KLy- 80, ry what is $ Pa=? = 60 the' column streugth design
Find the lightest S-shape for the beam shown if the working stress in bending is 175 MPa. What is the maximum bending stress in the beam selected? 50 kN/m 4 m -2.4 m· -x X- TABLE B-3 Properties of I-Beam Sections (S-Shapes): SI Units Flange Axis Y-Y Web X-X spxy Mass Area Width ıdə (wu) Thickness thickness Pl1 = S (10³ mm³) yli^ = 4 (10° mm*) (u) 27.7 27.7 22.1 Designation (10° mm*) (ww) 20.3 (ww) 38.9 („ww) (ww) (ww) 081 158 006 77 622 622 204 1 320 1 220 081 x 019S 4 230 240 34.5 338 × 158 x 149 x 134 007 184 15.7 3930 3 260 247 O'ZE 19.7 320 001 07 6'6E 32.3 006 81 00 I 15 200 149 6'81 15.9 166 937 677 234 019 215 134 181 22.1 090 € 2870 019 18.6 205 33.0 611 22.1 12.7 874 241 17.5 197 61I x 34.0 019 S510 × 143 143 18 200 16300 14 200 12 500 91S 91S 183 23.4 23.4 20.2 20.3 16.8 S69 653 2 700 20.8 228 x 128 x 112 × 98.2 128 112 33.8 34.5 29.5 179 961 2 540 007 194 19.4 80s 508 216 152 162 16.1 533 001 7 1950 12.3 98.2 20.2 12.8 495 6SI 11.4 144 30.2 661 S460 × 104 104 13 200 457 6SI…
The rubber band given below is subjected to the following tensile loading. Calculate the minimum thickness of the rubber (tr) and the minimum steel pin diameter (Dpin) so that the structure does not fail. Consider: Allowable tensile strength of the rubber= 20MPa Allowable shear strength of the steel = 200MPa Reflection: 1) How would you solve this problem if a Factor of Safety was given? 2) Are there any other dimensions worth calculating for the rubber belt?
The modulus of elasticity of mild steel is equal to .. .? .... .. a. 2 x 10ʻN/m? a. 2 x 10'N/m 2 b. 2 x 10° N/m 4 c. 2 x 10 N/mm 5 d. 2 x 10 N/mm Check
Photos 4 Crop & rotate Filters * Adjustments 2 Undo all Redo all Crop & rotate Straightening 0 ° Two round bars are joined and used as a tension member as illustrated. The modulus of elasticity is 200 GPa for steel and 70 GPa for aluminum. Rotate D4 Flip O Aspect ratio : Custom Aluminum 100 mm diameter Reset bar 50 mm diaineter P Steel bar +280 mm 400 mm Determine the magnitude of the force P (in kN) that willI lengthen the two bars 0.45 mm. * Your answer Save a copy Cancel 6:50 PM O Type here to search 日 28°C Partly cloudy 4)) ENG 23/09/2021
The light rigid bar ABCD shown is pinned at C and connected to two vertical rods. The bar was initially horizontal, and the rods were stress-free before the load P = 20 KN is applied. %3D Steel E200 GPa A-600mm L1 m 2.0 m 0.6 m 1.5 m Pa20 KN Aluminum E-70 GPa A=900mm L-1.5 m What is the ratio of the deformation of steel to deformation of aluminum?
Compute for the linear weight of 10 mm reinforcing bar. (Use the unit weight of the Steel). o 0.692 kg/m o 0.811 kg/m o 0.888 kg/m o 0.617 kg/m
(40 How A Rismatic steel bar of a Square cro ss section (looxlo0)mm is loaoled by Compressive force P= 1300 K, the bar has length 2.5 m, Eg=200GPay V=0.3,determine the shortening 8, the increase in the dimensions of cross Section, the change in Volume DV of the bar? loomm loomm
A rigid beam rest on two bronze bars 700mm long. Another aluminum bar is placed between them which is 699.93mm long. If the weight of the beam is 375KN, determine the stress in the aluminum bar in MPa. (Round your answer to 3 decimal places) (Exclude your unit in the answer box) Bronze Alum 850 1500 Ain mm E in GPa 70 83 bronz alum bronze
A bronze bar is fastened between a steel bar and an aluminum bar as shown in the figure. Axial loads are applied at the positions indicated. The allowable overall deformation is not to exceed 1.2 mm. Assume that the 2P Aluminum L- 1.5 m A= 550 mm E= 70 GPa Steel Bronze L= 2.0 m L= 1.0 m A= 600 mm A= 650 mm E= 200 GPa E= 83 GPa assembly is suitably braced to prevent buckling. Find the maximum permissible value of P.