Assuming the question and calculations below are correct, make another real-world case scenario where you would use the

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Assuming the question and calculations below are correct, makeanother real-world case scenario where you would use thesame of calculations such as calculating stress and strain,recreate it, and calculate it.
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Assuming The Question And Calculations Below Are Correct Make Another Real World Case Scenario Where You Would Use The 1 (168.91 KiB) Viewed 31 times

Assuming The Question And Calculations Below Are Correct Make Another Real World Case Scenario Where You Would Use The 2 (516.49 KiB) Viewed 31 times

As illustrated in the following scheme, a rigid beam is fixed at the top of three bars symmetrically arranged. Two of the = 200 GPa, asteel made of steel (Esteel (EA. = 73.1 GPa, αA. = 23x10-6 C0-1). The structure is initially at a temperature of 20 °C of circular cross section, which are and the other one is 12x10-6 C⁰-1) = bars are mm in the absence of loading on the beam. made of aluminium have a length of 250 Considering that the structure is then subjected to the action of a uniformly distributed load with and to a temperature increase of 60 °C, determine the stress and strain in each bar. 40 mm steel 300 mm 60 mm aluminium 300 mm 7 40 mm steel 150 kN/m and the bars an intensity of 150 kN/m
Solu Let Os be the stress in 2 steel beams (due to symmetrical arrangement both beams have some stress) and A. be stress in Aluminium beam K 300mm F₁ = Os F = 0₂ A₂ Balancing moment about с F₁ X 0.6 + √₂₂x0.3 O A₁ طا 2 Fi + F₂ DA OA = А - (150x 0.6) X/10.6) = 0 90 by let As be compression of steel beem As = D'A گی = (F) [1 JAL = = Ę F. LA A EA Fx0.25 хо Jx0.06² 4 = 12.096 X107 E₂ since beam is rigid 11 = : compression of Aluminium iJ ls श ܠ AtA A- 3 E Fix 0.25 JC (0.04)² x 200x106 4 9.947X107 Fi ۲۷۶ solving equation (1) e (2) 16-443 JC 4 B 122.886 J. -x0.062 4 ↑ F₁ 34.139 F₁ → F₁ and F₂ (compressive) I's and s'A be elongation due to temperature 7 A₁ = %&AT = 12×10 % ×0-25×60 = 1800 x10? m от XAA AT = 23 X 106x0.25 x 60 = ΔΑ + ΔΑ 0-25 x 73.1×106 x0.04² 9.947 F₁ - 12.096F₂ =-1650 24.182 F₁ + 12.056F₂ 7-85×10 (1) A₁ = ⇒ -9.947X167 F₁ + 1800×10²7 = -12.096 × 10²¹ F₂ +2450×10] 9.947 F₁ - 12.096F₂ = -1650 (2) -4 A₂: area of aluminium beam ( : area of steel beam = due to fi (Is: length of steel beem Es: Young modulus or steel (Compressive) =1088.64 -561.36 122.886 knj 300 mm J 0.25 7.85 x 10-4 (elongation) moment due to distributed load in kPa since F, ish kN = -16-443. KN Es = Asta's (strate in steel) ls (163.56 +1800) X107 la: Fi = wx² w: intensity x: length of beam : length of Alamini -um column 0.25 GA: Young modules of Aluminium 13084.92 kN/m² 13.084 MPa (elongation) 42205.86 kN/m² = 42.205 MPa (compressive) (tensile ) -due to compressive load. (-1486.43 +3480) x 107