b 10 t S M = 25 kNm 30° pp. y t Z b = 100 mm h = 200 mm t = 8 mm 20 2> Z Ch t IN
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In the bending of a U-profile beam, the load path passes through the torsional center C, causing a moment of 25 kNm at the cross-section under consideration. Additionally, the beam is subjected to an axial tensile force of 100 kN at the centroid.
Determine the maximum absolute normal stress.
(Check hint: approximately 350 MPa, but where?)
Step by step
Solved in 2 steps with 1 images
- Calculate the dimensions of the I-section of a connecting rod, stating any assumptions. using the data below: Maximum cylinder pressure = 3.15N/mm² Cylinder bore = 100mm Factor of Safety=6 Crank length 95mm Connecting rod length=380mm Take the constant a=7500.Hand written plzProblem 4 Press-Fit (3-116) Matlab Code is for (3-117) The table gives data concerning the shrink fit of two cylinders of differing materials and dimensional specification in inches. Elastic constants for different materials may be found in Table A-5. Identify the radial interference &, then find the interference pressure p, and the tangential normal stress on both sides of the fit surface. If dimensional tolerances are given at fit surfaces, repeat the problem for the highest and lowest stress levels. Problem Number Material 3-116 3-117 Inner Cylinder di Steel Steel 0 0 do 2.002 2.002 Outer Cylinder D; Material Steel Cast iron 2.000 2.000 Do 3.00 3.00
- QB2 SHEAR AND TORSION An aircraft engine develops 150kW at 1500rpm. The engine output shaft is steel which fails when the shear stress is 160N/mm². a) If the output shaft is solid, determine a suitable diameter to give a safety factor of three. b) If the shaft is hollow with an external diameter of 50mm, calculate a suitable internal diameter to give a safety factor of three. Also, determine the percentage saving in weight.See provided solutions.Draw V & M diagram 800 Ib 1000 Ib 4000 Ib 200 Ib 4' 6' 4' R R2
- Q4: A steel cantilever, the factor of safety is 2, the sensitivity factor 0.9, surface finish factor is 0.81, size factor is 0.8. Sut= 600 MPa and vield strength is 310 MPa .Calculate the diameter at the fillet. Neglect the transverse shear stress. 1:5 100mm A 0.2d d -50 kN 150kN 200mmGiven: bxh = 400 mm x 600 mm Ast=8 x 32 mm diameter bars Clear concrete cover to 12 mm diameter ties = 40 mm Concrete fc' = 28 MPa Steel fy = 415 MPa 400mm Mu = -420 kN-m; Vu = 370 kN; Nu = 580 KN (Compression) For WL to the -X direction: 600mm. Reduction Factor for shear = 0.75 Due to reversal lateral force, the design axial load due to the combined effect of DL, LL and WL change are follows: For WL to the +X direction: Mu = +420 kN-m; Vu = -370 kN; Nu = -450 kN (Tension) 1. Ans. 220 Determine the concrete shear strength (kN) for the positive x direction of WL by simplified calculation. 2. Ans. 0 Determine the concrete shear strength (kN) for the negative x direction of WL by simplified calculation. 3. Ans. 125 Calculate the required spacing (mm) of shear reinforcement. Apply code provisions on spacing limits of reinforcement where applicable.This is an old practice exam. The answers are OAB = 19.10 ksi OBC = 2.228 ksi OCD = −2.865 ksi v = 0.2792delta Ltot = 0.01585 in (increase) but why
- Material Sy Sut E AISI 4130 100 ksi 125 ksi 29 x 10^3 ksi AISI 1025 36 ksi 55 ksi 29 x 10^3 ksi Aluminum 2014 T6 53 ksi 60 ksi 10.8 x 10^3 ksi Aluminum 6061 T4 16 ksi 30 ksi 9.9 x 10^3 ksi Ti-6AL-4V 126 ksi 134 ksi 16 x 10^3 ksi You have a bar of AISI 4130 steel, a bar of Aluminum 6061 T4, and a bar of Ti-6AL-4V. All of the bars have identical cross section areas of 1 in^2. An axial stress of 20 ksi is placed on each of the bars. Which bar is the strongest? AISI 4130 Steel Aluminum 6061 T4 Ti-6AL-4V All of the bars have the same strengthMaterial Sy Sut E AISI 4130 100 ksi 125 ksi 29 x 10^3 ksi AISI 1025 36 ksi 55 ksi 29 x 10^3 ksi Aluminum 2014 T6 53 ksi 60 ksi 10.8 x 10^3 ksi Aluminum 6061 T4 16 ksi 30 ksi 9.9 x 10^3 ksi Ti-6AL-4V 126 ksi 134 ksi 16 x 10^3 ksi You have a bar of AISI 4130 steel, a bar of Aluminum 6061 T4, and a bar of Ti-6AL-4V. All of the bars have identical cross section areas of 1 in^2. An axial stress of 20 ksi is placed on each of the bars. Which bar will yield? AISI 4130 Steel Aluminum 6061 T4 Ti-6AL-4V All of the bars will yieldRepeat Problem 11.2-14 using L = 12 ft, ß = 0.25 kips/in., ßRl= 1.5ßL2, and ßR2= 2 ßR1.