9-5. Solve Prob. 9-4 using the stres equations developed in Sec. 9.2. Show the r 642002
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- Solve the preceding problem using the fol low-data: W 8 × 21 section, L = 84 in., P = 4.5 kips, a = 22.5°.P14-L In cach case, the state of stresa ,y Tay pruduces normal and shear stress companents along saction AB of the clement that have values of o, - -5 kPa and Tay -8 RPa when calculated uing the stres transformation equations Establish ther'and y axs for cach segment and specity the angle 6, then show these results acting on each segment. (a)nt 2 poa page 25 ol 40) - Google Orome nline.gmitie mod/quz/attempt.phplattempt-514100omid-4251718pages24 GMIT 25 A strel -200 Ga rod with a circular cross section k long Determine the minimum diameter reguired if the rod must transmit a tensile force of 50N withut eceeding an alowable stress of 180 M State your answer in mm without including the units and correct to two decimali places Answer ous page Nt pg 9ype here e ah
- A solid cylindrical shaft made of AISI 1020 steel(quenched and tempered at 870oC) rotates at a speedthat produces a safety factor of 2.5 against the stresscausing yielding. To instrument the shaft, a small hole isdrilled in its center for electric wires. At the same timethe material is changed to AISI 1080 steel that has beenquenched and tempered at 800oC. Find the safety factoragainst yielding of the new shaft.Delermine the orientation of principal planes af stress. Express your anewers ualng three significant figures eeparated by a comma. 8, ,- 19.3,72.24 Submit Prevlous Ancwere Requect Ancwer X Incorrect; Try Again; 3 attempts remaining Part C Determine the maximum in plane shear stress. Express your anawer to three slgnificant figuree and Include the approprlate unit Tmax in-plane 19.2 ksi Submit Previous Ancwers Correct Part D Determine the average normal stress. Express your anBwer to three slgnificant figuree and Include the approprlate unit Owg -15.0 ksi Submit Previous Ancwers Сorect • Part E Figure O 1 of 1 Delermine the orientation of the planes of maximum in plane shear stress. Express your anewers ualng three algnificant figures separated by a comma. ? 30 ksi Submit Requect Ancwar 12 ksi Provide FeedbackConsider the 2-D state of stress shown below. Using the provided scales graph the Mohr's Circle for the 2-D state of stress with "full' details Draw the Planes of Principal and Maximum Shear Stresses Provide "full" details and Use 3 Sig. Fig. in this problem Oy = 15 Ksi Tcw → S Txy= -5 ksi Ox = 10 Ksi Tyx = 5 Ksi 10 5 -20 -10 5 10 15 20 5 10 Tccw → S'
- Let's consider a solid circular shaft subjected to the torque loads as shown Determine the minimum required diameter "D" if the shaft is made of a material having a yielding shear stress Ty = 100 MPa, for a factor of safety FS = 2. The polar moment of inertia of a solid circular cross-section is given by Jr.R¹/2. 300 mm B 600 mm 100 N.m 500 N.m 600 mm 400 N.mFor the given question and solution, Why is taux'y' after the 25 degree rotation = positive Rsin(phi) = 1.05MA instead of negative Rsin(phi) ? how can i determine the sign convention after rotation? i get mixed up with the sign convention of the shear stress after rotation.The cubic element of the solid bady is subject to the stress Shaun in figure, where : State 6x = 60 Mpa t xys 10 MPa Gy = 50 MPa Determine principal stresses using both analy tical method and Mohr circle, as well as princifal directions (using Mohr circle only), Draw +he Mohr circle and physical plane. dy
- 7:04 10, O - • Z. Jay Patil Sir Samrud. 3 minutes ago JA force of 32 KN is required to Pumch a circular hole of 14 mm dicmeter in a metal plate 2 mm thich. calculate the compresi ve Stress developed in the pumching rod cmd shear streas in metal Plate.For the three stress states given, find all principal normal stresses using the eigenvalue method. σxx = 25 MPa, σyy = -50 MPa, ?xy = 25 MPa (clockwise)(4) Construct Mohr's Circle for the state of stress at Point B. Use the Mohr's Circle to construct a stress element (rotated properly) that illustrates the principal stresses at Point B. Ans. ₁16.4 ksi; o₂ = -0.061 ksi ау 2 in bi 5'-0" (5) Check your work on the prior problem by plugging into the stress transformation equations. Remember: these are best done as one, fluid calculator operation. Did you get the same answer? 0 30 x 180 kip 10 in. Oy² (0x,Oy, Txy, 6) = 0x + Oy 2 6 in. . У 0₂-Oy cos 20 2 10 in. B O₂² (√x,Oy,Txy,0) = 0x10 + 0x-O cos 20 + Txy sin 20 2 2 Txy sin 20 - Z - Txy³ (0x, Oy, Txy,0) = − (Ox-Oy) sin 20 + Txy cos 20 2