Find the stresss in the left rod after its lower end is attached to the support.
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Find the stresss in the left rod after its lower end is attached to the support.
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- Solve the preceding problem if F =90 mm, F = 42 kN, and t = 40°MPaA hollow, circular, cast-iron pipe (Ec =12,000 ksi) supports a brass rod (Ec= 14,000 ksi} and weight W — 2 kips, as shown. The outside diameter of the pipe is dc= 6 in. (a) If the allowable compressive stress in the pipe is S00O psi and the allowable shortening of the pipe is 0.02 in., what is the minimum required wall thickness trmm? (Include the weights of the rod and steel cap in your calculations.) (b) What is the elongation of the brass rod Srdue to both load Wand its own weight? (c) What is the minimum required clearance h?Solve the preceding problem if the mass of the tailgate is MT— 11 kg and that of the crate is hic— 6S kg. Use dimensions H = 305 mm, L = 406 mm, dc= 460 mm, and dT= 350 mm. The cable cross-sectional area is At= 11.0 mm'. (a) Find the tensile Force T and normal stress T in each cable. (b) IF each cable elongates
- The magnitude of force (F) if Al = 0.06 m: F = kx F = (500)(0.06) F = 30 N Homework A hollow cylinder 2 m long has an outside diameter of 50mm and inside diameter of 30mm. If the cylinder is carrying a load of 25 KN. FInd the stress in cylinder, also find deformation of the cylinder E=100 Gpa. (Ans: Stress=20.0 N/mm2, Deformation= 0.4 mm) A load of 5KN is to be raised with the help of a steel wire. Find the minimum diameter of wire if stress is not to exceed 100 MPa. For more solved problems: (Ans: d=7.98mm)a=2.1m (1) c = 2m b = 3.1m d = 1.8m P = 80 kN WHAT IS THE AXIAL STRESS IN ROD 1 ? ANSWER IN MPa ROUNDED-OFF TO 2 DECIMAL PLACES B dia. of rod 1: = 13mm dia of rod 2 = 11 mmIf a solid circular rod having an initial diameter D = 0.03 m and initial length L = 0.8 m, is subjected to a tensile force F = 300 kN, the new diameter will be: Given: The rod is made of steel having Young's modulus E = 200 Gpa and a Poisson's ratio = 0.3. Select one: a. Df= 2.993 mm Ob. Df = 10.92 mm Oc. Df= 3.007 mm Od. Df = 30.02 mm Oe. Df = 29.98 mm
- F2 Q1) Axial displacement of point C in the system shown on the left is 0.01 cm. Find the maximum elongation of the bar and the maximum normal stress. A F1 A. GIVEN: F2 = 100 kN, (1 = 240 cm, €2= 160 cm, bi = 5 cm, bz=10 cm, h=5 cm, E=2.107 N/em², a=20 cm Note: neglect stress concentration. h bị b2 A-A sectionR₁ $D₁ P = 20 kN Mv = 600 Nm. $D₂ R₂ D3 My f D₁ = 34 mm, D₂ = 40 mm, D3 = 36 mm, R₁ = 5 mm, R₂ = 2 mm, n = 0.45 -Calculate the greatest stress in the axle.Figure 1 shows a composite shaft ABCD acted upon horizontal forces. The moduli of elasticity and the cross-section areas of the segments of the shaft are also provided. Eal 70 GPa E = 126 GPa E = 200 GPa AAR=58 mm² ABC 77 mm2 Acp = 39 mm² Сopper 20KN 8 kN Aluminum Steel 7 kN B 20KN A 8 kN 450 mm -300 mm -- 400 mm Figure 1: Axially loaded composite shaft 2.1. Construct the diagram of internal normal forces for each cross-section of the (6) shaft (3) 2.2. Calculate the displacement of the point C with respect to the fixed-point A 2.3. If the shaft must have one uniform diameter determine the required minimum diameter of the shaft if the normal yield stress in the composite shaft must not (11) exceed 450 MPa with a factor of safety equal to 1.8.
- For the assembly, the temperature rises by 35°C after a load W = 140 kN is applied. Assume the horizontal bar will not deform. apronze=18.9µm/m "C and a steel=11.7µm/m°C. Determine the stress in each of the vertical rods. Bronze L= 3 m A = 1300 mm2 E = 83 GPa Steel L= 1.5 m A = 320 mm2 E = 200 GPa 1 m 2.5 m 1.5 m10/10 find the tensile stress in upper plate in section 3-3 5. 1 4 80 kN 80 kN 8-bolt DIA. 10 MM 80 kN 80 kN 8-bolt DIA. 10 MM 2) 300 mm 8 mmBD rod made of G1 = 78 GPa with a diameter of 27 mm attached to the CA tube at point B, the tube made of G2 = 133 GPa had an outer diameter of 59 mm and a wall thickness of 10 mm. If T1 = 935 Nm and T2 = 1683 Nm, answer the following questions: 1- It is the polar inertial moment of the bar BD. 2- Maximum shear stress BD. 3- It is the polar moment of inertia of the CA tube. 4- It is the maximum shear stress BD. 5- It is the maximum shear stress of the CA tube. 6- It is the maximum shear stress of the assembly. 7- is the torsion angle between D and B. 8- It is the torsion angle between C and A. -9 is the torsion angle at D. 10- is the torsion angle between C and B. 11- is the reaction at point A.