AN ALUMINUM SEGMENTS HAVING A DIAMETER AND LENGTHS SHOWN IN THE FIGURE. IF E = 85 KPa, COMPUTE THE TOTAL DEFORMATION. ASSUME THAT THE BARS ARE SUITABLY BRACED TO PREVENT BUCKLING. 12 mm diam 4 mm diam 6 mm diam 300 N A 1900 N 500 N C 200 N L = 800mm 700mm 1000mm
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- An A36 tension member shown in the figure is connected with three 19 mm diamter bolts. Diameter of hole = 22mm. Properties of 150 mm x 100 mm x 12.5 mm angle A = 3065 mm2 y = 50.55 mm Fy = 248 MPa Fu = 400 MPa Determine the design strength due to yielding in the gross section. (kN)Tempered rod with length of c= 8m and cross sectional dimensions a=297 mm is carrying an axial load P=75kN if the modulus of elasticity is 185gpa determine the elongationAn A36 tension member shown in the figure is connected with three 19 mm diamter bolts. Diameter of hole = 22mm. Properties of 150 mm x 100 mm x 12.5 mm angle A = 3065 mm2 y = 50.55 mm Fy = 248 MPa Fu = 400 MPa Determine the desire strength due to fracture in the net section. Use U = 0.85 kN
- The steel bar has the original dimensions shown in the figure. If it is subjected to an axial load of 50 kN, determine the change in its strength in mm at section a-a. Est = 200 GPa, vst = 0.29The 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?Before a load is applied, a rigid bar is fastened to two vertical rods. Calculate the vertical movement when P = 50Kn is used. Area(mm2) length E ALUMINUM 500 4m 70Gpa STEEL 300 3m 200Gpa
- A steel plate 375mm x 16mm thick is bolted by four bolts placed symmetrically to form a single lap joint. The bolts are spaced horizontally at a distance of 140mm and vertically at a distance of 150mm. If the diameter of bolt is 20 mm, calculate the design tensile strength in KN based on net section fracture considering a straight vertical path of failure Use Fy=248MPa and Fu=400MPa a) 1339.2 b) 1569.6 c)2092.8 d) 1488Given the class averages: F = 4.9 N, W= 0.0098 m, t= 0.0098 m, d =29 intervals (dial gauge reading) and L = 0.3 m Calculate the stiffness, in GPa, for the steel. (give your answer up to three significant figures).Answer all.a.A mild steel flat of width b= 150 mm, thickness t = 20 mm and length L= 6000 mm carries an axial pull P = 300 kN. If modulus of elasticity of mild steel is 200 GPa and Poisson’s ratio is 0.25, calculate the change in length, width and thickness of the flat. B.A stepped bar made of brass is subjected to axial loads as shown in the below figure. Determine the total change in length of the bar. Take E =100 GPa.figure is attached C.A thin cylinder of external diameter 820 mm, length 3 m and wall thickness of 10 mm is subjected to an internal pressure of 2.5 N/mm2 . Determine the change in diameter and length. Take Young’s modulus = 2×105 N/mm2 and Poisson’s ratio = 0.25.
- Question 29 An A36 angle bar is used as a tension member connected on both legs with 20 mm . Given that s = 60 mm, calculate the design strength (kN) by LRFD with the limiting state of tensile yielding in the gross area. Write your answer to the nearest whole number. Ag = 1,852 mm2 t = 8 mmAN ALUMINUM SEGMENTS HAVING A DIAMETER AND LENGTHS SHOWN IN THE FIGURE. IF E = 75 KPa, COMPUTE THE TOTAL DEFORMATION. ASSUME THAT THE BARS ARE SUITABLY BRACED TO PREVENTBUCKLING. ALSO, DRAW THE AXIAL FORCE DIAGRAM.Axial loads are applied to the compound rod that is composed of an aluminum segment rigidly connected between steel and bronze segments. What is the stress (in Mpa) in the steel given that P= 96.3 kN? Express your answer in two decimal places.