The minimum factor of safety i able load P that may be applied
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- The structure shown supports a distributed load of 700 lb/ft. Assume a -41 ft, b-1.0ft,c-2.1 ft, and d-3.0ft. The 0.500-in- diameter bolts at A and Bare each used in double shear connections. The cross-sectional area of axial member (1) is 0.75 in Axial member (1) has a yield strength of 39 ksi, and each of the pins has an ultimate shear strength of 51 ksi. Determine (a) the factor of safety with respect to the yield strength for axial member (1). (b) the factor of safety with respect to the ultimate shear strength for pin A (c) the factor of safety with respect to the uitimate shear strength for pin B. Answers: (a) FS, - (b) FS. - ( FSaThe axial stresses are 12 MPa C in the wood post B and 150 MPa T in the steel bar A as shown below. Pins at A, C, and D are smooth. Determine (a)The load P.(b)The minimum diameter for pin C if it is in single shear and the cross-shearing stress is limited to 70 MPa.(c)The minimum diameter for pin D if it is in double shear and the cross-shearing stress is limited to 70 MPa.The AB bond must be made of a steel for which the normal stress limit is 450 MPa. Determine the cross-sectional area for AB for the which the safety factor is 3.50. Suppose the link is adequately reinforced around the pins at A and B. What is the deformationof bar AB when subjected to the load applied to it considering the modulus of elasticity of steel E = 210 Gpa and the length L of thebar equal to 0.75m (All dimensions are in meters)?
- Link 3 shown acts as a brace to support the 270-lb load. For buckling in the plane of the figure, the link may be regarded as pinned at both ends; for out of plane buckling both ends are fixed. If the link is to be made of CD 1020 steel and is 1-in wide, and considering only out-of-plane buckling, determine a suitable standard thickness using a design factor of 3. F= 270 lbf 3 ft 60 SOLUTION:The bell-crank mechanism shown below is supported by a single-shear pin connection at B and a roller support at C. A load P acts at an angle of θ=40° at joint A. The pin at B has a diameter of 0.375 in, and the bell crank has a thickness of 0.188 in. The ultimate shear strenght of the pin material is 28 ksi, and the ultimate bearing strenght of the crank material is 52 ksi. A minimum factor of safety of 3.0 with respect to shear and bearing is required. Assume that a = 9.0 in. and b =5.0 in. What is the maximum load P that may be applied at joint A of the bell crank?Two plates each with thickness of 18 mm are bolted together with 6 – 20mm ø bolts forming a lap connection. Bolt spacing are as follows: X1 = 50 mm, X2 = 120 mm and X3 = 100 mm FvRivet = 130 MPa. Calculate the the pemissible tensile load P under the following conditions. a. Based on the effective net area. b. Based on block shear. c. Based on the safe value of P.
- PROBLEM 1 The two plates are joined together using a rectangular pin. What is the smallest width of the plates and pin if the load applied is 45 KN. The allowable tensile stress is 180MPA and the allowable shear and bearing stress are 120MPa and 380MPA respectively. rectanguler pin 1omm 10 mm 45k P.Given:a. the force in BD=25KNb. vertical reaction at point C=-10KNc. horizontal reaction at point C=15KNd. resultant of reaction at point C=18.0KNe. allowable tensile (normal) stress in member BD =83.33MPaf. the allowable shear stress in the pin at C =72.7MPaPlease help me solve for:1. What is the required cross-section area and minimum thickness t of member BD?2. What is the required SHEAR cross-section area at point C?3. What is the minimum diameter for the pin at C?(Solve for a value with the combination of mm and N)s after the decimal po 200mm b18010 300mm L2 Joint D b18010 L1 "p double AR shear b180106003-1892615 B 50mm b18010 - 30mm b180106003-18926151 A frame is given above and an external load R is applied at point C. Pin connection Joint B b18010 b180106003- 18926151126 dg b1801000s Calculate the maximum normal stress for member BD by taking into account the reduced sectional areas at the pin connections. b180106003- 18926151126 (dD= 8mm, single shear b18010 b180108003-18926151126 at Dis in double shear, B is in single shear. b180106003-18926151126 b180106003 151126 24,05 MPa A b18010 18926151126 31,37 MPa b180106003 -18926151126 18926151126 01000s-18926151126 37,04 MPa b180108003-18926151126 b18010 828 18926151126 43,73 MPa bi8010 b180108003- 18928151126 b180106003-18926151126 18926151126 b180106003- 18926151126 b180100003-18926151126 b180106003- 18926151126 b180100003-18926151126 b18010 b180108003-18926151126 b180108003-18926151126 b18010
- The one end of a hollow square tube whose side is 10 in with 1 in thickness is under a tensile stress 102,500 psi and the other end is connected with a U bracket using a double-pin system. Find the minimum diameter of pin is used according to shear strength. Take the factor of safety as 1.5 and σ(allowable stress) = 243 ksi for pin material.• Bars (1) have a cross-sectional area of A₁ = 0.65 in.2 and a length, L₁= 7 ft. • Bars (1) are made of cast iron with an elastic modulus of E₁ = 24,000 ksi, a coefficient of thermal expansion, a₁ = 12.3 x 10-6/°F and yield strength of oy = 120 ksi. • Bar (2) has a cross-sectional area of A₂ = 1.45 in.² and a length, L₂ = 5.5 ft. • Bar (2) is made of stainless steel with a coefficient of thermal expansion, a₂ = 10.6 × 10-6/°F. • The stress-strain diagram provided below presents the results of the stainless steel's (i.e., Bar (2)) bar tension test. • There is a gap of A = 0.1 in. in the connection at B and a = 3 ft. • All bars are unstressed before a load P = 12 kips is applied and the temperature increases by AT = 40°F. Stress-strain diagram for stainless steel bar: (1) a Stress (ksi) L2 Rigid bar 120 100 80 60 40 20 0 0 0 B (2) P a (1) C Upper scale L₁ Connection details at node C Lower scale. C 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.006 0.008 0.010 0.012 0.014 0.002 0.004 Strain…Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diameter of 13 mm and an allowable normal stress of 135 MPa. Pipe (2) has an outside diameter of 56 mm, a wall thickness of 6 mm, and an allowable normal stress of 150 MPa. Assume x₁ = 3.7 m, x₂ = 1.5 m, and y₁ = 3.0 m. Determine the maximum load Pmax that can be supported by the structure without exceeding either allowable normal stress. B P 3₁ (1) Answer: Pmax X1 i C kN