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The structure shown consists of members ABD and CEF, and link DE. Link DE is 0.75 in. wide, and 0.125 in. thick (into the page). Determine the normal stress in the central section of that link if = 25° and no buckling occurs. Verify that no buckling occurs in link DE if it is made of steel with a Young’s modulus of E = 30×106 psi.
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- What would be the resultant force F of the triangular distributed load and its location d measured from point A. 150 lb/ft -3 ft < 5 ft 2 ft Select.one: a F-375 Ib and d = 1.67 ft b.F=300 lb and d = 2.5 ft OcF=37S Ib and d = 3.67 ft id F=750 lb and d = 3.67 ftPravinbhaiThe 50-ft measuring tape weighs 2.4 lb. Compute the span L of the tape to four significant figures.
- 4) A simply-supported beam has a shear diagram shown in Figure 2. a) Draw the moment diagram based on this shear diagram. b) Draw the load diagram based on this shear diagram. c) Locate the inflection poit (or the point where the moment diagram crosses the horizontal tine) from the left support, in meters.X Your answer is incorrect. A flanged wooden shape is used to support the loads shown on the beam. The dimensions of the shape are shown in the second figure. Assume LAB - 8 ft, Lec-2 ft, Lco-3 ft. LDE-3 ft, Pc- 1870 lb. PE-2190 lb, WAE-650 lb/ft, b₁-9 in., b2-2 in., b3-6 in., d₁-2 in..d₂-7 in., d3-2 in. Consider the entire 16-ft length of the beam and determine: (a) the maximum tension bending stress o at any location along the beam, and (b) the maximum compression bending stress oc at any location along the beam. Answers: WAB LAB (a) GT = (b) σ, = i i 402.3 558.1 B Pr LBC by C IH by LCD b₂ psi. psi. D d₁ LDE dz dy РЕ EAs per the crane shown in the Figure. Determine the force member at CE.
- Find the greatest weight that can be applied without the front wheels lifting off the ground. The answer is 30,043.43 N show solutionCalculate forces (moments if applicable) for all points shown and write them down in a table. 1200N 100cm B- 100cm to OD- 100cm 100cm E O -2.4m-The force system shown consists of the couple C and four forces. The resultant of this system is a 800-lb-in. counterclockwise couple. Determine the magnitude of the couple C (in Ib-in). * Your answer
- Suppose that P1 = 1500 lb and P2 = 1800 lb Determine force in members: CD, KJQ2. Rigid member DEF is supported by three rods, AD, BE, and CF, and the rods are of the same original dimension. Rods AD and CF are of the same material, with Young's modulus of 10.6x103 ksi, and rod BE is made of a different material, with Young's modulus of 17.0×103 ksi. Points B, E, and G are collinear, and the lengths are /₁ = 3.6 ft and /2 = 3.9 ft. Determine the force in rod CF in the unit of lb. A D 4₁ G B 1 kip -1₂- 2 C F 4i li. In. O MMecnaniCS / مطلوب Untitled Section F1 F2 a F3 B. d F4 F5 Consider the following values: - a - 25 m; b - 5 m; a - 25 m; d - 5 m; F1 = 6 KN; F2 = 7 kN; F3 = 8 KN; F4 = 9 kN; F5 = 10 KN; a = 60°, 0 = 45°, B== 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 15.8 kN.m b) 74 kN.m c) 88 kN.m d) 59 kN.m e) 241 kN.m 0- 47 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 218.8 kN.m b) – 117.8 kN.m c) - 131.6 kN.m d) 98.9 kN.m e) 125.4 kN.m f) - 224.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 186.7 kN.m b) - 143.6 kN.m c) 114.6 kN.m d) - 217.9 kN.m e) 224.1 KN.m ) - 135.3 KN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 217.8 kN.m b) – 139.5 kN.m c) 111.9 kN.m d) 206.9 kN.m e) - 127.1 kN.m f) 99.1 kN.m Activate 5] What is the moment of the force F2 about point E? Go to Setting f) 321.3 kN.m a) 121.7 kN.m…