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- A cylindrical brick chimney of height H weighs w = 825 lb/ft of height (see figure). The inner and outer diameters are d1= 3 ft and d2= 4 ft, respectively. The wind pressure against the side of the chimney is p = 10 lb/ft2 of projected area. Determine the maximum height H if there is to be no tension in the brickwork.Repeat Problem 11.2-3 assuming that R= 10 kN · m/rad and L = 2 m.An L-shaped reinforced concrete slab 12 Ft X 12 ft, with a 6 Ft X 6 ft cut-out and thickness t = 9.0 in, is lifted by three cables attached at O, B, and D, as shown in the figure. The cables are are combined at point Q, which is 7.0 Ft above the top of the slab and directly above the center of mass at C. Each cable has an effective cross-sectional area of Ae= 0.12 in2. (a) Find the tensile force Tr(i = 1, 2, 3) in each cable due to the weight W of the concrete slab (ignore weight of cables). (b) Find the average stress ov in each cable. (See Table I-1 in Appendix I for the weight density of reinforced concrete.) (c) Add cable AQ so that OQA is one continuous cable, with each segment having Force T, which is connected to cables BQ and DQ at point Q. Repeat parts (a) and (b). Hini: There are now three Forced equilibrium equations and one constrain equation, T1= T4.
- I need help please Two gondolas on a ski lift are locked in the position shown in the figure while repairs are being made elsewhere. The distance between support towers is L = 100 ft. The length of each cable segment under gondola weights WB = 450 lb and WC = 650 lb are DAB = 12 ft, DBC = 70 ft, and DCD = 20 ft. The cable sag at B is ΔB = 3.9 ft and that at C(ΔC) is 7.1 ft. The effective cross-sectional area of the cable is Ae = 0.12 in2.(a) Find the tension force in each cable segment; neglect the mass of the cable.(b) Find the average stress (s) in each cable segment.Find the support reacions and joint force of the construction shown on the figure, if; P=12 kN q=9 kN/m M=12 kN and L=2 mFind the internal force system acting on section 3 for the pin-connected frame. a P = 255 N (C), V = 0, M = 29.8 N · m b P = 255 N (T), V = 0, M = 29.8 N · m c P = 255 N (C), V = 0, M = 92.8 N · m d P = 525 N (C), V = 0, M = 29.8 N · m
- Problem 1.12: A journal 25 mm in diameter and supported in sliding bearing has maximum end reaction of 25 kN Assuming allowable pressure 5 N/mm. find length of sliding bearing Solution: Given data:d25 mm, W== 25 KN = 2.5x 10¹ N.Draw the free body diagram of this problemA beam of mass Mv = 40 kg and length L = 3.0 m is attached to a hinge in the vertical wall. On the bar there is a block, of mass mb = 10 kg, which can be in different positions “x” on the wall. The cable that secures the end of the bar to the upper horizontal slab makes an angle θ = 45o with the bar, that is, with the horizontal (Figure).a) What are the minimum and maximum tensions in the cable as the block moves along the beam?b) If the block is at the center of the beam, what force will the hinge exert?
- Two gondolas on a ski lift are locked in the postion show in the figure while repairs are being made elsewhere.The distance between support towers is L = LOO ft.The length of each cable segement under gondolas weighing WB= 450 lb and Wc=650 lb are DAB=12 ft, DBC=70 a , and DCB=20 ft . The cable sag at B is AB = 3.9 ft and that C is A = 7.1 ft.THe effective cross-sectional area of the cables is Ae=0.12 in". (a) Find the tension force in each segment; neglect the mass of the cable. (b) Find the average stress(σ) in each cable segment.A truck of mass 3 tonnes runs into a buffer stop having three buffer springs each of 40 kN/cm stiffness connected in parallel. Find the maximum compression of the springs, and the load if the wagon is travelling at 60 kmph. 1-Equivalent Stiffness of spring in kN/mm is 2-Kinetic Energy in kNmm is 3-The maximum compression of the spring in mm is 4-Load acting on the spring in kN isThe simplest truss consists of two rods and is loaded by the block. Given: block weight Q = 9,4 kN. Angle α = 31 degrees. Find: the algebraic (positive or negative) value (in kN) of normal force N1 in the rod 1, considering that N1>0 for tension and N1<0 for compression.