For a W14 x 53 column of A992 steel, 45 ft tall, determine the available load capacity when the ends are pinned and the column is braced at mid-height in the weak axis.
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- * Required For nos. 38-40, satisfy the conditions of the given problem below: The internal drag truss of the wing of a light airplane is subjected to the forces shown in Determine the force in figure 3. each member, and state if the members are in tension or compression. 38. Load on member BC * 1 point H 0.61 m 0.61 m- 0.61 m 356 N O A. 579.23 N (T) B. 579.32 (C) O C. 580.20 N(T) D. 580.20 N (C) 356 N 0.61 m -0.46 m- 267 N 178 Na b c d F 5 m 1.5 m 2.1 m 2.4 m 26 kN The structure above is subjected to a load at point C. Using the table below, determine the forces acting at the support pins and at pin B on member ABC. Forces: (Denote tension forces with positive values; compressive forces with negative values.) FAX = ? FAY = ?Q) For the shown truss, determine the stiffness matrix of member 1 (1 to 3) and member 2 (2 to 3)? Let A = 50X10 m² and E-210 GPa for all elements. 50 KN 30 m - 3 20 m 100 KN 30 m 40 m
- PROBLEM 3. The aluminum rod has a solid circular cross-section and supports the loading shown. Drawing the nomal force diagram and keeping the weight minimum, detemine the diameter of the rod at each segment AB, BC, CD and DE. Compare the weight of the rod if the magnesium alloy AZ91D was used as the rod material. 8 kN 4 kN 6 kN 2 kN -4 m -2 m- -2 mPROBLEM 3. The aluminum rod has a solid circular cross-section and supports the loading shown. Drawing the normal force diagram and keeping the weight minimum, determine the diameter of the rod at each segment AB, BC, CD and DE. Compare the weight of the rod if the magnesium alloy AZ91D was used as the rod material. 8 kN E 4 kN 6 kN 2 kN 4 m -2 m- 2m-2 m-Question 3: Given properties of the system are qr=15 kN/m, q=20KN/m, hi=30m, L=30m and L;=15m. If the force at column BD is only in the axial direction, calculate the sag at cable BC. (h=?) (There are hinges at both ends of the column BD.) 92 91 B A h2 C L2 D Good Luck...
- S ! 160-44 Fx= Required information In the structure shown, cable segment ED and member ECF are horizontal. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. By= Bx= Cx= lb 8 in. radius 21 in. 84 in. D E B W 才 G A 60 in. lb FW= 600 lb, determine the force supported by members ABCD, ECF, and GBF. (Include a minus sign if necessary.) The force supported by members ABCD, ECF, and GBF is as follows: Fy= Cy= F T 28 in. 80 in. L lb (Round the final answer to four decimal places.) lb (Round the final answer to four decimal places.) lb (Round the final answer to four decimal places.) lb (Round the final answer to four decimal places.)44) With the given information, calculate the compressive load at the right hip to solve for R. Given that wt. = 80 kg, distance of center of mass is 15cm from the hip joint center, and Fm is 5cm from the joint center.. FBW= FmThe fabricated steel members are being assembled into a truss on-site in the municipality of Alfonso, Cavite. The assembly started at the elevated ground at both sides and is to meet at the midspan. Nearing to the end of construction, it was found that the last member, CD, was fabricated too short by A D= 4.450 mm. The engineer then came up with an idea of applying a force P at joints C and D as shown to close the gap. P. с c' P 3 m -3 m- -3 m- 3 m- Use E = 200 GPa and cross-sectional area of members BC, DE, CD each 4500 mm2 while AC and DF being 6363.961 mm2 a) Determine the necessary force, P, to close the gap in kN. wwww b) If force P will then be removed after being assembled, calculate the force in CD in kN.
- A ladder with a uniform cross-section has a length of 4.80-foot long. It weighs W Ib. It is arranged such that one end is on the ground and the other against a vertical wall. The angle of triction at all contact surtaces is 20 degrees. What is the value of the coefficient of triction? What is f, in terms of variable Na? What is the normal force at A in terms of variable W? Answer in three decimal placesThe timber column has a length of 18 ft and is fixed connected at its ends. Use the NFPA formulas to determine the largest axial force P that it can support.TDE= TBE Required information The structure consists of pin-connected members. Take X = 4 kN. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. = 2m Determine the force supported by all members. Enter forces as positive numbers if the member is in tension and as negative numbers if the member is in compression. (Round the final answers to four decimal places.) The force supported by all members are as follows: TAB -7.0711 KN TAD= -5 KN TBD= 2.2361 kN KN |1m-1m-1m-1m4 2 kN 3 KN X 3 KN D TEF= TBF= 4.4721 TCF= TBC= -5.6569 kN KN E KN KN F C