TAILIEUCHUNG - Theory and Problems of Strength of Materials Part 3

Tham khảo tài liệu 'theory and problems of strength of materials part 3', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 56 STATICALLY INDETERMINATE FORCE SYSTEMS TENSION AND COMPRESSION CHAP 2 The ultimate load thus corresponds to the situation when Ff FỊ Ơ PA and this load is found from Ơ as Ơ PA I 2cos J This limit load should be divided by some safety factor to obtain a working load. . Suppose the three-bar system of Problem is to withstand a load p - 200 kN. Compare the bar weights required if the design is based upon u the peak stress just reaching the yield point and i ultimate load analysis. Assume that all bars are of identical cross section that Ớ - 45 and take the yield point of the material to be 250 MPa. ú According to the elastic theory of Problem the force in the vertical bar becomes 2P F _ 7 - 117 kN 2 f V2 If the stress in that bar is equal to the yield point we have a required cross-sectional area of F Aị t p. Hence 117 X 10 4 250 or A 468 mm2 6 If the ultimate load analysis of Problem is employed the stresses in all three bars are equal to the yield point and from . of Problem we find a cross-sectional area of 200 X 103 250441 2 or A2 331 mm2 Ultimate load analysis thus implies a 29 percent saving in cross-sectional area and the same weight saving. . The frame shown in Fig. 2-34 consists of three pinned end bars AD HD and CD. The bars are of identical material and cross section and the ultimate load-carrying capacity of each is 30 kN. Determine the ultimate vertical load p that may be applied to the system at point D. Let us assume that bars BD and CD have reached yield. Examination of a free-body diagram for the node D as shown in Fig. 2-35 leads to 2 F 30 sin 35 - PAI sin 70 - 0 PAl - kN CHAR 2 STATICALLY INDETERMINATE FORCE SYSTEMS TENSION AND COMPRESSION 57 Thus bar AD does not yield since the bar force for equilibrium is less than the 30 kN required for yield. Summing vertically for equilibrium we have IF s -Pu 30 30cos35c 0 p kN . A system composed of a rigid horizontal member AB supported by four bars .

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