PDF vdoc.pub_statics-and-strength-o X + File C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf C Google M Gmail Electronics, Cars, Fa... Pandora One - Liste... My TurboTax® - G... EasyBib: Free Biblio... GE News Because the critical stress is greater than the proportional limit, Euler's column formula does not apply. An allowable load based on the Euler formula would be meaningless. Homebanking milConnect - TAP PROBLEMS 17.1 Compute the minimum column length for which Euler's formula applies if the column is pinned at both ends and consists of a. A4 X 4 American Standard Timber made of Southern pine (Tables A.9 and A.10 of the Appen- dix) for which the proportional limit is 6100 psi. b. Nominal 4-in. standard weight pipe (Table A.8 of the Appendix) for a proportional limit of 33,000 psi. 17.2 A W14 X 74 steel column has a length of 30 ft. The steel has a modulus E = 29 x 103 ksi and a propor- tional limit σp = 33 ksi. Use a factor of safety of 2. Determine the critical or Euler load and the allowable load if (a) the ends are pinned, (b) one end is pinned and the other is fixed, and (c) both ends are fixed. For section properties, see Table A.3 of the Appendix. 17.3 A W360 x 110 rolled steel column has a length of 9 m. The steel has a modulus of E = 200 × 10° kPa and a proportional limit σp = 230 × 103 kPa. Use a factor of safety of 2. Determine the critical or Euler load and the allowable load if (a) the ends are pinned, (b) one end is pinned and the other is fixed, and (c) both ends are fixed. For section properties, see Table A.3 (SI Units) of the Appendix. 17.4 A boom consists of a 100-mm-diameter steel pipe AB and a cable BC as shown in the figure. The modulus of 432 CHAPTER SEVENTEEN B 3.6 m- A P PROB. 17.4 1.5 m 50 mm 100 mm 50 mm <-50 mm 50 mm 100 mm' PROB. 17.7 Type here to search steel E 200 × 10° kPa. Determine the capacity P of 17.5 TANGENT MODULUS A CAD ? 42°F 8:02 PM 4/2/2024 O + Homebanking milConnect - TAP C PDF vdoc.pub_statics-and-strength-ox Google Gmail + File C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf Electronics, Cars, Fa... Pandora One - Liste... My TurboTax® - G... Southern pine (Tables A.9 and A.10 of the Appen- dix) for which the proportional limit is 6100 psi. b. Nominal 4-in. standard weight pipe (Table A.8 of the Appendix) for a proportional limit of 33,000 psi. 17.2 A W14 X 74 steel column has a length of 30 ft. The steel has a modulus E = 29 x 103 ksi and a propor- tional limit σp=33 ksi. Use a factor of safety of 2. EasyBib: Free Biblio... GE News and a proportional limit σp = 230 x 10 kPa. Use a factor of safety of 2. Determine the critical or Euler load and the allowable load if (a) the ends are pinned, (b) one end is pinned and the other is fixed, and (c) both ends are fixed. For section properties, see Table A.3 (SI Units) of the Appendix. 17.4 A boom consists of a 100-mm-diameter steel pipe AB and a cable BC as shown in the figure. The modulus of Type here to search 432 CHAPTER SEVENTEEN C B -3.6 m A PROB. 17.4 1.5 m 50 mm 50 mm 100 mm 50 mm |||| -50 mm 100 mm' PROB. 17.7 steel E = 200 x 100 kPa. Determine the capacity P of the boom. Use the Euler formula with a factor of safety of 3. Neglect the weight of the pipe. Section properties of pipe: A = 2.05 x 103 mm² and the radius of gyra- tion r 38.4 mm. 17.5 A boom consists of a 4-in.-diameter steel pipe AB and a cable as shown. The modulus of steel E = 29 × 103 ksi. Determine the capacity P of the boom. Use the Euler formula with a factor of safety of 2.5. Neglect the weight of the pipe. Section properties of pipe: A = 3.174 in.² and the radius of gyration r = 1.51 in. 17.5 TANGENT MODULUS THEORY The range of the Euler column formula can be extended if we define a new modulus called the tangent modulus ET. The tangent modulus is the slope of the stress-strain curve at any point. It represents the instantaneous relationship between stress and strain for a particular value of stress. The stress-strain curve for a ductile material, together with the stress versus the tangent modulus Er curve for the same ma- terial, is shown in Fig. 17.7. The slope of the stress-strain curve is a constant up to the proportional limit. Therefore, the tangent modulus is also a constant-shown in the figure as ET, Above the proportional limit, the slope drops rapidly ne the etrace and strain incronco Thie ie chown in the fimiro A CAD ? 42°F 8:02 PM 4/2/2024 O + C PDF vdoc.pub_statics-and-strength-o X Google Gmail + File C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf Electronics, Cars, Fa... Pandora One - Liste... My TurboTax® - G... or pipe: A-2.05 x 10 mm and the radius of gyra- tion r 38.4 mm. 17.5 A boom consists of a 4-in.-diameter steel pipe AB and a cable as shown. The modulus of steel E = 29 × 10³ ksi. Determine the capacity P of the boom. Use the Euler formula with a factor of safety of 2.5. Neglect the weight of the pipe. Section properties of pipe: A = 3.174 in.² and the radius of gyration r = 1.51 in. EasyBib: Free Biblio... G= News Homebanking we define a new modulus called the tangent modulus ET. The tangent modulus is the slope of the stress-strain curve at any point. It represents the instantaneous relationship between stress and strain for a particular value of stress. The stress-strain curve for a ductile material, together with the stress versus the tangent modulus ET curve for the same ma- terial, is shown in Fig. 17.7. The slope of the stress-strain curve is a constant up to the proportional limit. Therefore, the tangent modulus is also a constant-shown in the figure as ET. Above the proportional limit, the slope drops rapidly as the stress and strain increase. This is shown in the figure for one value as Erg. Substituting the tangent modulus ET for the elastic modulus E in Eq. (17.5), we write milConnect - TAP Type here to search 9 ft σC= πΕτ (KL/r)² (17.6) B -12 ft PROB. 17.5 17.6 The cross section of an 18-ft-long column is shown in the figure. Determine the slenderness ratio if (a) the ends are pinned, and (b) one end is fixed and the other end is free. This equation is the tangent modulus or Engesser formula. Engesser first proposed the formula in 1889. The Euler and tangent modulus formulas, Eqs. (17.5) and (17.6), have been plotted for the same ductile material in Fig. 17.8. Because the tangent modulus and the elastic modulus are identical below the proportional limit, the two curves for critical stress versus the slenderness ratio coin- cide. Extensive test results indicate close agreement with the 0 VS. € 2 in. 4 in. 2 in. 2 in. 1/4 in. 2 2 in. PROB. 17.6 Stress, a Proportional limit = arctan ETo 17.7 The cross section of a 5.5-m-long column is shown in the figure. Determine the slenderness ratio if (a) the ends are pinned, and (b) one end is fixed and the other end is free. Op=arctan ETP ETQ ETP Strain and tangent modulus, e and ET FIGURE 17.7 0 vs. ET A CAD ? 42°F 8:03 PM 4/2/2024 O + C PDF vdoc.pub_statics-and-strength-o X Google Gmail + File C:/Users/baude/Downloads/Important%20Files/EGR%20254/vdoc.pub_statics-and-strength-of-materials.pdf Electronics, Cars, Fa... Pandora One - Liste... My TurboTax® - G... EasyBib: Free Biblio... GE News Homebanking milConnect - TAP Type here to search The allowable load P₂ =σA = 19.04 kip/in.2 (1.44 in. 2) = 27.4 kip Answer PROBLEMS 17.8 The bracket shown supports a force P. Assume that both members have their ends pinned in the plane of the bracket ABC and fixed in a perpendicular plane. The modulus E = 207 x 10 kN/m², and the yield stress σy = 375 x 103 kN/m². Determine the allowable load if the factor of safety is 2. Use Eq. (17.14) or (17.15). 17.9 In the mechanism shown, called a toggle joint, a small force can be used to exert a much larger force F. As- sume that both members have their ends pinned in the plane of the mechanism ABC and fixed in a perpen- dicular plane. The modulus E = 30 × 103 ksi, and the yield stress σy = 50 ksi. Determine the allowable load if the factor of safety is 2.5. Use Eq. (17.14) or (17.15). P 30° 20 mm Columns 437 -4 L152 x 152 x 12.7 Lacing 500 mm 60° 150 mm 150 mm 10 mm Cross section of both members PROB. 17.8 Cross section of both members 500 mm- PROB. 17.14 cross section shown. E = 200 x 106 kN/m², and σy = 250 x 10³ kN/m². Use AISC formulas-Eq. (17.10) or (17.11). For section properties, see Table A.6 of the Appendix. 17.15 Solve Prob. 17.14 if the column is aluminum alloy 6061-T6. Use Eq. (17.16), (17.17), or (17.18). For sec- tion properties, see Table A.6 of the Appendix. 17.16 Member BC of the pin-connected truss has a cross A CAD ? 42°F 8:03 PM 4/2/2024 O +/n INSTRUCTIONS 17.2, 4, 6 & 8 using Euler Column Formula Eqn. 17.6 & following homework format Need to provide typed solutions for this work