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QuizBKey - Seat Name E VWEO 0 2 C 1 General examination...

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Unformatted text preview: Seat# Name E VWEO ) 0 2 C 1 General examination rules: 1) Do not put your completed work on a desk or on the floor next to you or anywhere it can be seen by others. If any part of your work can be seen by others it will be confiscated and you will not be permitted to rework those problems. Place it face down on our desk under our existin work.. 2) Please remove your hat. If it is part of your head, turn it around backwards. 3) Please believe this: if your work not legible, or if I cannot follow your logic at a glance, or if you use a #9 nail for a pencil with 2 point font, it will receive no credit. This paper will be written to acceptable engineering standards or it will receive no credit. 4) Please read the problem very carefully. Giving me the correct answer to something I didn’t ask is worth zero pomts. 5) Do not unstaple this exam. You are permitted to work on the back of any page. Simply so state, so we won’t miss any of your work. 6) Extra sheets are included in the exam if needed. Ethical Standards: Upon accepting admission to Texas A&M University, a student immediately assumes a commitment to uphold the Honor Code, to accept responsibility for learning, and to follow the philosophy and rules of the Honor System. Students will be required to state their commitment on examinations, research papers, and other academic work. Ignorance of the rules does not exclude any member of the TAMU community from the requirements or the processes of the Honor System. "On my honor, as an Aggie, I have neither given nor received unauthorized aid on this exam." Signature of student Please do not open this exam until you are told to do so . Miscellaneous equations for circles: Area = pi*r"2 Moment of inertia = pi*r"4/4 Polar moment of inertia = pi*r/‘4/2 726 APPENMX c: PROPEH'HES or SELECTED STRUCTURAL-STEEL SHAPES Table (1.1 Continued 47341.23 41.4..»441M0wc7 7 74746 W. 4 7 173%; .. 24: 4 4 .494 Web Flange Depth to Width Thickness— 2 . 6441440.. -nmmun— W10x 45 13.3 10.10 10% 0.350 3 {358 6 0.620 3 .246 . . >< 39 11.5 9.92 9%; .m 1% % 7.966 8 0.530 6 209 45.0 11.3 1.96 x 32 9.71 . 9% 0.290 77".; 736 7.960 8 0.435 T"; 170 36.6 9.20 1.94 W10 x 30 6.34 10.47 10% 0.300 3% 735 6.210 6% 0.610 6 170 16.7 3.75 1.37 x 26 7.61 10.33 10% 0.260 % 4 6.770 5% 0.440 1% 144 14.1 4.89 1.36 x 22 6.49 10.17 10% 0.240 9 3 5.750 5; 0.360 g 118 11.4 3.97 1.33 W10 x 19 6.62 10.24 10% 0.260 9 ~35 4.020 4 0.395 g 96.3 429 2.14 0.374 x 17 4.99 10.11 10% 0.240 9 3 4.010 4 0.330 1% 31.9 3.56 1.73 0.644 x 15 4.41 9.99 10 0.230 § 4 4.000 4 0.270 9 68.9 2.69 1.45 0.610 x 12 3.54 9.37 9% 0.190 3% g 3.960 4 0210 % 538 2.18 1.10 0.765 W6 x 67 19.7 9.00 9 0.570 % % 6.230 3% 0.935 {—2 272 68.6 21.4 2.12 x 56 17.1 2.75 3% 0.510 9 i 3.220 6; 0310 {—3- 226 76.1 16.3 2.10 x 48 14.1 8.50 8% 0.400 3 4% 3.110 8% 0.665 4% 164 609 15.0 2.06 x 40 11.7 6.26 6% 0.360 3 «1% 3.070 8% 0.660 4% 146 49.1 12.2 2.04 x 36 10.3 6.12 8% 0.310 1—5.. 1% 6.020 8 0.496 3 127 ,- . 42.6 10.6 2.03 X 31 9.13 6.00 8 0.285 1% 1% 7.995 8 0.435 74 110 27.6 3.47 37.1 927 2.02 W8 x 26 8.25 6.06 3 0.265 3% 1% 6.535 6% 0.465 175 93.0 24.3 3.46 21.7 6.63 1.62 x 24 7.06 7.93 7% 0.245 i- ; 6.495 6% 0.400 g 62.6 20.9 3.42 18.3 5.63 1.61 W6 x 21 6.16 328 3% 0.250 9 g 6.270 5% 0.400 g 75.3 12.2 3.49 9.77 3.71 1.26 x 13 6.26 6.14 6}; 0.230 9 4 5.250 5; 0330 15—4 61.9 16.2 3.43 7.97 3.04 1.23 W8 x 1.5 4.44 3.11 851; 0.245 3 g 4.016 4 0.316 6 430 11.3 3.29 3.41 1.70 0.876 x 13 3.84 7.99 8 0.230 37 6 4.000 4 0.256 1 396 991 3.21 2.73 137 0.343 x 10 2.96 7.69 73 0.170 1% 3 3.940 4 0.206 3% 30.8 7.61 3.22 2.09 1.06 0.341 Problem 1a) Draw a shear and moment diagram for the beam shown. Use w = 10 kN/m, P = 120 kN, M = '600 kNm, L = 2 m. = . ; fiN RlvrimzmgéfiMew lp :. IZO 2: l 65 Mlhllivivliffiwm MM :(9 i \\\ \\\\ l , M “#33“ . - g L L [ 2L L L 1’ R1575fi“ R2:t(=54w (8 3 EM R. :0 \ =QOWWY/ZMY‘W3 + 120(2m\ +(oOOM/m‘ 32 m R2 =— Masha 2M R11 0= (\OK flV‘\ (12M\<4M\ +on “News $00 - R (8‘4 ; A F .ZOVpJJA/YA : AZ: Jrqo .A3=—4ZO A4 =*Z7O 'ZEOhM/m 350 Problem 2: ) Use the stress transformation equations to determine the maximum principal shear and normal stresses, in or out of plane, due to the state of stress shown. Magnitudes of the stresses shown are 0'X =30 ksi, 6y: 20 ksi, ”EX X), — -10 ksi. Directions of the applied stresses are as shown. 20 Lw —i 30 T Gin/Va: Egfl’:w CY“; 3:7;Hifz Problem 3) Using the beam tables given on the second page of this exam, design the lightest simply supported beam pinned on each end to carry a uniform load of 2 k/ft which includes the weight of the beam. Use a maximum allowable bending stress equal to 24 ksi, and a maximum shearing stress of 18 ksi. The beam is 18 feet long. . 2 WFT‘ ’ 61¢ S (,0 4" Twin/low? r—‘W TWW 8x48 Problem 4) Two plates are bolted together as shown with 1% inch diameter bolts. The holes in the plate were drilled 1/16” larger than the bolt so that the bolts would fit when the plates were assembled. The top plate is 1%” thick by 8” wide. The bottom plate is 2” thick by 10” wide. 150 kip tensile loads are applied as shown. 3) Determine the maximum shear stress in the bolts, b) the maximum tensile stress in the plates, and c) the maximum bearing stress in the plates. I 00000 150K 00000 = E-—->150K % OOOOO Problem 5) Determine the maximum state of stress in a tubular steel pipe loaded as shown. No principal stresses are requested. Just a final “postage stamp” placed at the point of critical stress on the pipe, with the final stresses (a single sigma x, a single sigma y, and a tau xy) shown on the stamp. The pipe is loaded with P = 2 kips, F = 40 kips, T = 10 kip ft, internal pressure = 500 psi. The pipe is 4 feet long with an ID = 8” and an OD = 8.4”. Extra blank sheet on next page. MTWM + ANN— / Y“ Z. G” = + £‘ _ :0 time— ZC J2: = 9'92" 6&0 STRESS 2+ fix: C 73' R"3/2 @R 55URE a; 1:9 Stresses at the wall for a cantilever tube with axial force, concentrated load at end of pipe, twisted, and internal pres Problem 1 Lcantilever = 4 feet OD pipe = 8.4 inches Tension = positive stress ID pipe = 8 inches Top Bottom Bending force on end = 2 kips Bending stress x = 9.31 -9.31 ksi Axial force on end = 40 kips Axial stress x = -7.76 -7.76 ksi Twisting force on end = 10 kip feet Torsional stress shear: 5.82 5.82 ksi pressure = 500 psi Pressure stress x = 5.25 5.25 ksi Pressure stressy: 10.50 10.50 ksi Area: 5.15221 inA2 = 86.6601 inA4 Final stresses: l = 43.3301 in/\4 Sigma x = 6.79 ~11.82 ksi t = 0.2 Sigma y = 5.25 5.25 ksi 9 Tau xy = 5.82 5.82 ksi Problem 2? Lcantilever = 6 feet OD pipe = 10.2 inches Tension = positive stress ID pipe = 10 inches Top Bottom Bending force on end = 4 kips Bending stress x = 36.30 -36.30 ksi Axial force on end = 20 kips Axial stress x = -6.30 -6.30 ksi Twisting force on end = 30 kip feet Torsional stress shear = 22.69 22.69 ksi pressure = 500 psi Pressure stress x = 12.75 12.75 ksi Pressure stress y = 25.50 25.50 ksi Area = 3.17301 in/‘2 = 80.9275 in/‘4 Final stresses: = 40.4638 in/‘4 Sigma x = 42.75 ,-29.85 ksi = 0.1 Sigma y = 12.75 12.75 ksi Tau xy= 22.69 22.69 ksi _fi“ 7777/7777 Seat # Name 2 General examination rules: 1) Do not put your completed work on a desk or on the floor next to you or anywhere it can be seen by others. If any part of your work can be seen by others it will be confiscated and you will not be permitted to rework those problems. Place it face down on your desk under your existing work. 2) Please remove your hat. If it is part of your head, turn it around backwards. 3) Please believe this: if your work not legible, or if I cannot follow your logic at a glance, or if you use a #9 nail for a pencil with 2 point font, it will receive no credit. This paper will be written to acceptable engineering standards or it will receive no credit. 4) Please read the problem very carefully. Giving me the correct answer to something I didn’t ask is worth zero points. 5) Do not unstaple this exam. You are permitted to work on the back of any page. Simply so state, so we won’t miss any of your work. 6) Extra blank pages are included in the exam if you need them. Ethical Standards: Upon accepting admission to Texas A&M University, a student immediately assumes a commitment to uphold the Honor Code, to accept responsibility for learning, and to follow the philosophy and rules of the Honor System. Students will be required to state their commitment on examinations, research papers, and other academic work. Ignorance of the rules does not exclude any member of the TAMU community from the requirements or the processes of the Honor System. "On my honor, as an Aggie, I have neither given nor received unauthorized aid on this exam." Signature of student Please do not open this exam until you are told to do so . Miscellaneous equations for circles: Area = pi*r"2 Moment of inertia = pi*r"4/4 Polar moment of inertia = pi*r"4/2 726 APPENDlX c: 78098717129 or 62156750 STRUCTURAL-STEEL SHAPES . Tabla 0.1 Comment W49“ '49.,“ ”932539-4143. 061462;..3'9174 0%. 7. by W”; ; 24518340 .19 "5.4424 A .sAjAw .. 4141231 111191.211; ’z- 11.1%; v; ‘a 27'3"» Flange Eiasfic Properties H I r And—XX Axis YY Thickness _.., Width Thickness— nun—- “munn—mm- in Designation - max 45 13.3 10.10105 0.350 g 13 6020 8 0620 E; 49.1 4.32 13.3 2.01 x 39 11.6 9.92 9; 0.315 355 1% 7.985 8 0.530 4 209 42.1 4.27 235.0 113 1.98 x 32 9.71 9.73 9% 0.290 3% 4% 7.960 8 0.435 3% 170 35.0 4.19 36.6 9.20 1.94 W10 x 30 8.84 10.47 10% 0.300 3% 1% 5.310 5% 0.510 4 170 $ 4.33 16.7 5.75 1.37 x 26 7.61 10.33 10% 0.260 % 2 5.770 5% 0.440 {g 144 7.9 4.35 14.1 4.89 1.36 x 22 6.49 10.17 10% 0.240 % ~§ 5.750 5; 0.360 g 118 23.2 4.27 11.4 3.97 1.33 W10 x 19 5.62 10.24 10% 0.250 9 3g 4.020 4 0.395 9 96.3 18.8 4.14 4.29 2.14 0.874 x 17 4.99 10.11 10% 0.240 4 9 4.010 4 0.330 1—5;. 819 16.2 4.05 356 1.78 0.844 x 15 4.41 9.99 10 0.230 3 4 4.000 4 0.270 § 68.9 13.8 3.95 2.89 1.45 0.810 x 12 3.54 9.87 9% 0.190 736 g 3.960 4 0210 1% 53.8 10.9 3.90 2.18 1.10 0785 WS x 67 19.7 9.00 9 0.570 % % 8.280 8% 0.935 {92 272 60.4 3.72 88.6 21.4 2.12 x 58 17.1 8.75 8% 0.610 3 9 8.220 8;- 0.810 9% 228 52.0 3.65 751 18.3 2.10 x 48 14.1 8.50 8% 0.400 9 3%,.» 8.110 8%, 0685 44 184 43.3 3.61 609 15.0 2.08 x 40 11.7 8.25 8% 0.360 2 75;; 8.070 8% 0560 9% 146 $33.53 491 12.2 2.04 x 35 10.3 8.12 8% 0.310 3% 1% 8.020 8 0.495 4 127 31.2 3.51 42.6 10.6 2.08 x 31 9.13 8.00 8 0.285 {’5 135 7.995 8 0.435 9 110 27.5 3.47 87.1 927 2.02 W8 x 28 8.25 8.06 8 0.285 3% 1%- 6.535 64 0.465 T71 98.0 24.3 3.45 21.7 6.63 1.62 x 24 7.08 7.93 7% 0.245 1 g- 6.495 6% 0.400 9 82.8 20.9 3.42 18.3 5.63 1.61 W8 x 21 6.16 8.28 8% 0.250 % ;— 5270 5% 0.400 g 75.3 18.2 8.49 9.77 3.71126 x 18 6.26 8.14 8% 0.230 .1; 9 5250 5; 0330 % 61.9 15.2 3.43 7.97 3.04 1.28 W8 x 15 4.44 8.11 3% 0.245 9 i 4.016 4 0.316 1% 48.0 11.8 3.29 3.41 1.70 0.876 x 13 3.34 7.99 8 0.230 9 .1. 4.000 4 0.255 7 39.6 9.91 3.21 2.73 1.37 0.843 x 10 2.96 7.89 7% 0.170 £1 4 3.940 4 0.205 1% 30.8 7.81 3.22 2.09 1.06 0.841 Problem 1) Use Mohr’s circle to determine the maximum prizcipal shear and normal stresses, in or out of plane, due to the state of stress shown. Magnitudes of the str sses shown are 0x = 40 ksi, 0y = 30 ksi, Txy = 10 ksi. Directions of the applied stresses are as shown. Means CIRCLE Problem 2) Draw a shear and moment diagram for the beam shown. Use M = 100 kNm, L = 1 In, W = 5 kN/m, P = 60 kN. R1 is given as 55 kN and R2 = 35 kN. Extra blank page on next sheet. Problem 3) For the two plates bolted together as shown, determine a) the maximum tensile stress in the plates, b) the maximum bearing stress in the plates, and c) the maximum shear stress in the bolts. Q; ch diameter bolts are used and the holes in the plate were drilled 1/8” larger than the bolt so that olts would fit when the plates were assembled. The top plate is 1%” thick by 10” wide. The bottom plate is 2” thick by 12” wide. 150 kip tensile loads are applied as shown. Problem 4) For the tubular steel pipe loaded as shown, determine the maximum state of stress. N0 principal stresses are requested. Just a final “postage stamp” placed at the point of critical stress on the pipe, with the final stresses (a single sigma X, a single sigma y, and a tau xy) shown on the stamp. The pipe is loaded with P = 4 kips, F = 20 kips, T = 30 kip ft, internal pressure = 500 psi. The pipe is 6 feet long with an ID = 10.2” and an OD = 10”. Extra blank page on next sheet. // / 77/ Stresses at the wall for a cantilever tube with axial force, concentrated load at end of pipe, twisted, and internal pres Problem 1 Lcantilever = 4 feet OD pipe = 8.4 inches Tension = positive stress ID pipe = 8 inches Top Bottom Bending force on end = 2 kips Bending stress x = 9.31 -9.31 ksi Axial force on end = 40 kips Axial stress x = -7.76 -7.76 ksi Twisting force on end = 10 kip feet Torsional stress shear: 5.82 5.82 ksi pressure = 500 psi Pressure stress x = 5.25 5.25 ksi Pressure stress y = 10.50 10.50 ksi Area = 5.15221 inA2 J = 86.6601 in/‘4 Final stresses: l = 43.3301 inA4 Sigma x = 6.79 -11.82 ksi t = 0.2 Sigma y = 5.25 5.25 ksi Tau xy = 5.82 5.82 ksi Problem 2 Lcantilever = 6 feet OD pipe = 10.2 inches Tension = positive stress lD pipe = 10 inches Top Bottom Bending force on end = 4 kips Bending stress x = 36.30 -36.30 ksi Axial force on end = 20 kips Axial stress x = -6.30 -6.30 ksi Twisting force on end = 30 kip feet Torsional stress shear: 22.69 22.69 ksi pressure = 500 psi Pressure stress x = 12.75 12.75 ksi Pressure stressy: 25.50 25.50 ksi Area: 3.17301 inA2 = 80.9275 in/\4 Final stresses: = 40.4638 in/‘4 Sigma x = 42.75 .-29.85 ksi = 0.1 Sigma y = 12.75 12.75 ksi Tau xy= 22.69 22.69 ksi 7777777777 Problem 5) Using a maximum allowable bending stress equal to 20 ksi, and a maximum shearing stress of 16 ksi, use the beam tables given on the second page of this exam to design the lightest simply supported beam pinned on each end to carry a uniform load of 3 k/ft which includes the weight of the beam. The beam is 12 feet lon . ...
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