Study Guide. Given Formulae:

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1 E311 EXM 1 STUDY GUIDE Exam 1 Date: Thursday, September 21 st, 2017, 7 to 9:30pm This Guide Last Updated on ugust 21 st, 2017 Study Guide This study guide compliments the Past Exam Problems. Students should study both. nswers at the end. Given Formulae: Law of osines: b a Law of Sines: c Stress = E = G ending Flexural (normal) Stress M/EI = 1/ = -y/ =My/I M=EI I = ( /4)r 4 for a solid circular cross-section eam Shear Stress = VQ/It q = VQ/I q = F/s I = ( /4)r 4 for a solid, circular shape of radius r lso: = PL/E

2 Exam Format Exam designed for 2 hours. 2.5 hour time limit. You will be allowed to use your steel manual, a calculator, and pencils. Some questions will be purely conceptual in nature, while others will be computational. pproximately 30% of the exam will be multiple choice, short answer, true/false, etc and these will often be conceptual in nature, while approximately 70% will be fully worked problems. Recommended Problems From the Segui Textbook: hapter 2: 2.3, 2.5 hapter 4: 4.3-1, 4.3-3, 4.7-1, Note: Segui covers the use of the Stiffness Reduction Factor, but E311 will not cover this. overage Lesson Objectives from Lessons 1 to 9 and Lab 1. You are expected to be able to do Strength of Materials problems involving direct normal stresses, direct shear stresses, bending normal stresses, beam shear stresses, and the combination of axial and bending effects. Define strength and serviceability limit states. 1. (1 point) Situation: a structural engineering firm was sued because a building it designed shook so excessively in high winds that the exterior glass fell out, landing in the street. Is this an example of a strength limit state or a serviceability limit state, exceeded? 2. (1 point) Situation: a hospital sued a structural engineering firm because excessive vibrations in the building were inappropriate for an operating room. Is this an example of a strength limit state or a serviceability limit state, exceeded? 3. (1 point) Situation: a structural engineering firm was sued because an agricultural building roof it designed collapsed under a record snow loading (more than the region had ever experienced), killing 16 hogs. Is this an example of a strength limit state or a serviceability limit state, exceeded? 4. (1 point) TRUE or FLSE. strength limit state is only considered to have been reached if any parts of the structure (e.g., beams, columns, connections, braces, etc.) fracture. 5. (1 point) TRUE or FLSE. buckled column is considered an example of a Serviceability Limit State. Distinguish between strength limit states of material failure and instability. 6. (2 points) Using sketches and descriptions, distinguish between strength limit states of material failure and instability. Distinguish between member instability (member buckling) and overall structural instability 7. (1 point) TRUE or FLSE. If the loads P were increased until failure occurred, the structure below would fail because the entire structure is unstable. P P P P Indicates pin (or momentrelease) Indicates rigid (or momentresisting) connection. Define Structural nalysis and Steel Design. 8. (1 point) TRUE OR FLSE. Structural nalysis is the selection of structural members that have adequate strength and stiffness. Define Normal and Shear stress, list example occurrences, and state the applicable formulae. 9. (1 point) TRUE or FLSE. The maximum normal stress on this beam is equal to P/. Given: The beam shown has a cross-sectional area of.

3 P h rea Moment of Inertia I L/2 L/2 ross-section 10. (1 point) TRUE or FLSE. For previous beam, the maximum normal stress on this beam is equal to P/I. 11. (1 point) TRUE or FLSE. For previous beam, the maximum normal stress on this beam is equal to Mc/I, where M=(P/2)(L/2), c = h/2, and I is given. 12. (1 point) True or False. For the previous beam, the average shear stress at the left support is equal to P/(2). 13. (10 points) Determine the required pin diameter for the pin at point if it is subjected to double shear and has an allowable shear stress allow = 6 ksi. ompute internal normal force P, bending moment M, and shear force V, normal stress, and shear stress in determinate structures by cutting and applying equilibrium. 14. (10 points). The beam below has a 6 x 6 cross-section. Determine maximum compressive stress due to bending occurring anywhere in the beam and specify the location at which this occurs. Given: Point is supported by a pin, while Point is a roller support. 5 kips 5 kips 1 kip/ft 5 ft 12 ft 5 ft 15. (10 points). For the previous beam, determine the maximum magnitude of shearing stress that occurs anywhere in the beam and specify the location at which this occurs. 16. (1 point). TRUE or FLSE. For the previous beam, the internal moment at point places compression on the top-side of the beam. 17. (10 points). Determine the internal forces and moments at Point. Given: The beam has a fixed support at Point and is free at Point.

4 18. (1 points). TRUE or FLSE. For the previous beam, the internal moment at point places compression on the top-side of the beam. 19. (10 points) Determine the magnitude of the internal axial force at point F, specifying either compression or tension as well as the magnitude of the internal moment M at point F, specifying whether the moment causes compression on the top or the bottom of the member. Given: Member F is a continuous member that is connected to member DE by a pin at. There are pinned supports at and E. There is a uniformly-distributed loading of 1 kip/ft from to and a uniformly-distributed loading of 2 kips/ft from to D. 2 kip/ft 1 kip/ft F D 10 ft E 5 ft 5 ft 10 ft 20. (5 points) The steel I-beam (E=29000 ksi) below is made from three 8 x1 plates, resulting in a total depth of 10 and a moment of inertia I = 368 in 4. If the bending stress at the top of the flange is 2.9 ksi, determine the strain at the top of the web (point a ). 8 x 1 Flange a 8 x 1 Web 8 x 1 Flange I-EM ROSS- SETION SIDE VIEW OF ENT I-EM

5 21. (5 points) Determine the internal moment on the section of 10 x10 wooden beam below if the internal normal stress distribution that is shown can be resolved into two internal resultant forces R = 1 kip each. M =? 10 in. 10 in. ROSS-SETION OF WOODEN EM SIDE VIEW OF ENT EM R = 1kip R = 1kip 22. (20 points) Determine the resultant horizontal normal force on the flange of the Tee beam at section 1-1, located 2 feet to the right of the left support, as shown. Given: The Tee-beam is built from an 8 x1 flange and an 8 x1 web. It is subjected to a 1 kip/ft uniformly distributed load over its 10 foot simple span. 8 x 1 Flange 1 w = 1 kip/ft 8 x 1 Web 1 10 ft. TEE-EM ROSS- SETION 2 ft. SIDE VIEW OF TEE EM 23. (10 points) Determine the normal stress, due to flexure, at point. Given: the beam is simply-supported by a pinsupport at and a roller support at. The beam is center-point-loaded by a 100-kip load, as shown. The cross-section is 6 x12, as shown. 100 kips 50 inches 50 inches Side View of eam 10 in. 5 in. ompute the moment of inertia of a composite section (such as an I-shape). (with other objectives) ompute the extreme fiber stress for a member subjected to combined bending and axial effects. 24. (12 points) Determine the required dimension b of the beam cross section if the allowable bending stress allow = 1.40 ksi and the distributed load w =200 lb/ft. Given: ssume the support at acts as a pin and that the support at can be treated as a roller. 12 in. 6 in. eam ross Section

6 25. (17 points) Determine the magnitude of normal stress on the top of the beam at point F and specify whether the stress is compressive or tensile. Given: ll members have a cross-section that is a 6 x6 solid. Member F is a continuous member that is connected to member DE by a pin at. There are pinned supports at and E. There is a uniformly-distributed loading of 1 kip/ft from to and a uniformly-distributed loading of 2 kips/ft from to D. 2 kip/ft 1 kip/ft F D 10 ft E 26. (2 points) For the frame below, circle the correct answer: a. x = 0 and the axial force in is zero. b. x 0 and the axial force in is zero. c. x = 0 and the axial force in is non-zero. d. x 0 and the axial force in is non-zero. 5 ft 5 ft 10 ft 1 kip/ft D 15 ft x E E x y 20 ft 20 ft E y 27. (20 points) building frame is made with rigidly-connected W8x24 steel shapes, oriented as shown below (note the flange positions). The frame is supported by pinned foundations at points and E. Positions and D are rigid connections. There is an internal pin connection at point. The frame is subjected to a live load of 8 kips at pin. Determine the maximum compressive stress due to live load at a position that is immediately to the right of Point and indicate whether this maximum compressive stress occurs on the top or on the bottom side of beam.

7 8 kips D 15 ft. Orientation of W8x24 E 20 ft. 20 ft. onstruct shear and moment diagrams for beams, using the Strength of Materials sign convention. 28. (7 points) Write the function for the internal shear force V(x) and the internal bending moment M(x) of beam due to the concentrated moment at where the origin for x is at pinned support, directed toward roller support. Write these functions in terms of M, L and x. Given: oncentrated moment M at position. L M x 29. (15 points) Write the shear function V(x) and the moment function M(x) for the beam shown. It is subjected to a distributed load that increases linearly from 0 at to 1 kip/ft at 1 kip/ft 10 feet x 30. (13 points) Draw the moment diagram for the beam below. Label the magnitude and location of each local maximum moment. Given: Support is a roller, support is fixed, position is an internal pin. 1.2 kip/ft 2 kips onstruct moment diagrams for frames (plotting the diagram on the compression side of the members), computing and labeling min and max points. 31. (15 points) Draw the moment diagram for member D, only. Use the usual Strength of Materials onvention of plotting the diagram on the compression side of the members. Report all minimum and maximum values.

8 1 kip/ft D 15 ft E 20 ft 20 ft ompute shear stresses at any position within beams and frames. Illustrate the powerful influences of span length and structural depth on structural stiffness. 32. (5 points). The trussed girder shown was constructed by welding diagonal members to the top and bottom chords such that the distance from the centers of the top and bottom chords is 5. Under a centerpoint load of P, the given girder was found to deflect 1 at midspan. If the distance from the centers of the two chords was increased to 12, all other factors staying the same, determine the expected midspan deflection. P Illustrate the powerful influence of unbraced length on column buckling. 33. (3 points). olumns and have the same cross-section. olumn is 10 -long, while olumn is 6 -long. If olumn buckles elastically at an applied axial force P=100 kips, at what force does olumn buckle at, assuming it also buckles elastically? P=100 kips P=? 10 6 olumn olumn Estimate truss deflections using the moment of inertia of the top and bottom chords, applying beam deflection formulae. Identify real pinned-connection and rigid connections. 34. (2 points) Is the following beam-column connection simple (non-moment-resisting) or rigid (moment resisting) for strong-axis bending? Given: the wide-flange beam is connected to the column-face with a bolted angle. There is an unconnected space between the beam flange and the face of the column.

9 nswer (ircle or ) olted angle connection. Simple (non-moment-resisting). Rigid (moment resisting) ompute the degree of static indeterminacy (DOI) for 2D frames and beams using fundamental principals (Statics), ompute the degree of static indeterminacy (DOI) for 2D frames and beams using the loop method, and/or ompute the degree of static indeterminacy (DOI) for 2D frames and trusses. Determine the degree of indeterminacy of the structures below (3 points each) 35. ll members pin-ended, as shown

10 Show why the structure below is indeterminate, but unstable.

11 Draw the line diagrams (FD s) for a fill beams, girders, columns, and 1 -wide slab strips for a simple (pin-connected) framed building structure, given the dead weights of the structure, Draw moment diagrams from line diagrams, ompute the bending stresses in a beam, given the internal moment and the section properties. onsider the building below for the following problems. ll floors, including the roof consist of 6 thick concrete slabs (concrete unit weight=150 lb/ft 3 ), uniformly-spaced W16x26 fill beams, W18x35 girders, and W12x45 columns. Girders and fills are in strong-axis bending. Slab-strips are considered simplysupported. typical floor framing plan and 3D image are shown. 1 D Roof 3 rd Flr 2 nd Flr Ground D (2 points) TRUE or FLSE. ll of the interior fill beams on the 2 nd floor between column line and have the same maximum bending moment. 47. (2 points) TRUE or FLSE. Interior fill beams on the 2 nd floor between column line and have greater maximum bending moment then interior fill beams on the 3 RD floor between column line and. 48. (2 points) TRUE or FLSE. The column at grid location 4 resists a greater axial force at the ground than at a position just above the 3 rd floor. 49. (10 points) Draw the line diagram for fill beam (see plan view), determine the maximum shear force, the maximum bending moment, and the maximum normal stress due to bending.

12 50. (12 points) Draw the line diagram for point-loaded girder (see plan view), determine the maximum shear force, the maximum bending moment, and the maximum normal stress due to bending. 51. (12 points) Determine the column force for the column at grid location D4 at the foundation level. 52. (25 points). Problem: Determine whether the column at grid location 3 is adequate using the SD method, between the ground and 2 nd floor, considering all load combinations. Given: The 12 -long column at grid location 3 is an 992 (F y =50 ksi) W10x39. The building is a stable simple-braced frame where columns are considered to be pinned at floor levels (bracing and floor diaphragms omitted for clarity). olumns are all 12 -long. Dead Loads 2 nd floor dead load : consists of a 4.5 thick concrete slab (unit weight of concrete = 150 pcf), supported by W16x26 fill beams and W18x35 girders. The roof dead load: consists of a built-up roof deck that weighs 14 psf, supported by W14x22 fill beams and W16x36 girders. The column weight should be accounted for. Live and Snow Loads: 2 nd floor live load is 80 psf. Live load reduction factor is applicable. Roof snow load is 100 psf (building is located in Northern Manitoba) Roof live load L r = 20 psf (this accounts for the loads present during roof repair, etc.) No Other Loads (e.g., this column does not participate in the lateral system, so it takes no wind or earthquake) Roof 2 nd Flr Ground Flr (25 points). Problem: Referring to the previous building, determine whether the column at grid location 4 between the ground and the 2 nd floor is adequate using the LRFD method, considering all load combinations (refer to previous building) Given: The 12 -long column is an 992 W8x31. The building is a stable simple-braced frame (bracing omitted for clarity). ll loads are the same as previous. 54. (15 points) Problem: Referring to the previous building, determine the LRFD moment M u that the 2 nd floor girder on column line 2, between D and E must be designed for. 55. (15 points). Problem: referring to the previous building, determine the SD shear force V that the roof fill beam on line D, between 1 and 2 must be designed for. 56. (12 points) 20 -long 992 W16x26 shape is embedded in concrete at its base, but is free at its top, supporting a water tank. What is the maximum load that may be placed in the tank at the top of the column, without actually causing D 30 E

13 buckling? Use theoretical K-values. Tank Tank (3 points) True or False. column with F y =50ksi and a controlling slenderness ratio KL/r=150 will buckle elastically, when loaded to failure. 58. (3 points) True or False. column with F y =50ksi and a controlling slenderness ratio KL/r=50 will buckle elastically, when loaded to failure. 59. (2 points) True or False. The Euler expression is a good predictor (within 15% most of the time) of the column buckling stress when the column is very slender. 60. (2 points). True or False. The SD method will generally require larger member sizes than the LRFD method, if the applied live loads are very high, relative to the applied dead loads. 61. (2 points). True or False. The SD and LRFD methods will always result in the same member sizes 62. (2 points) What does LRFD stand for? 63. (2 points) Which design method, LRFD or SD, uses two different safety factors, accounting for both load uncertainty and resistance uncertainty? 64. (2 points) What is the standard SD safety factor for a ductile failure? 65. (2 points) What is the standard SD safety factor for a brittle failure? 66. (3 points) In LRFD, why is there (typically) a higher load factor for live load than for dead load? 67. (3 points) In LRFD, why is there a higher factor for buckling or yielding than for fracture? 68. (2 points) TRUE or FLSE. The SD method will always lead to more conservative (larger, safer) member selection than LRFD. 69. (2 points) TRUE or FLSE. The LRFD method will lead to more conservative (larger) member selection than SD when live loads are relatively high and dead loads are relatively low. 70. (3 points) haracterize or describe the kind of event (i.e., give a name to this event) that is represented by the LRFD load combination U=1.2D + 1.6S + 0.5L 71. (3 points) For the LRFD load combination U=1.2D + 1.6S + 0.5L, explain why the load factor on live load less than 1? 72. (4 points) Fill in the blank, using the words in the box on the right Every in a structure must be able to sustain all effects (forces, Words moments, etc.) of each. Each load combination represents a particular kind of, which may be critical for some members of the structure, but not roofbeam event all. For example, a may be most vulnerable to the U=1.2D + 1.6S + 0.5L combo because snow is weighted high. Meanwhile, of the building are dead and live load combination more likely to sustain maximum bending due to the U=1.2D + 1.6L + 0.5S combo because loads are weighted heavily. floor beams member 73. (2 points) TRUE or FLSE. The LRFD design buckling load P n is larger than the SD allowable buckling load P n / P allow because the LRFD method is less conservative. 74. (2 points) TRUE or FLSE. In both LRFD and SD, the computation of the nominal column buckling strength P n is the same procedure with the same result. 75. (2 points) TRUE or FLSE. In both LRFD and SD, the nominal column buckling strength P n = F cr, where F cr is the buckling stress and is the cross-sectional area. 76. (2 points) TRUE or FLSE. The SD method relies on the Euler expression to compute the buckling stress, while the LRFD method relies on two different expressions to compute the buckling stress. 77. (2 points). TRUE or FLSE. The frame shown is considered to be braced because the beams are infinitely stiff and the beam-column connections are rigid.

14 Infinite EI Infinite EI 78. (2 points) TRUE or FLSE. K is theoretically always less than or equal to one if the frame is braced. 79. (2 points) TRUE or FLSE. K is always greater than or equal to one if the frame is not braced. 80. (2 points) TRUE or FLSE. For the rigid frame shown below, the K value for the columns is less than one. 81. (2 points) What is the K for the column in the frame below (all connections are simple): 82. (2 points) TRUE or FLSE. For the previous frame, column is considered braced, but column is considered unbraced. 83. (2 points) TRUE or FLSE. It is possible for a compression member to have rotational restraint at one end, but not at the other end. 84. (2 points) TRUE or FLSE. In a rigid building frame, columns receive greater rotational restraint when the beams that they are rigidly connected to have greater flexural stiffness. 85. (2 points) TRUE or FLSE. In a simple-braced building, columns receive greater rotational restraint when the beams that they are pin-connected to are larger. 86. (2 points) What is the theoretical effective length factor K for olumn in the frame below? Given: is a pinned foundation, is a rigid beam-column connection, is a pinned connection, D is a pinned foundation. eam is an ordinary beam (neither very large nor very small). D ircle the best answer: ). K = 1 ). 1 K 2 ). K 1 D). K 1

15 87. (2 points) Estimate the effective length factor K (circle one answer) for the columns shown, assuming that they are rigidly connected to a beam of infinite stiffness. There is no bracing present. Infinitely Stiff eam olumn olumn a). K = 0 b) 0.5 < K < 1.0 c) K = 1 d) 1.0 < K < 2.0 e) K = 2 f) K > (12 points) Determine the SD allowable load P n / if the W-shaped column is an 992 (F y =50 ksi) W12x40. Given: The column is 18 feet long, but has a mid-height brace that prevents movement along one axis, as shown below. P n / =? P n / =? (12 points) Determine the LRFD design strength P n if the column is an 36 (F y = 36 ksi) S12x50. Given: The column is 12 feet long, but has 5 equally-spaced intermediate braces that prevent movement along one axis, as shown, above. Pn =? Pn =? 12 races 90. (18 points) ompute the SD allowable axial load P n / for the column shown. It is made out of Grade 50 (F y =50 ksi) material; a W12x65 section is strengthened by the addition of ¾ thick plates, as shown. It is fixed on one end and pinned on the other, about both axes. Use the Recommended design value when ideal conditions are approximated for the K-value of the pinned-fixed condition.

16 91. (18 points) ompute the LRFD design strength P n for the previous. 92. (22 points) Use SD to assess the adequacy of the W10x60 column at location 3, between the foundation and the second floor. Given: ll beams, columns, and braces are 992 (F y =50 ksi). ll connections are simple (pinned connections), as are the connections to the foundation. ll bays are 40 and all story heights are 12. s shown in the isometric, the building is braced with diagonal bracing at the corners of the building. This bracing effectively braces every column in the building, at its top and bottom. Loads: Roof Loads: 50 psf dead load (includes all roof dead load slab, beams, etc), 50 psf snow load, 20 psf roof live load 2 nd Floor Loads: 100 psf dead load (includes all floor dead load slab, beams, etc), 100 psf live load. Live load reduction factor is applicable. North 1 North D Roof 12 2 nd Flr 12 Foundation 1 race, typ. 2 40, typ , typ. 4 D Isometric View (fill beams omitted, for clarity) Plan View Framing Plan

17 nswers 1. (NSWER: Serviceability) 2. (NSWER: Serviceability) 3. (NSWER: Strength) 4. (NSWER: False) 5. (NSWER: False) (NSWER: False) 8. (NSWER: False) 9. (NSWER: False) 10. (NSWER: False) 11. (NSWER: True) 12. (NSWER: True) (NSWER: 8.33 ksi at supports, bottom fiber) 15. (NSWER: 0.25 ksi at supports, neutral axis) 16. (NSWER: False) 17. (NSWER: V=288 lbs, M = 1152 ft-lb) 18. (NSWER: False) 19. (NSWER: P=7.5 kips, M= 25 ft-kips, on bottom) 20. (NSWER: Strain is in/in) 21. (NSWER: M=6.67 kip-in) 22. (NSWER: Resultant on flange is 13.9 kips) 23. (NSWER: Normal stress is ksi) 24. (NSWER: b=4.02 ) 25. (NSWER: Max normal stress is ksi (tension) ) 26. (NSWER: d) 27. (NSWER: Max. compressive stress is 46.8 ksi on bottom) 28. (NSWER: V(x) = -M/L, M(x) = -Mx/L) 29. (NSWER: V(x) = x^2, M(x)=1.66x x^3) 30. (NSWER: Neg: Pos: 11.1 at 6.93 left of. ) (NSWER: ) 33. (NSWER: 278 kips) 34. (NSWER: ) 35. (NSWER: 1) 36. (NSWER: 0) 37. (NSWER: 1) 38. (NSWER: 6) 39. (NSWER: 0) 40. (NSWER: 2) 41. (NSWER: 5) 42. (NSWER: 22) 43. (NSWER: 0) 44. (NSWER: 0) 45. (NSWER: Put horz force on beam: Sum of forces in x is not zero. ) (NSWER: FLSE) 48. (NSWER: TRUE) 49. (NSWER:V=6.02kips, M=45.1 kip-ft, =14.1 ksi) 50. (NSWER: V=28.1kips, M=252 kip-ft, =52.5 ksi) 51. (NSWER:55.8 kips) 52. olumn is OK. ontrolling applied load is 166.6kips (ombination 4). llowable load is 234 kips 53. olumn is OK. ontrolling applied load is 45.5kips (ombination 4). llowable load is 188 kips 54. Factored Moment Mu=287.7 kip-ft. 55. D+S combination controls. V=13.16 kips 56. Pn=10.49 kips

18 57. TRUE 58. FLSE 59. TRUE 60. FLSE 61. FLSE 62. Load and Resistance Factor Design 63. LRFD 64. 5/ Live load is more variable than dead load 67. Fracture is a higher risk 68. FLSE. It depends on the dead-live ratio 69. TRUE. LRFD fears and respects the variability of live loads 70. lizzard 71. It is unlikely that the largest live load event coincides with a blizzard. 72. Member, load combination, event, roof beam, floor beams, dead+live 73. FLSE 74. TRUE 75. TRUE 76. FLSE 77. FLSE 78. TRUE 79. TRUE 80. FLSE FLSE 83. TRUE 84. TRUE 85. FLSE E kips kips 90. Using parallel axis theorem, Ix=756in 4, Iy=915in 4. K=0.8. Pn/ =1001 kips kips 92. ombination 4 controls. pplied load P=353.9 kips. Pn/ =421.3 kips. dequate.