EXAM1-Solution-Fall-2023-10-23

10/23/2023 1 Online Distributed Midterm Exam Kurt McMullin CE164 – Fall 2023 A 15 ft The only load on the beam is a Dead Load which is the three point loads of 10 kips each. The two point loads are halfway between Point A and Point B and halfway between Point B and Point C. There is no Live Load on the frame. Determinant Continuous Beam P 15 ft B C The internal hinges are 3 feet from Points B and C Points A and B are pin supports. Point C is a fixed support. Calculate three things. The Reaction at B. The Shear 1’ to the right of B. The Moment at B. P P Engr Fundamentals RB = (1/15){10*7.5 + 10*15 + 5*18} = 21 kips VB = 5 kips MB = 5*3 = 15 kip-ft.

10/23/2023 2 A 15 ft The only load on the beam is a Dead Load which is the three point loads of 5 kips each. The two point loads are halfway between Point A and Point B and halfway between Point B and Point C. There is no Live Load on the frame. Determinant Continuous Beam P 15 ft B C The internal hinges are 3 feet from Points B and C Points A and B are pin supports. Point C is a fixed support. Calculate three things. The Reaction at B. The Shear 1’ to the right of B. The Moment at B. P P Engr Fundamentals RB = (1/15){5*7.5 + 5*15 + 2.5*18} = 10.5 kips VB = 2.5 kips MB = 2.5*3 = 7.5 kip-ft. A 15 ft The only load on the beam is a Dead Load which is the three point loads of 5 kips each. The two point loads are halfway between Point A and Point B and halfway between Point B and Point C. There is no Live Load on the frame. Determinant Continuous Beam P 15 ft B C The internal hinges are 3 feet from Points B and C Points A and B are pin supports. Point C is a fixed support. Calculate three things. The Reaction at B. The Shear 1’ to the left of B. The Moment at B. P P Engr Fundamentals RB = (1/15){5*7.5 + 2.5*18} = 5.5 kips VB = 2.5 kips MB = 2.5*3 = 7.5 kip-ft.

10/23/2023 4 TOP VIEW BOLTED CONNECTION 3x8 2x10 SIDE VIEW The Dead Load force applied to the assembly is 6 kips. Each bolt can resist 2 kips parallel to the grain and 1.5 kips perpendicular to the grain. Calculate the net area of the 2x10. Calculate the tension stress on the 2x10. Calculate the strength, Z’, of the assembly if we only consider the bolt strength of the 2x10. SAWN LUMBER DEAD LOAD 30 degrees Anet = 13.88 – 2(7/8 + 1/8)(1.5) = 10.88 sq inch ft = 6000 / 10.88 = 551 psi Z’ = 0.9(6)(2) = 10.8 kips 7/8” Diameter Bolts TOP VIEW BOLTED CONNECTION 3x8 2x10 SIDE VIEW The Dead Load force applied to the assembly is 8 kips. Each bolt can resist 2.5 kips parallel to the grain and 1 kip perpendicular to the grain. Calculate the net area of the 2x10. Calculate the tension stress on the 2x10. Calculate the strength, Z’, of the assembly if we only consider the bolt strength of the 2x10. SAWN LUMBER – All bolts are 3/4” diameter DEAD LOAD 30 degrees Anet = 13.88 – 2(3/4 + 1/8)(1.5) = 11.23 sq inch ft = 8000 / 11.23 = 712 psi Z’ = 0.9(6)(2.5) = 13.5 kips

10/23/2023 5 TOP VIEW BOLTED CONNECTION 3x8 2x10 SIDE VIEW The Dead Load force applied to the assembly is 8 kips. Each bolt can resist 2.5 kips parallel to the grain and 1 kip perpendicular to the grain. Calculate the net area of the 2x10. Calculate the tension stress on the 2x10. Calculate the strength, Z’, of the assembly if we only consider the bolt strength of the 2x10. SAWN LUMBER – All bolts are 1/2” diameter DEAD LOAD 30 degrees Anet = 13.88 – 2(1/2 + 1/8)(1.5) = 12.01 sq inch ft = 8000 / 12.01 = 666 psi Z’ = 0.9(6)(2.5) = 13.5 kips TOP VIEW NAILED CONNECTION 2x8 4x10 SIDE VIEW SAWN LUMBER The stars represent 16d nails. The nails are 0.22 inch diameter and 3.5 inches long. The yield stress is 90 ksi. DEAD LOAD Calculate the strength of a single nail (Z) considering the bearing strength of both wood pieces is 3000 psi. Consider only modes I and IV. Rd = KD = 10(0.22) + 0.5 = 2.7 Mode I Main = (0.22)(2)(3000)/2.7 = 489 lbs. Side = (0.22)(1.5)(3000)/2.7 = 367 lbs. Mode IV - for Re = 1 Z = (0.22)^2/2.7 * sqrt{2*3000*90000/(3*2)} = 170 lbs Z = 170 lbs .

10/23/2023 7 TOP VIEW NAILED CONNECTION 2x8 4x10 SIDE VIEW SAWN LUMBER The stars represent 16d nails. The nails are 0.18 inch diameter and 3.2 inches long. The yield stress is 90 ksi. DEAD LOAD Calculate the strength of a single nail (Z) considering the bearing strength of both wood pieces is 2000 psi. Consider only modes I and IV. Rd = KD = 10(0.18) + 0.5 = 2.3 Mode I Main = (0.18)(1.7)(2000)/2.3 = 266 lbs. Side = (0.18)(1.5)(2000)/2.3 = 235 lbs. Mode IV - for Re = 1 Z = (0.18)^2/2.3 * sqrt{2*2000*90000/(3*2)} = 110 lbs Z = 110 lbs . TOP VIEW BOLTED CONNECTION 2x8 SIDE VIEW Calculate the tabulated strength of the assembly (Z) for the 2x8’s bolted together using the allowable strengths of the table. Assume bolts are spaced so that no reduction is required. ALL BOLTS ARE 1/2” DIAMETER SAWN LUMBER 2x8 DEAD LOAD Bolts are loaded perpendicular to grain of post. Z = 4*0.48 = 1.92 kips = 1920 pounds.

10/23/2023 8 TOP VIEW BOLTED CONNECTION 2x8 SIDE VIEW Calculate the tabulated strength of the assembly (Z) for the 2x8’s bolted together using the allowable strengths of the table. Assume bolts are spaced so that no reduction is required. ALL BOLTS ARE 5/8” DIAMETER SAWN LUMBER 2x8 DEAD LOAD Bolts are loaded perpendicular to grain of post. Z = 4*0.54 = 2.16 kips = 2160 pounds. TOP VIEW BOLTED CONNECTION 2x8 SIDE VIEW Calculate the tabulated strength of the assembly (Z) for the 2x8’s bolted together using the allowable strengths of the table. Assume bolts are spaced so that no reduction is required. ALL BOLTS ARE 3/4” DIAMETER SAWN LUMBER 2x8 DEAD LOAD Bolts are loaded perpendicular to grain of post. Z = 4*0.60 = 2.40 kips = 2400 pounds.

10/23/2023 10 BEAM DESIGN 18’ 2x10 Douglas Fir/Larch – Select Structural Beam • The beam has a uniform Dead Load of 0.15 klf (includes self weight). • The beam has two Roof Live Load point loads of 2 kips each located at one- third of the span (6’ from each end). Determine the critical Demand/Capacity ratio for bending (fb/Fb’) and shear (fv/Fv’). All factors are 1.0 except Load Duration Factor. SAWN LUMBER Load Duration Factors Dead Load = 0.9 Roof Live Load = 1.25 Snow Load = 1.15 RLL RLL DL DL M = (1/8)(0.15)(18*18) = 6.075 k-ft Vcrit = (9-0.8)(0.15) = 1.23 kips RLL M = 2*6 = 12 kp-ft Vcrit = 2 kips Shear D/C DL alone fv = 3(1.23)/(2*13.88) = 133 psi Fv’ = 0.9*60 = 54 psi D/C = 2.46 DL + RLL fv = 3(3.23)/(2*13.88) = 349 psi Fv’ = 1.25*60 = 75 psi D/C = 4.65 CRITICAL D/C = 6.76 The tabulated strength of the lumber is F b = 1200 psi and F v = 60 psi. BEAM DESIGN 18’ 2x10 Douglas Fir/Larch – Select Structural Beam • The beam has a uniform Dead Load of 0.15 klf (includes self weight). • The beam has two Roof Live Load point loads of 2 kips each located at one- third of the span (6’ from each end). Determine the critical Demand/Capacity ratio for bending (fb/Fb’) and shear (fv/Fv’). All factors are 1.0 except Load Duration Factor. SAWN LUMBER Load Duration Factors Dead Load = 0.9 Roof Live Load = 1.25 Snow Load = 1.15 RLL RLL DL DL M = (1/8)(0.15)(18*18) = 6.075 k-ft Vcrit = (9-0.8)(0.15) = 1.23 kips RLL M = 2*6 = 12 kp-ft Vcrit = 2 kips Bending D/C DL alone fb = (6.075*12)/21.39 = 3408 psi Fb’ = 0.9*1350 = 1215 psi D/C = 2.80 DL + RLL fb = (18.075*12)/21.39 = 10140 psi Fb’ = 1.25*1350 = 1688 psi D/C = 6.01 The tabulated strength of the lumber is F b = 1350 psi and F v = 180 psi.

10/23/2023 11 BEAM DESIGN 18’ 2x10 Douglas Fir/Larch – Select Structural Beam • The beam has a uniform Dead Load of 0.15 klf (includes self weight). • The beam has two Roof Live Load point loads of 2 kips each located at one- third of the span (6’ from each end). Determine the critical Demand/Capacity ratio for bending (fb/Fb’) and shear (fv/Fv’). All factors are 1.0 except Load Duration Factor. SAWN LUMBER Load Duration Factors Dead Load = 0.9 Roof Live Load = 1.25 Snow Load = 1.15 RLL RLL DL DL M = (1/8)(0.15)(18*18) = 6.075 k-ft Vcrit = (9-0.8)(0.15) = 1.23 kips RLL M = 2*6 = 12 kp-ft Vcrit = 2 kips Shear D/C DL alone fv = 3(1.23)/(2*13.88) = 133 psi Fv’ = 0.9*180 = 162 psi D/C = 0.82 DL + RLL fv = 3(3.23)/(2*13.88) = 349 psi Fv’ = 1.25*180 = 225 psi D/C = 1.55 CRITICAL D/C = 6.01 The tabulated strength of the lumber is F b = 1350 psi and F v = 180 psi. BEAM DESIGN 18’ 2x10 Douglas Fir/Larch • The beam has a uniform Dead Load of 0.2 klf (includes self weight). • The beam has two Roof Live Load point loads of 1.5 kips each located at one- third of the span (6’ from each end). Determine the critical Demand/Capacity ratio for bending (fb/Fb’) and shear (fv/Fv’). All factors are 1.0 except Load Duration Factor. SAWN LUMBER Load Duration Factors Dead Load = 0.9 Roof Live Load = 1.25 Snow Load = 1.15 RLL RLL DL The tabulated strength of the lumber is F b = 1500 psi and F v = 80 psi. DL M = (1/8)(0.2)(18*18) = 8.10 k-ft Vcrit = (9-0.8)(0.2) = 1.64 kips RLL M = 1.5*6 = 9 kip-ft Vcrit = 1.5 kips Bending D/C DL alone fb = (8.10*12)/21.39 = 4544 psi Fb’ = 0.9*1500 = 1350 psi D/C = 3.37 DL + RLL fb = (17.1*12)/21.39 = 9593 psi Fb’ = 1.25*1500 = 1875 psi D/C = 5.12

10/23/2023 13 BEAM DESIGN 18’ 2x10 Douglas Fir/Larch • The beam has a uniform Dead Load of 0.2 klf (includes self weight). • The beam has two Roof Live Load point loads of 1.5 kips each located at one- third of the span (6’ from each end). Determine the critical Demand/Capacity ratio for bending (fb/Fb’) and shear (fv/Fv’). All factors are 1.0 except Load Duration Factor. SAWN LUMBER Load Duration Factors Dead Load = 0.9 Roof Live Load = 1.25 Snow Load = 1.15 RLL RLL DL The tabulated strength of the lumber is F b = 1400 psi and F v = 90 psi. DL M = (1/8)(0.2)(18*18) = 8.10 k-ft Vcrit = (9-0.8)(0.2) = 1.64 kips RLL M = 1.5*6 = 9 kip-ft Vcrit = 1.5 kips Shear D/C DL alone fv = 3(1.64)/(2*13.88) = 177 psi Fv’ = 0.9*90 = 81 psi D/C = 2.18 DL + RLL fv = 3(3.14)/(2*13.88) = 339 psi Fv’ = 1.25*90 = 113 psi D/C = 3.00 CRITICAL D/C = 5.48 BEAM DESIGN 18’ 2x10 Douglas Fir/Larch • The beam has a uniform Dead Load of 0.2 klf (includes self weight). • The beam has two Roof Live Load point loads of 1.5 kips each located at one- third of the span (6’ from each end). Determine the critical Demand/Capacity ratio for bending (fb/Fb’) and shear (fv/Fv’). All factors are 1.0 except Load Duration Factor. SAWN LUMBER Load Duration Factors Dead Load = 0.9 Roof Live Load = 1.25 Snow Load = 1.15 RLL RLL DL The tabulated strength of the lumber is F b = 1200 psi and F v = 60 psi. DL M = (1/8)(0.2)(18*18) = 8.10 k-ft Vcrit = (9-0.8)(0.2) = 1.64 kips RLL M = 1.5*6 = 9 kip-ft Vcrit = 1.5 kips Bending D/C DL alone fb = (8.10*12)/21.39 = 4544 psi Fb’ = 0.9*1200 = 1080 psi D/C = 4.21 DL + RLL fb = (17.1*12)/21.39 = 9593 psi Fb’ = 1.25*1200 = 1500 psi D/C = 6.40

10/23/2023 14 BEAM DESIGN 18’ 2x10 Douglas Fir/Larch • The beam has a uniform Dead Load of 0.2 klf (includes self weight). • The beam has two Roof Live Load point loads of 1.5 kips each located at one- third of the span (6’ from each end). Determine the critical Demand/Capacity ratio for bending (fb/Fb’) and shear (fv/Fv’). All factors are 1.0 except Load Duration Factor. SAWN LUMBER Load Duration Factors Dead Load = 0.9 Roof Live Load = 1.25 Snow Load = 1.15 RLL RLL DL The tabulated strength of the lumber is F b = 1200 psi and F v = 60 psi. DL M = (1/8)(0.2)(18*18) = 8.10 k-ft Vcrit = (9-0.8)(0.2) = 1.64 kips RLL M = 1.5*6 = 9 kip-ft Vcrit = 1.5 kips Shear D/C DL alone fv = 3(1.64)/(2*13.88) = 177 psi Fv’ = 0.9*60 = 54 psi D/C = 3.28 DL + RLL fv = 3(3.14)/(2*13.88) = 339 psi Fv’ = 1.25*60 = 75 psi D/C = 4.52 CRITICAL D/C = 6.40 Determine the value of I about the X-X axis for the shape shown. Determine the value of Q to be used to determine the max shear stress when bending is about the X-X axis. All four rectangles are 5 inches by 1 inch – true dimensions. For a rectangle, I = (1/12)bd 3 And PAT says I x ’ = I x + Ad 2 Q = A’y’ X X Y Y STRESS MECHANICS QX = (1*5)(3) + 2(1*2.5)(1.25) = 15 + 6.2 = 21.2 in^3 IX = (2/12)(1)(5^3) + 2(1/12)(5)(1^3) + 2(1*5)(3^2) = 20.8 + 0.8 + 90 = 111.6 in^4

10/23/2023 16 Determine the value of I about the X-X axis for the shape shown. Determine the value of Q to be used to determine the max shear stress when bending is about the X-X axis. All three rectangles are 5 inches by 1 inch – true dimensions. For a rectangle, I = (1/12)bd 3 And PAT says I x ’ = I x + Ad 2 Q = A’y’ X X Y Y STRESS MECHANICS QX = (1*5)(3) + (1*2.5)(1.25) = 15 + 3.1 = 18.1 in^3 IX = (1/12)(1)(5^3) + 2(1/12)(5)(1^3) + 2(1*5)(3^2) = 10.4 + 0.8 + 90 = 101.2 in^4 The 3/4” bolt is loaded with a force 30 degrees away from vertical. Determine the allowable force (Z) the bolt can resist if the lumber is 4 inches thick. P a = {P x P y } / {P sin 2 Theta + P y cos 2 Theta} P = Allow load parallel to grain P y = Allow load perpendicular to grain 30 degrees Z = (2.89)(1.49) / {2.89 sin^2(60) + 1.49 cos^2(60)} = 4.31 / {2.17 + 0.37} = 1.70 kips per bolt

10/23/2023 17 TJI Beam Design The TJI shown is simply supported and sits on two 2x4 stud walls. The TJI supports two concentrated forces (DL and LL) at third points. The maximum moment applied is 2700 ft-lbs and the maximum shear is 1025 lbs. Rim joists (parallel to the wall) are on each end of the joist. TJI TJI JOIST 2x10 RIM JOIST – Each End Bending – M = 2700 ft-lbs – need 210 Shear – V = 1025 lbs – need 110 Reaction – R = 1025 lbs – 230 USE 9.5 TJI 230 Shear and Moment Diagrams V(k) 10 5 ft A shear and moment diagram has been drawn for a beam but the drawing of the beam and the loading is not shown. Answer the questions about the original loading on the beam. 5 ft 5 ft 5 ft 5 ft 5 ft M(k-ft) 5 6 2 3 13 7 17 7 T T C C No calculations required. You can answer the questions without drawing the original beam and loading. I could have asked these questions without providing you the numbers on the diagram. 17 3 13 If the supports are provided at the ends of the beam it is simply-supported – no matter what the loading looks like.

10/23/2023 19 BEAM DESIGN SPAN 2x10 Eastern Hemlock – Select Structural Beam In CE164, we use a value of 1 for the Beam Stability Factor, because: SAWN LUMBER Load Duration Factors Dead Load = 0.9 Roof Live Load = 1.25 Snow Load = 1.15 We expect the beam to be braced as required. TOP VIEW NAILED CONNECTION 2x8 4x10 SIDE VIEW SAWN LUMBER What can be said about the nailed connection? DEAD LOAD The nails are in single shear.

10/23/2023 20 TOP VIEW BOLTED CONNECTION 3x8 2x10 SIDE VIEW What can be said about the type of stress that will result in the 3x8? SAWN LUMBER DEAD LOAD 30 degrees The stress will be both tension and bending. BEAM DESIGN ITEM A in the picture is usually called what? ITEM A BRIDGING