1. Given the shown built-up tee-shape, determine the following for both strong- and weak-axis bending:

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Lucinda McLaughlin
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Radius of gration Solution: (a) Location of the ENA: Strong Ais Bendg (X-Ais): Total Area, A= 8" 1 ( 1" ) + 11" ( ") = 1.

1 1. Given the shown built-up tee-shape, determe the followg for both strong- and weak-ais bendg: a. Location of the neutral ais b. Moment of ertia c. Section Moduli d. Radius of gration Solution: (a) Location of the ENA: Strong Ais Bendg (X-Ais): Total Area, A= 8" 1 ( 1" ) + 11" ( ") = A 8" ( 1" )( ") "( 11" )( 6.5" ) A = = =.9 (Distance measured from the top of the TFL downward.) Weak Ais Bendg (Y-Ais): Sce there is a le of smmetr along the -ais, it follows that the ENA lies on the - ais. CE 56 Assignment 1 Solution Page 1 of 1

2 (b) Moment of Inertia: Strong Ais Bendg (X-Ais): 1 I = bh + ad I = ( 8)( 1) + ( 8 1)(.9 1 ) + ( 1 )( 11) + ( 11 1 )( ) 1 1 I = 17.6 Weak Ais Bendg (Y-Ais): 1 1 I = ()( 1 8) + ( 11)( 0.5) = = (c) Elastic Section Moduli: Strong Ais Bendg (X-Ais): I 17.6 = = = 9.15 S,bot 1 cbot I 17.6 = = = S,top 5 ctop.9 Weak Ais Bendg (Y-Ais): I.8 S = = = c (d) Radius of Gration: Strong Ais Bendg (X-Ais): I 17.6 ρ = = = A Weak Ais Bendg (Y-Ais): I.8 ρ = = = 1.78 A 1.5 CE 56 Assignment 1 Solution Page of 1

3 . Given the shown built-up I-shape with coverplates, determe the followg for both strong- and weak-ais bendg: a. Location of the neutral ais b. Moment of ertia c. Section Moduli d. Radius of gration Solution: (a) Elastic Neutral Ais Location: Strong Ais Bendg (X-Ais): Part Area, A ( ) () A ( ) Coverplate Bottom Flange Web Top Flange Σ CE 56 Assignment 1 Solution Page of 1

4 A = = = 9.51 A.6875 measured up from the bottom. Weak Ais Bendg (Y-Ais): Sce there is a le of smmetr along the -ais, it follows that the ENA lies on the - ais. (b) Moments of Inertia: Moment of Inertia about Strong Ais (X-Ais): Part b () h () 1 a ( ) d () I = bh ( ad ( ) ) 1 Coverplate 1 5/ Bottom Flange Web 1/ Top Flange Σ I X X 1 = bh + ad 1 I = I = 95.6 X Moment of Inertia about Weak Ais (Y-Ais): Part b () h () 1 a ( ) d ( ) I = bh ( ad ( ) ) 1 Coverplate 5/ Bottom Flange Web 0.75 ½ Top Flange Σ I Y Y 1 = bh + ad 1 I = CE 56 Assignment 1 Solution Page of 1

5 (c) Section moduli: Strong Ais Bendg (X-Ais): I 95.9 S = = = 1.77,top ctop ( ) I 95.6 S,bot = = = 19.1 cbot 9.51 Weak Ais Bendg (Y-Ais): I S = = = 9.6 c 6 (d) Radii of gration: Strong Ais Bendg (X-Ais): I 95.9 ρ = = = A Weak Ais Bendg (Y-Ais): I ρ = = =.61 A 1.5 CE 56 Assignment 1 Solution Page 5 of 1

6 . Determe all of the reactions and draw the shear and bendg moment diagrams for the followg loaded beam: Solution: + F = 0= 0 + A B + F = 0 A k k k ( ) ( ) ( ) ( ) + CCW M = 0 = B B = 1.67 k A = 9. k k ft k ft k ft ft CE 56 Assignment 1 Solution Page 6 of 1

7 . Determe all of the reactions and draw the shear and moment diagrams for the followg loaded beam: Solution: 1) ( ) kip ft ft kips + F = 0= A + B + C ) + F = 0 10 ft k ) ft ft ft k ft ft + CCW MA = 0 = 1 ( 10 ) + B( 16 ) 15 ( ) + C ( ) We have more unknowns than we have equations of equilibrium. Pass a section through the hge, where we know moments are zero. This etra piece of formation gives us an equation of condition. 15 kips 8 ft C ft C ( 8 ) ) + CCW Mhge = 0 = Mhge C = 0 B A k k = 5.65, positive, k = 0.65, negative, CE 56 Assignment 1 Solution Page 7 of 1

8 CE 56 Assignment 1 Solution Page 8 of 1

9 5. On Tuesda, August 1 st, we will test a Gr. A6 steel coupon tension to failure. Usg the data from that test, a. Plot Stress (ksi) vs. Stra (/), b. Determe the Elastic Modulus of the steel specimen, c. Use the 0.% offset method to determe the ield strength of the specimen, and d. Determe the ultimate strength of the specimen. (a) Usg a cross-sectional area, A=0.5 and a stra value calculated usg the etensometer readg divided b the gage length, the followg stress vs. stra plot was found: Stress vs. Stra Stress (ksi) Stra (/) CE 56 Assignment 1 Solution Page 9 of 1

10 (b) To fd the elastic modulus of the tensile coupon, data the elastic region were plotted. A lear trendle was fit to the data. The slope of the elastic region was founf to be 7,106 ksi, therefore, E = 7,106 ksi Stress vs. Stra - Elastic Region = R = Stress (ksi) Stra (/) CE 56 Assignment 1 Solution Page 10 of 1

11 (c) The equation = was plotted on the stress-stra diagram. That equation was then shifted b.%: = Stress vs. Stra Stress (ksi) Stra (/) Stress vs. Stra Stress (ksi) Stra (/) The pot at which the 0.% offset le crossed the stress-stra curve is the ield strength. F =.7 ksi. CE 56 Assignment 1 Solution Page 11 of 1

12 (d) The ultimate strength of the tensile coupon is the maimum stress sustaed. F u = 70.1 ksi. CE 56 Assignment 1 Solution Page 1 of 1

This procedure covers the determination of the moment of inertia about the neutral axis.

327 Sample Problems Problem 16.1 The moment of inertia about the neutral axis for the T-beam shown is most nearly (A) 36 in 4 (C) 236 in 4 (B) 136 in 4 (D) 736 in 4 This procedure covers the determination