CIV 207 Winter For practice

Transcription

1 CV 07 Wnter 009 Assgnment #6 Frday, February 13 th Complete the frst three questons. Submt your work to Bo #5 on the th floor of the MacDonald buldng by 1 noon on Tuesday March 3 rd. No late submssons wll be accepted. The second set of three questons are for practce only. Groups of up to 3 members are permtted. Clearly prnt your names and student numbers on your submsson. Ensure that the submsson s properly stapled The cross-sectonal dmensons of a beam are shown. The nternal bendng moment about the centrodal as s M = +70 lb-ft. Determne: (a) the mamum tenson bendng stress n the beam. (b) the mamum compresson bendng stress n the beam. 8.. A flanged wooden shape s used to support the loads shown on the beam. The dmensons of the shape are shown n Fg. b. Consder the entre 18-ft length of the beam and determne: (a) the mamum tenson bendng stress at any locaton along the beam, and (b) the mamum compresson bendng stress at any locaton along the beam Two steel plates, each n. wde and 0.5 n. thck, renforce a wood beam that s 3 n. wde and 8 n. deep. The steel plates are attached to the vertcal sdes of the wood beam n a poston such that the composte shape s symmetrc about the as, as shown n the sketch of the beam cross secton. Determne the mamum bendng stresses produced n both the wood and the steel f a bendng moment of M = +50 kp-n s appled about the as. Assume E wood =,000 ks and E steel = 30,000 ks. For practce

2 8.16 The cross-sectonal dmensons of a beam are shown n Fg. P8.16. The nternal bendng moment about the centrodal as s M = +70 lb-ft. Determne: (a) the mamum tenson bendng stress n the beam. (b) the mamum compresson bendng stress n the beam. Soluton Centrod locaton n y drecton: Shape Area A (from bottom) y A y (n. ) (n.) (n. 3 ) bottom flange left web left top flange rght web rght top flange n n. 3 3 ΣyA n. y = = = n. ΣA n. (measured upward from bottom edge of bottom flange) Moment of nerta about the as: Shape C d = y y d²a C + d²a (n. ) (n.) (n. ) (n. ) bottom flange left web left top flange rght web rght top flange Moment of nerta about the as (n. ) = Fg. P8.16 (a) Mamum tenson bendng stress: For a postve bendng moment of M = +70 lb-ft, the mamum tenson bendng stress wll occur at the bottom surface of the cross secton (.e., y = n.). From the fleure formula, the bendng stress at the bottom of the cross secton s: My (70 lb-ft)( n.)(1 n./ft) = 3,66.6 ps = 3,70 ps (T) n. (b) Mamum compresson bendng stress: The mamum compresson bendng stress wll occur at the top surface of the cross secton (.e., y =.50 n n. = 1.77 n.). From the fleure formula, the bendng stress at the top of the cross secton s: My (70 lb-ft)(1.77 n.)(1 n./ft), ps =,700 ps (C) n.

3 8.17 Two vertcal forces are appled to a smply supported beam (Fg. P8.17a) havng the cross secton shown n Fg. P8.17b. Determne the mamum tenson and compresson bendng stresses produced n segment BC of the beam. Fg. P8.17a Fg. P8.17b Soluton Centrod locaton n y drecton: Shape Area A (from bottom) y A y (mm ) (mm) (mm 3 ) top flange 3, ,500.0 stem 1, ,00.0,0 mm 617,700 mm 3 3 ΣyA 617,700 mm y = = = ΣA,0 mm mm (measured upward from bottom edge of stem) Moment of nerta about the as: Shape C d = y y d²a C + d²a (mm ) (mm) (mm ) (mm ) top flange 56, ,15,997.08,7,7.08 stem 3,07, ,033,37.5 8,105,37.5 Moment of nerta about the as (mm ) = 10,577,57.3

4 Shear-force and bendng-moment dagrams: The mamum moment occurs between B and C. The moment magntude s 1 kn-m. Mamum tenson bendng stress: For a postve bendng moment, the mamum tenson bendng stress wll occur at the bottom surface of ths cross secton. From the fleure formula, the bendng stress at the bottom of the tee stem s: My (1 kn-m)( mm)(1,000 N/kN)(1,000 mm/m) = MPa (T) mm Mamum compresson bendng stress: The mamum compresson bendng stress wll occur at the top of the flange: My (1 kn-m)(175 mm mm)(1,000 N/kN)(1,000 mm/m) mm 0.7 MPa = 0.7 MPa (C)

5 8. A flanged wooden shape s used to support the loads shown on the beam n Fg. P8.a. The dmensons of the shape are shown n Fg. P8.b. Consder the entre 18-ft length of the beam and determne: (a) the mamum tenson bendng stress at any locaton along the beam, and (b) the mamum compresson bendng stress at any locaton along the beam. Fg. P8.a Fg. P8.b Soluton Centrod locaton n y drecton: Shape Area A (from bottom) y A y (n. ) (n.) (n. 3 ) top flange web bottom flange n. 1.0 n. 3 3 ΣyA 1.0 n. y = = = n. ΣA 5.0 n. (from bottom of shape to centrod) = n. (from top of shape to centrod) Moment of nerta about the as: Shape C d = y y d²a C + d²a (n. ) (n.) (n. ) (n. ) top flange web bottom flange Moment of nerta about the as (n. ) = 1,85.056

6 Shear-force and bendng-moment dagrams Mamum bendng moments postve M = 1,851 lb-ft negatve M = 8,00 lb-ft Bendng stresses at ma postve moment (1,851 lb-ft)( n.)(1 n./ft) 1, n. = 83 ps (C) (1,851 lb-ft)( n.)(1 n./ft) 1, n. = 1, 099 ps (T) Bendng stresses at ma negatve moment ( 8,00 lb-ft)( n.)(1 n./ft) 1, n. = 77 ps (T) ( 8,00 lb-ft)( n.)(1 n./ft) 1, n. = 6 ps (C) (a) Mamum tenson bendng stress = 1,099 ps (T) (b) Mamum compresson bendng stress = 83 ps (C)

7 8.36 A cantlever tmber beam (Fg. P8.36a) wth a span of L = 3 m supports a unformly dstrbuted load of w. The beam wdth s b = 300 mm and the beam heght s h = 00 mm (Fg. P8.36b). The allowable bendng stress of the wood s 6 MPa. Determne the magntude of the mamum load w that may be carred by the beam. Fg. P8.36a Fg. P8.36b Soluton Secton modulus for sold rectangular secton 3 bh /1 bh (300 mm)(00 mm) S = 10 mm c = h/ = 6 = 6 = 6 3 Mamum allowable bendng moment: M Mallow S = (6 N/mm )( 10 mm ) = 1 10 N-mm S Mamum bendng moment n cantlever span: The mamum bendng moment magntude n the cantlever beam occurs at support A: wl M ma = Mamum dstrbuted load: wl M allow 6 M allow (1 10 N-mm) wallow = =.67 N/mm =.67 kn/m L [(3 m)(1,000 mm/m)]

8 8.5 An alumnum [E = 10,000 ks] bar s bonded to a steel [E = 30,000 ks] bar to form a composte beam (Fg. P8.5b). The composte beam s subjected to a bendng moment of M = +300 lb-ft about the as (Fg. P8.5a). Determne: (a) the mamum bendng stresses n the alumnum and steel bars. (b) the stress n the two materals at the jont where they are bonded together. Fg. P8.5a Fg. P8.5b Soluton Denote the alumnum as materal (1) and denote the steel as materal (). The modular rato s: E 30,000 ks n = = = 3 E1 10,000 ks Transform the steel bar () nto an equvalent amount of alumnum (1) by multplyng ts wdth by the modular rato: b, trans = 3(1.50 n.) =.50 n. Thus, for calculaton purposes, the 1.50 n n. steel bar s replaced by an alumnum bar that s.50-n. wde and 0.50-n. thck. Centrod locaton of the transformed secton n the vertcal drecton Shape Wdth b Heght h Area A (from bottom) y A y (n.) (n.) (n. ) (n.) (n. 3 ) alumnum bar (1) transformed steel bar () ΣyA n. y = = = ΣA 3.00 n n. (measured upward from bottom edge of secton) Moment of nerta about the horontal centrodal as Shape C d = y y d²a C + d²a (n. ) (n.) (n. ) (n. ) alumnum bar (1) transformed steel bar () Moment of nerta about the as = n.

9 (a) Mamum bendng stress n alumnum bar (1) From the fleure formula, the mamum bendng stress n alumnum bar (1) s: My (300 lb-ft)( n.)(1 n./ft) 1 = 11, 080 ps (T) n. (a) Mamum bendng stress n steel bar () The bendng stress n the transformed materal must be multpled by the modular rato n. Therefore, the mamum bendng stress n steel bar () s: My (300 lb-ft)(1.000 n n.)(1 n./ft) (3) = 19,90 ps (C) n. (b) Bendng stress n alumnum bar (1) at nterface My (300 lb-ft)(0.50 n n.)(1 n./ft) 1 =, 0 ps (T) n. (b) Mamum bendng stress n steel bar () at nterface My (300 lb-ft)(0.50 n n.)(1 n./ft) (3) = 6, 650 ps (T) n.

10 8.50 Two steel plates, each n. wde and 0.5 n. thck, renforce a wood beam that s 3 n. wde and 8 n. deep. The steel plates are attached to the vertcal sdes of the wood beam n a poston such that the composte shape s symmetrc about the as, as shown n the sketch of the beam cross secton (Fg. P8.50). Determne the mamum bendng stresses produced n both the wood and the steel f a bendng moment of M = +50 kp-n s appled about the as. Assume E wood =,000 ks and E steel = 30,000 ks. Fg. P8.50 Soluton Let the wood be denoted as materal (1) and the steel plates as materal (). The modular rato s: E 30,000 ks n = = = 15 E1,000 ks Transform the steel plates () nto an equvalent amount of wood (1) by multplyng the plate thcknesses by the modular rato: b, trans = 15(0.5 n.) = 3.75 n. (each). Thus, for calculaton purposes, each n. 0.5 n. steel plate s replaced by a wood board that s -n. tall and 3.75-n. wde. Centrod locaton: Snce the transformed secton s doubly symmetrc, the centrod locaton s found from symmetry. Moment of nerta about the centrodal as Shape C d = y y d²a C + d²a (n. ) (n.) (n. ) (n. ) wood beam (1) two transformed steel plates () Moment of nerta about the as = 168 n. Bendng stress n wood beam (1) From the fleure formula, the mamum bendng stress n wood beam (1) s: Mc (50 kp-n.)( n.) 1 = = = ks = 1,190 ps 168 n. Bendng stress n steel plates () The bendng stress n the transformed materal must be multpled by the modular rato n. Therefore, the mamum bendng stress n the steel plates () s: Mc (50 kp-n.)( n.) = n = (15) = 8.93 ks = 8,930 ps 168 n.