Truss Analysis for Vertical Displacement at Joint C
The vertical displacements at joint C in two steel trusses are determined under different loading conditions and fabrication errors. The calculations involve virtual forces, real forces, member characteristics, and structural properties to find the final displacements accurately.
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Example 8-15 The cross-sectional area of each member of the truss shown in the figure is A = 400 mm2 and E = 200 GPa. (a)Determine the vertical displacement of joint C if a 4-kN force is applied to the truss at C. (b) If no loads act on the truss, what would be the vertical displacement of joint C if member AB were 5 mm too short? (c)If 4 kN force and fabrication error are both accounted, what would be the vertical displacement of joint C. C 4 kN 3 m A B 4 m 4 m t
SOLUTION Part(a) Virtual Force n. Since the vertical displacement of joint C is to be determined, only a vertical l kN load is placed at joint C. The n force in each member is calculated using the method of joint. Real Force N. The N force in each member is calculated using the method of joint. 1 kN C 4 kN C 4 kN A 0.667 2 0 B B A n (kN) N(kN) 0.5kN 0.5kN 1.5kN 1.5kN a
1 kN C C 4 kN C B BA B A A 0.667 2 8 n (kN) N (kN) = L (m) C B A 10.67 nNL (kN2 m) CV) = nNL (1kN)( AE 10.67 kN m 1( 10.41+10.41+10.67) = = CV (400 10 6m2)(200 106kN) AE m2 CV = +0.133mm, 1
Part (b): The member AB were 5 mm too short 1 kN C A 0.667 B n (kN) 5 mm (1)( CV ) = n( L) =(0.667)( 0.005) CV CV = -3.33mm, Part (c): The 4 kN force and fabrication error are both accounted. CV = 0.133 - 3.33 = -3.20mm CV = -3.20mm, 8
Example 8-16 Determine the vertical displacement of joint C of the steel truss shown. The cross-section area of each member is A = 400 mm2 and E = 200 GPa. F E 4 m D A C B 4 m 4 m 4m 4 kN 4 kN a
SOLUTION Virtual Force n. Since the vertical displacement of joint C is to be determined, only a vertical l kN load is placed at joint C. The n force in each member is calculated using the method of joint. Real Force N. The N force in each member is calculated using the method of joint. F -0.333 E F E -4 0.333 1 0.667 4 m 4 m 4 4 0A 0A 0.333 4 4 4 0.667 D D C C B B 4 m 4 m 4 m 4 m 4 m 4 m 0.333kN 1 kN0.667 kN n (kN) 4 kN 4 kN 4 kN 4 kN N(kN) zo
F E F E F-0.333 E 4 -4 0.333 1 0.667 4 4 4 4 - 4 4 4 4 4 4 0.333 0.667 B C C B B C D A D A DA L(m) n (kN)1 kN 4kN N(kN) 4 kN = F 5.33 E 5.33 16 10.67 5.33 10.67 B C A D nNL(kN2 m) CV) = nNL (1kN)( AE 72.4 kN m 1[(15.07+3(5.33)+2(10.67)+16+30.18)]= = CV (400 10 6m2)(200 106kN) AE m2 = 1.23 mm, z CV
Example 8-17 Determine the vertical displacement of joint C of the steel truss shown. Due to radiant heating from the wall, members are subjected to a temperature change: member AD is increase +60oC, member DC is increase +40oC and member AC is decrease -20oC.Also member DC is fabricated 2 mm too short and member AC 3 mm too long. Take = 12(10-6) , the cross-section area of each member is A = 400 mm2 and E = 200 GPa. wall D C 10 kN 3 m B A 20kN 2 m zz
SOLUTION Due to loadingforces. 1 kN 1 kN 20 kN 10 kN C 0.667kN 23.33kN D 0.667 D 23.33 C D 2 C 1 20 3 m 3 m 3 0 20 3 0 0 2 B B B 2 m A 2 m A 0.667kN 13.33kN A 20 kN L (m) N (kN) n (kN) D 31.13 C nNL AE (1kN)( ) = CV 60 0 1 = (60+31.13+104.12) 0 CV (400)(200) B A CV= 2.44 mm, nNL(kN2 m) zs
1 kN C D D D 0.667 D +40 2 C C C -2 1 +60 3 3 0 2 0 B B B B A A A A T (oC) L (m) n (kN) Fabrication error (mm) Due to temperature change. (1kN)( CV) = n ( T)L = (12 10 6)[(1)(60)(3)+(0.667)(40)(2)+( 1.2)( 20)(3.61)] = 3.84 mm, CV Due to fabrication error. (1kN)( CV) = n( L) = (0.667)( 0.002)+( 1.2)(0.003) = 4.93mm, CV Total displacement . z 4 ( CV)Total= 2.44+3.84 4.93=1.35 mm,
Method of Virtual Work : Bending d M w M C A B d C dx RA RB Virtualloadings ds = d 1 = (m )(d ) = (m ) Mdx L 1 M d = ds C dx EI EI Real displacements zt
Method of Virtual Work : Beams and Frames w w C C A B A B C C RA RB RA RB Virtual loadings Virtualloadings 1 = (m)(d ) = (m ) Mdx 1 = (m )(d ) = (m ) Mdx L L C C EI EI Real displacements Real displacements za
Method of Virtual Work : Beams and Frames Vertical Displacement Realload Virtual unit load w x2 x1 C C A B B A C x2 x1 RA RB 1 RA RB w x1 x2 M2 m 2 M1 m 1 B B x1 V2 v 2 x2 V1 v 1 RB RA RA RB 1 = (m ) Mdx L C EI z 1
Slope Realload Virtual unit couple w x2 x1 1 C C A B B A Cx x1 RA RB 2 RA RB w x2 M2 m 2 M1 m 1 B B x1 V2 v 2 x2 V1 v 1 RB RA RA RB 1 = (m ) Mdx L C EI z 8
Example8-18 The beam shown is subjected to a load P at its end. Determine the slope and displacement at C. EI is constant. P B C A C 2a a za
Displacement at C SOLUTION Virtual Momentm Real MomentM 1 kN P x2 x2 x1 x1 B B A A C C 2a 2a a a 1 2 m P 2 M 3 2 3P 2 = x1 Px m 2 =-x2 m M = -a 1 M2 =-Px2 1 1 2 2 -Pa a ( x2)( Px2)dx2 0 1 C= dx = 2a ( 1 1 m M x 2 Px 2 1)( 1 )dx1+ EI EI EI L 0 2a a 8Pa3 Pa3 Pa3 Px3 Px3 3EI +( 2) = C = ( + = 1 ) 12EI 12EI 3EI 3EI 0 a s o
Slope at C Virtual Momentm x1 Real MomentM x2 P x2 1 kN m x1 B B A A C C 2a 2a a a 1 1 P 2 M 3P 2a m 2a 2 = x1 Px m M11 = 1 1 1 -1 M2 =-Px2 2a m 2 = 1 2 -Pa a ( 1)( Px2 )dx2 (1 kN m)( C ) = dx = 2a ( L m M 1 1 x 2a a Px 2 1)( 1)dx1+ EI EI EI 0 x3 3 0 0 2a 2 1 8a3 3 1 1 1 7 Pa2 ( P 4a P 4a Px Pa2 2 C = ( +( = ( )+( ) = )( 2) )( )( )( ), )( )(1) 2 6 EI EI EI EI EI 0 0 s
Conclusion P B C A C C 2a a Pa3 C =3EI 2 =7 Pa ( ), C 6 EI sz
