4 kN 2 kN 3m 2m Influence line for F BG 1 Load at D 025 075 3m A B F 3m 4m X Y

4 kn 2 kn 3m 2m influence line for f bg 1 load at d

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4 kN 2 kN 3m 2m Influence line for F BG : 1* Load at D: 0.25* 0.75* 3m A B F 3m 4m X Y F FG F BC F BG 0.5* 0 y F 0.25 + F BG (4/5) = 0 F BG = -0.3125* (c)
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CIVL 228 Introduction to Structural Engineering, Spring 2015 29 Example 1.5 kN 3m A B C D E F G H 3m 3m 3m 4m X Y What should be the minimum design force for element F BG ? 4 kN 2 kN 3m 2m Influence line for F BG : 1* Load at E: 0* 1* 3m A B F 3m 4m X Y F FG F BC F BG 0* F BG = 0
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CIVL 228 Introduction to Structural Engineering, Spring 2015 30 Example1.5 kN3mABCDEFGH3m3m3m4mXYWhat should be the minimum design force for element FBG?4 kN2 kN3m2m :
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CIVL 228 Introduction to Structural Engineering, Spring 2015 31 Example 1.5 kN 3m A B C D E F G H 3m 3m 3m 4m X Y What should be the minimum design force for element F BG ? 4 kN 2 kN 3m 2m Influence line for F BG : 3m 1m 3m 3m 0.3125* -0.625* compression tension 2m F BG
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CIVL 228 Introduction to Structural Engineering, Spring 2015 32 Example 3m 1m 3m 3m 0.3125* -0.625* compression tension 2m x When x = 0: F BG = 1.5(0) + 4(0) + 2(0) F BG F BG = 0 kN 1.5 kN 4 kN 2 kN 3m 2m
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CIVL 228 Introduction to Structural Engineering, Spring 2015 33 Example 3m 1m 3m 3m 0.3125* -0.625* compression tension 2m x When x = 6: F BG = 1.5(-0.625) + 4(-0.625/6*4) + 2(-0.625/6*1) F BG F BG = -2.8125 kN 1.5 kN 4 kN 2 kN 3m 2m
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CIVL 228 Introduction to Structural Engineering, Spring 2015 34 Example 3m 1m 3m 3m 0.3125* -0.625* compression tension 2m x When x = 7: F BG = 1.5(-0.625/2*1) + 4(-0.625/6*5) + 2(-0.625/6*2) F BG F BG = -2.96875 kN 1.5 kN 4 kN 2 kN 3m 2m
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CIVL 228 Introduction to Structural Engineering, Spring 2015 35 Example 3m 1m 3m 3m 0.3125* -0.625* compression tension 2m x When x = 8: F BG = 1.5(0) + 4(-0.625) + 2(-0.625/6*3) F BG F BG = -3.125 kN 1.5 kN 4 kN 2 kN 3m 2m
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CIVL 228 Introduction to Structural Engineering, Spring 2015 36 Example 3m 1m 3m 3m 0.3125* -0.625* compression tension 2m x When x = 9: F BG = 1.5(0.3125) + 4(-0.625/2*1) + 2(-0.625/6*2) F BG F BG = -1.2 kN 1.5 kN 4 kN 2 kN 3m 2m
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CIVL 228 Introduction to Structural Engineering, Spring 2015 37 Example 3m 1m 3m 3m 0.3125* -0.625* compression tension 2m x When x = 11: F BG = 1.5(0.3125/3*1) + 4(0.325) + 2(-0.625) F BG F BG = 0.21 kN 1.5 kN 4 kN 2 kN 3m 2m
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CIVL 228 Introduction to Structural Engineering, Spring 2015 38 Example 3m 1m 3m 3m 0.3125* -0.625* compression tension 2m x When x = 13: F BG = 1.5(0) + 4(0.325/3*1) + 2(0) F BG F BG = 0.43 kN 1.5 kN 4 kN 2 kN 3m 2m
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CIVL 228 Introduction to Structural Engineering, Spring 2015 39
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