CIVL 228 Introduction to Structural Engineering Spring 2015 67 Truss Zero force

# Civl 228 introduction to structural engineering

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CIVL 228 Introduction to Structural Engineering, Spring 2015 67 Truss: Zero force members Example: A B D E F G I H J Represents hinge connection Fix support 3’ 3’ 3’ C 10 kips E D
CIVL 228 Introduction to Structural Engineering, Spring 2015 68 Truss: Zero force members Example: A B C D E F G I H J Represents hinge connection Fix support 3’ 3’ 3’ 10 kips
CIVL 228 Introduction to Structural Engineering, Spring 2015 69 Truss: Zero force members Example: A C D H 3’ 3’ 4’ 10 kips 0 Y F ¦ X Y 3 10 0 5 CH F § · ´ ¨ ¸ © ¹ 16.67kips(t) CH F 0 X F ¦ 4 0 5 CH CD F F § · ³ ¨ ¸ © ¹ 13.33kips(c) CD F ´ C 10 kips F CD F CH
CIVL 228 Introduction to Structural Engineering, Spring 2015 70 Truss: Zero force members Example: A D H 3’ 3’ 4’ 0 Y F ¦ 3 16.67 0 5 Y A § · ´ ¨ ¸ © ¹ 10kips Y A 0 X F ¦ 4 13.33 16.67 0 5 X A § · ³ ´ ¨ ¸ © ¹ 0 kips X A 16.67 kips 13.33 kips X Y + M V + A + - 13.33 kips 0 A M ¦ ± ² ± ² 4 13.33 3 16.67 6 0 5 A M § · ´ ³ ¨ ¸ © ¹ 40 k-ft A M ´ A Y A X M A -40 kip-ft Shear Moment Axial -10 kips
CIVL 228 Introduction to Structural Engineering, Spring 2015 71 Truss: Method by sections Example: 40 kips 15’ A B C D E F G H 15’ 15’ 15’ 20’ 40 kips 40 kips 30 kips Find F FG , F FC and F BC 0 A M ¦ Î E Y = 70 kips Î -20*30 40*15 40*30 40*45 + 60*E Y = 0 X Y E A 0 y F ¦ Î A Y 40*3 + E Y = 0 Î A Y = 50 kips 0 x F ¦ Î A X + 30 = 0 Î A X = 30 kips 30 kips 50 kips 70 kips
CIVL 228 Introduction to Structural Engineering, Spring 2015 72 Truss: Method by sections Example: 40 kips 15’ A B F 15’ 30 kips Find F FG , F FC and F BC 0 F M ¦ Î F BC = 67.5 kips (t) Î -30*20 50*15 + F BC *20 = 0 X Y 0 y F ¦ Î 50 40 - F FC (4/5) = 0 Î F FC = 12.5 kips (t) 0 x F ¦ Î 30 -30 + F BC + F FC (3/5)+ F FG = 0 Î F FG = -F BC - F FC (3/5) = -75 kips (c) F FG F BC F FC 20’ 30 kips 50 kips = 67.5 kips (t) Î BC = 12.5 kips (t)
CIVL 228 Introduction to Structural Engineering, Spring 2015 73 Truss: 3D trusses Example: Find all element forces. , 0 x C F ¦ Î F BC = 0 , 0 x B F ¦ Î F BE = 0 , 0 x E F ¦ Î F FE = 0 , 0 x F F ¦ Î F FD = 0 , 0 x D F ¦ Î F AD = 0 , 0 z B F ¦ Î F FB sin(45) -800 = 0 Î F FB = 800/sin(45) = 1131.4 lb (t) Î FB 500 lb 800 lb 4’ 4’ 3’ x z y A B C D F E A z A x A y D z D y E y F y
CIVL 228 Introduction to Structural Engineering, Spring 2015 74 Truss: 3D trusses Example: Find all element forces. 500 lb 800 lb 4’ 4’ 3’ x z y A B C D F E , 0 y B F ¦ Î -F FB cos(45) - F AB = 0 Î F AB = -F FB cos(45) = -800 lb (c) , 0 z F F ¦ Î -F FB cos(45) - F AF = 0 Î F AF = -F FB cos(45) = -800 lb (c) , 0 z C F ¦ Î F CE sin(45) - 500 = 0 Î F CE = 500/ sin(45) = 707.1 lb (t) , 0 y C F ¦ Î -F CE cos(45) - F CD = 0 Î F CD = -F CE cos(45) = -500 lb (c) , 0 z E F ¦ Î -F CE cos(45) - F DE = 0 Î F DE = -F CE cos(45) = -500 lb (c) A z A x A y D z D y E y F y
CIVL 228 Introduction to Structural Engineering, Spring 2015 75 Truss: Equilibrium matrix Same as Method by joints, except you are solving for all element forces at the same time.
CIVL 228 Introduction to Structural Engineering, Spring 2015 76 Truss: Equilibrium matrix Example: X Y q 1 4 8 4 8 4 1 P 2 P 3 P 4 P 5 P 6 P 7 P 8 P 9 P 10 P 11 P q 2 q 3 q 4 q 5 q 6 q 7 q 11 q 8 q 9 q 10
CIVL 228 Introduction to Structural Engineering, Spring 2015 77 Truss: Equilibrium matrix 1 P 2 P 5 P 4 P 3 P 6 P 7 P 8 P 9 P 10 P 11 P Example: q 1 4 8 4 8 4 1 P 2 P 3 P 4 P 5 P 6 P 7 P 8 P 9

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