P.E. Civil Exam Review: Structural Analysis J.P. Mohsen Email: jpm@louisville.edu
Structures Determinate Indeterminate
STATICALLY DETERMINATE
STATICALLY INDETERMINATE
Stability and Determinacy of Trusses 300 lb. 400 lb. B C D 7.5 ft A 10 ft H 10 ft G 10 ft F 10 ft E R L R R 2j = m + r Truss is determinate 2j m + r indeterminate J = number of joints m= number of members r = number of reactions 2j m + r Unstable
Determine the force in members BH, BC, and DG of the truss shown. Note that the truss is composed of triangles 7.5 ft : 10.0 ft : 12.5 ft, so that they are 3:4:5 right angles. 300 lb. 400 lb. B C D 7.5 ft 10 ft H 10 ft G 10 ft F 10 ft E R L R R
Member BH. 300 lb. 400 lb. B C D A 10 ft H 10 ft G 10 ft F 10 ft E R L R R
Analysis of Member BH. 300 lb. 400 lb. B C D A 10 ft H 10 ft G 10 ft F 10 ft E R L R R F BH Applying Equation of Equilibrium to Joint H + F y 0 Fbh 0 F AH H F HG
Member BC. 300 lb. 400 lb. B C D A 10 ft H 10 ft G 10 ft F 10 ft E R L R R
Analysis of Member BC. 300 lb. 400 lb. B C D A 10 ft H 10 ft G 10 ft F 10 ft E R L R R = 275 lb. + M 0 20 R 7.5 F 0 G R BC 400 lb. F BC 275(20) 7.5 733 lbs ( compression) B C D F BC 12.5 ft 7.5 ft F BG F HG G 10 ft F 10 ft E R R
Member DG. 300 lb. 400 lb. B C D A 10 ft H 10 ft G 10 ft F 10 ft E R L R R
Analysis of Member DG. 300 lb. 400 lb. B C D A 10 ft H 10 ft G 10 ft F 10 ft E R L R R C F CD F DG F GF D 12.5 ft 7.5 ft G F 10 ft E R R
Analysis of Member DG. 300 lb. 400 lb. B C D A 10 ft H 10 ft G 10 ft F 10 ft E R L R R + Y DG Y DG F 0 R DG 0 3 5 R Y DG Y DG 458 lbs tension DG Y 275 lbs C F CD F DG F GF D 12.5 ft 7.5 ft G F 10 ft E R R
Draw the shear and moment diagrams for the beam shown. Indicate the maximum moment. 20 kn/m 60 kn 120 kn-m A B C D E 2 m 2 m 2 m 2 m
Draw the Free Body Diagram (FBD). (Note: The horizontal force at point B is equal to zero.) 20 kn/m 60 kn 120 kn-m A F B C D B F E 2 m 2 m 2 m 2 m
Solve for the reactions at supports B and E. 20 kn/m 60 kn 120 kn-m A F B = 100 kn C D F E = 40 kn 2 m 2 m 2 m 2 m + M B = 0 60(2) + 120 6F E = 0 F E = 40 kn + F Y = 0-60 80 + F E + F B = 0-100 + F B = 0 F B = 100 kn
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Draw the Shear Diagram for segment AB. 2 m 20 kn 40 kn m 0 0 V (kn) -40
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Show the change in Shear at B. 60 100 kn 0 0 V (kn) -40
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Draw the Shear Diagram for segment BC. 60 20 0 0 V (kn) 2 m 20 kn 40 kn m -40
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Show the change in Shear at C. 60 60 kn 0 0 20 V (kn) -40-40
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Draw the Shear Diagram for segment CE. 60 20 0 0 V (kn) 4 m 0 kn 0 kn m -40-40 -40
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Show the change in Shear at E. 60 40 kn 0 0 20 V (kn) -40-40 -40
20 kn/m 60 kn 120 kn-m Completed Shear Diagram A 2 m 2 m C 2 m D 2 m 100 kn 60 40 kn 0 0 20 V (kn) -40-40 -40
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Draw the Moment Diagram for segment AB. 60 0 0 20 V (kn) -40-40 -40 1 2 2 m 40 kn 40 kn m 0 0 2-40 M (kn-m)
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Draw the Moment Diagram for segment AB. 60 0 0 20 V (kn) -40-40 -40 1 2 2 m 40 kn 40 kn m 0 0 M (kn-m) -40
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Draw the Moment Diagram for segment BC. 60 0 0 20 V (kn) -40-40 -40 1 2 2 m 40 kn 2 m 20 kn 80 kn m 2 40 0 0 2 2-40 M (kn-m)
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Draw the Moment Diagram for segment CD. 60 0 0 20 V (kn) -40-40 -40 2 m 40 kn 80 kn m 2 40 0 0 2 2-40 -40 M (kn-m)
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Show the change in bending moment at D. 60 0 0 20 V (kn) -40-40 -40 120 kn m 2 40 80 0 0 2 2-40 -40 M (kn-m)
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Draw the Moment Diagram for segment DE. 60 0 0 20 V (kn) -40-40 -40 2 m 40 kn 80 kn m 2 40 80 0 0 2 2-40 -40 M (kn-m)
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Completed Moment Diagram. 60 20 0 0 V (kn) -40-40 -40 80 2 40 0 0 2 2-40 -40 M (kn-m)
A 60 kn 20 kn/m 120 kn-m 2 m 2 m C 2 m D 2 m 100 kn 40 kn Find the maximum moment. 60 20 0 0 V (kn) -40-40 -40 M max 80 kn m 2 40 80 0 0 2 2-40 -40 M (kn-m)
Find the force in the truss members shown.
What are the vertical and horizontal components of deflection at the 30K Load? All members have a cross sectional area of 1 square inch and modulus of elasticity of 29000 ksi.
S u L A E S = member force with proper sign A= Cross-sectional area of each member L= Length of each member E= modulus of elasticity of materials
These are internal member forces due to original loading
These are internal forces due to a vertical unit load at L2
These are internal member forces due to a horizontal unit load at L2
Please find all member forces and specify whether in tension or compression
What are the support reactions for the beam shown? K AB =4EI = I K BC =4EI = I L 10 L 20
Moment Distribution 1) calculate the fixed end moments 2) Calculate distribution of moments at the clamped ends of the members by the rotation of that joint 3) Calculate the magnitude of the moments carried over to the other ends of the members 4) The addition or subtraction of these latter moments to the original ) g fixed ends moments
Fixed End Moments P FEM = PL 8 L/2 L/2 FEM = PL 8 w FEM= wl 2 12 L FEM= wl 2 12 a P b FEM= Pb2 a L 2 L FEM= Pa 2 b L 2
Lock the joint B. FEM -25 + 25-50 + 50
K AB =4EI = I K BC =4EI = I L 10 L 20 Distribution ib ti Factor = K Sum of K for all members at the joint K1 DF1 K K 2 DF2 K
K AB =4EI = I K BC =4EI = I L 10 L 20 Distribution ib ti Factor = K Sum of K for all members at the joint Distribution Factor = K BA _ K BA + K BC K1 DF1 K K 2 DF2 K
K AB =4EI = I K BC =4EI = I L 10 L 20 1 10 = 2/3 1 + 1_ 10 20 D.F. at B for BA 1 20 = 1/3 1 + 1_ 10 20 D.F. at B for BC
Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B
Joint B Released Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B + 16.67 +8.33
Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B + 16.67 +8.33 C.O.M. + 8.33
Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B C.O.M. +16.67 +8.33 + 8.33 +4.17
Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B C.O.M. + 16.67 +8.33 + 8.33 +4.17 Final Moments - 16.67
Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B C.O.M. + 16.67 +8.33 + 8.33 +4.17 Final Moments - 16.67 + 41.67
Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B C.O.M. + 16.67 +8.33 + 8.33 +4.17 Final Moments - 16.67 + 41.67-41.67
Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B C.O.M. + 16.67 +8.33 + 8.33 +4.17 Final Moments - 16.67 + 41.67-41.67 + 54.17
Stiffness K K AB =4EI = I K BC =4EI = I L 10 L 20 D. F. 2/3 1/3 FEM -25 + 25-50 + 50 Balancing Joint B C.O.M. + 16.67 +8.33 + 8.33 +4.17 ---------- ------------ ---------- ---------- Final Moments - 16.67 + 41.67-41.67 + 54.17
20 k 1.5 K/FT -16.67 FT.K 41.67-41.67 54.2 20 k 1.5 K/FT 7.5 k 12.5 k 14.4 k 15.6 k
References Hibbeler, C. R., Structural Analysis, 3 rd Edition, Prentice Hall, 1995. Chajes, Alexander, Structural Analysis, Prentice Hall, 1982.
Thank You! Any Questions? Good Luck!