Analysis of 2D Frames

Fritz Canono, mece-1Herbert Diapolet, mece-1Lemuel Al Fernandez, mece-1Analysis of 2D Frames2D FramesFrames provide flexural rigidity unlike the truss.Frames include axial deformation at each end in addition to the beam deformations. Unlike beam which as 2 degrees of freedom, frames have 3 degrees of freedom at each node.2D FramesFrames provide flexural rigidity unlike the truss.Frames include axial deformation at each end in addition to the beam deformations. Unlike beam which as 2 degrees of freedom, frames have 3 degrees of freedom at each node.1P1, u1P2, u2P3, u3P4, u4P5, u5P6, u622D FramesBeam elements have their longitudinal axis aligned but for frames, the axis of the 2D frame can have any orientation in the plane.There is no coordinate transformation required for the beam elements but for frames, the development of the [] transformation matrix will necessary.Elemental Stiffness MatrixWe have previously derived the elemental stiffness matrix for the beam, which is as follows

1P1, u1P2, u2P3, u3P4, u4P5, u5P6, u62

Transformation Matrix123,u3u2u156,u6u5u44yx

Transformation Matrixyx

Transformation Matrix

2Example ProblemConsider the frame shown. Solve for the:1. Displacements2. Member forcesM=12k-ft112ft7 ftFor both members:E=30x106 PsiI=100 in4A=20 in2Flow chart28792146513For both members:E=30x106 PsiI=100 in4A=20 in2M=12k-ft7ft.12ft.

Keeping the units consistent to lb and in,

28792146513M=12k-ft

=0 (restrained)=0 (restrained)=0 (restrained)For member 1, = 90C = 0 S=128792146513M=12k-ft

C = 0 S=128792146513M=12k-ft

=0 (restrained)=0 (restrained)=0 (restrained)For member 2, = 0C = 1 S=028792146513M=12k-ft

C = 1 S=028792146513M=12k-ft

Combining, 28792146513M=12k-ft

Substituting to our equation,

We can now solve our displacements.

28792146513M=12k-ftTo solve for member forces,

28792146513M=12k-ftFor member 1,

28792146513M=12k-ftFor member 1,

28792146513M=12k-ftFor member 2,

1612#556#1612#4.45ft-k556#2.22ft-k12 ft-k557#1612#7.61 ft-k

1612#557#3.73ft-k7.61 ft-k4.45 ft-kNon-nodal loads2115ft10 ftFor both members:E=29x106 PsiI=100 in4A=20 in2320k10 ftNon-nodal loads2115ft10 ft320k10 ft10k10k300 in-k300 in-kEquivalent nodal forces17.88k8.94k17.88k8.94k8.94k25ft-k25ft-k8.94k4.47k4.47kFixed end forces for member 2

=0 (restrained)=0 (restrained)=0 (restrained)=0 (restrained)=0 (restrained)=0 (restrained)

To solve for member forces, we use again our formula,

Member 1 Member 2Member 3Member 2.21k.21k8.24ft-k10.58ft-k1.89k1.89k8.94k8.94k25ft-k25ft-k4.47k4.47k

Member 29.15k8.73k16.76ft-k14.42ft-k2.58k6.36k20kThermal EffectsTavyT2T1Temperature variation over the depth of the beamStrain at any point of the beam:(y)= T(y)

Stress for fixed end conditions:=ET(y)hWe assume T2 is larger than T1 , the fixed end moments will act on the beam in either clockwise and counterclockwise directions at the left and the right ends.

The equivalent nodal forces will act in the opposite directions.Example Problem28792146513For both members:E=30x106 PsiI=100 in4A=20 in27ft.12ft.Calculate the member forcesT=70F , h = 10 in.Flow chartCalculating the forces exerted by the member due heat expansionIn this problem, we had to assume that only member 2 has a temperature variation of +70 F (T) over its depth of 10 in.

Keeping the units consistent to lb and in,

28792146513

=0 (restrained)=0 (restrained)=0 (restrained)For member 1, = 90C = 0 S=182792146513M=12k-ft

C = 0 S=128792146513

=0 (restrained)=0 (restrained)=0 (restrained)For member 2, = 0C = 1 S=028792146513M=12k-ft

C = 1 S=028792146513

Combining,

21The force caused by heat expansion in member 2 is transmitted to member 1 as F4 =-Pt , F5 = 0 , F6 = Mt .

Note: F5=0 due to very small value of cross sectional area.Mt F4Assumptions+F+F+M

= x u

P = kglobal x

361#1714#361#40171 #-in.171411770#-in.361#1714#40171#-in1714#361#184140#-in.40171 #-in.40171 #-in.The illustration shows how the member attain its structural equilibriumSupport Movements28792146513For both members:E=30x106 PsiI=100 in4A=20 in27ft.12ft.Assume that node 1 move 0.3 in. downwardAnd node 2 to 0.5in. to the right