Effect of Shear and Axial Effects on Bending Deformation

Steven VukazichSan Jose State University

Lab 12 Exercise

The statically indeterminate steel moment frame on Line B of our CE 160 Lab Building is subjected to the seismic load found in Lab #10. This frame was analyzed in found to have the support reactions, moment, shear, and axial force diagrams shown below.Using the moment (MP), shear (VP), and axial force (FP) diagrams found in Lab #10; find the horizontal displacement, in inches, at the point where the load P is applied using the principle of virtual work.

20 ft

22 ft

P

"2

"2

1011"

1011"

P&W18x65

W14x82

W14x82

Real Moment Diagram (Mp )

10%

10%

0 0

Moment Diagramson the Compression Side

Real Shear Diagram (Vp )

+

+Positive Shear

$2

$2

−1011$

Real Axial Force Diagram (Fp )

Tension Positive

−$2

−1011$1011$

Need to find MQ , VQ , and FQ diagrams (use consistent sign convention)

Virtual System to Measure Horizontal Displacement at Point of Load Application

20 ft

22 ft

1

12

12

1110

1110

Bending Shear Axial force

The effect of bending, shear, and axial force internal work can be found by evaluating the following expression:

! = 1$% &'

()*)+,- +

1/01

&'

(2*2+,- +

1$034*4+5

The appropriate section properties (I, A, As) for each steel W section can be found in the tables of section properties in the AISC Manual of Steel Construction.

Note that the shear area for a W section is the web area:

Cross Sectional Properties of Beam and Columns

!" = $%&

%&$

Cross Section Properties for W18x65 Beam From the AISC Steel Manual

Cross Section Properties for W14x82 ColumnsFrom the AISC Steel Manual

The material properties, E and G, are related to Poisson’s ratio and can be found from:

E = 29,000 ksi for steel

ν = 0.3 (Poisson’s ratio for steel)

Material Properties of Beam and Columns

! = #2 1 + ' = 29,000

2 1 + 0.3 = 11,153.846 ksi

Use Statics to find Moment and Shearfunctions and evaluate product integrals(Recall Lab 10 exercise)

Evaluate Internal Work terms for Bending, Shear, and Axial Force

! = 1$% &'

()*)+,- +

1/01

&'

(2*2+,- +

1$034*4+5

Sign Convention and Coordinates

Sign convention and coordinates to use for functions

20 ft

22 ft

P

"2

"2

+

+

+

+

Shear

Moment

Axial

Tension is positive

yL yR

x

1011"

1011"

SAP 2000 Output Showing Joint Displacement