Beam-2-span-shear-release-UNIF-load EG V-M-diagrams PDF

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Summary

Example of a beam with Shear-Release under UNIFORM load. Draw Shear and Moment Diagrams. ...

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B. Goodno, Georgia Tech Spring 2019, p. 1

V & M diagram E.g.'s

beam-2-span-shear-release-EG_V-Mdiagrams.xmcd

> Beam with shear release - draw shear (V) and moment (M) diagrams

P  2400  lb

Draw FBD's - solve for reactions at supports Σ Fy  0

RHFB

Entire FBD

Dy  q  (10ft)  4000  lb 1

Σ MA  0

Cy



Σ Fy  0

Ay

C y

18ft

q  400

lb ft

< cut through shear release to create RHFB

Dy 28ft  P 10ft  q  28ft 28





ft  2



 3822 lbf

 Dy  P  q  (28ft)  5778 lbf

Shear Force Diagram V( x ) 

Ay  q  x if 0ft  x  10ft

 Ay  q x  P

if 10ft  x  18ft

Dy  q  (28ft  x) if 18ft  x  28ft 210

5

110

5

< shift to RHFB for segment CD

Ay

18

V( 0 )  5778 lbf V(18ft)  0 lbf

V ( x)

0  110

5

 210

5

NOTE that shear is zero at shear release  Dy 0

10

20 x

B. Goodno, Georgia Tech Spring 2019, p. 2

V & M diagram E.g.'s

beam-2-span-shear-release-EG_V-Mdiagrams.xmcd

Moment Diagram M (x) 

 A  x  q  x x  if 0ft  x  10ft y 2  A  x  q  x x  P (x  10ft) if 10ft  x  18ft 2  y  2  Dy  (28ft  x )  q (28ft  x )  if 18ft  x  28ft 2  

< shift to RHFB for segment CD

6

1.5 10

18

M (10ft)  37778 ft  lb

37778ft lb 6

1 10

M (18ft)  20000 ft  lb

M ( x)

20000ft lb 5 5 10

NOTE that max. moments occur when V = 0 0

0

10

20 x

> REPEAT for Beam with moment release - draw shear (V) and moment (M) diagrams

P  2400  lb

Draw FBD's - solve for reactions at supports RHFB

Σ M mrel  0

Entire FBD

Dy



q  400

lb ft

q  (10ft) 10ft   2000  lb < cut through shear release to create RHFB 10ft   2  1

1

Dy  28ft  P 10ft  q  28ft 28

Σ MA  0

Cy



Σ Fy  0

Ay

C y  Dy  P  q  (28ft)  4667 lbf

18ft





ft  2



 6933 lbf

B. Goodno, Georgia Tech Spring 2019, p. 3

V & M diagram E.g.'s

beam-2-span-shear-release-EG_V-Mdiagrams.xmcd

Shear Force Diagram V( x ) 

Ay  q  x if 0ft  x  10ft

Ay  q x  P 

if 10ft  x  18ft

Dy  q  (28ft  x) if 18ft  x  28ft

< shift to RHFB for segment CD

5 2 10

18

23

Ay

V( 0 )  4667 lbf

5

1 10 V ( x)

V(10ft)  1733 lbf

0

V (17.999ft)  4933 lbf

5

 1 10

V(18ft)  2000 lbf  4933lb

5

 2 10

0

10

20

 Dy  q  (28ft  x) = 0 solve x  23.0  ft

x

Moment Diagram M (x) 

 A  x  q  x x  if 0ft  x  10ft y 2  x A  x  q  x  P (x  10ft) if 10ft  x  18ft 2 y  2  Dy  (28ft  x )  q  (28ft  x )  if 18ft  x  28ft 2  

6 1 10

< shift to RHFB for segment CD

18

23 26667ft lb

5 8 10

M (18ft)  0 ftlb

5 6 10

M ( x)

M (23ft)  5000 ft  lb

5 4 10 5 2 10

0

M (10ft)  26667 ft  lb

5000ft lb 0

10

20 x

NOTE that moment is zero at moment release AND that max. moments occur when V = 0...