Shear and Moment Diagrams
MD. INTISHAR RAHMAN

Ambrose - Chapter 6
Beam Shear
• Vertical shear:
tendency for one part
of a beam to move
vertically with respect
to an adjacent part
– Section...
Beam Shear
• Magnitude (V) = sum of vertical forces on
either side of the section
– can be determined at any section along...
Beam Shear
• Why?
– necessary to know the maximum value of the
shear
– necessary to locate where the shear changes
from po...
Beam Shear – Example 1 (pg. 64)
• Simple beam
– Span = 20 feet
– 2 concentrated loads

• Construct shear
diagram
Beam Shear – Example 1 (pg. 64)
1) Determine the reactions
(1) ( + ↑ )∑ Fx = 0
( 2) ( + ↑ )∑ Fy = 0 = R1 − 8000lb. − 1200l...
Beam Shear – Example 1 (pg. 64)
•

Determine the shear at
various points along the beam
V( x =1) = 5,480 − 0 = +5,480lb.
V...
Beam Shear – Example 1 (pg. 64)
•

Conclusions
–

max. vertical shear = 5,840 lb.
•

–

max. vertical shear occurs at grea...
Beam Shear – Example 2 (pg. 66)
• Simple beam
– Span = 20 feet
– 1 concentrated load
– 1 uniformly distr. load

• Construc...
Beam Shear – Example 2 (pg. 66)
•

Determine the reactions
(1) ( + ↑)∑ Fx = 0
( 2) ( + ↑)∑ Fy = 0 = R1 − (1,000lb ft. ×12'...
Beam Shear – Example 2 (pg. 66)
•

Determine the shear at
various points along the beam
V( x =1) = 11,000 − (1× 1,000) = 1...
Beam Shear – Example 2 (pg. 66)
•

Conclusions
–

max. vertical shear = 11,000 lb.
•

–

at left reaction

shear passes th...
Beam Shear – Example 3 (pg. 68)
• Simple beam with
overhanging ends
– Span = 32 feet
– 3 concentrated loads
– 1 uniformly ...
Beam Shear – Example 3 (pg. 68)
•

Determine the reactions

(1) ( + ↑)∑ Fx = 0
( 2) ( + ↑)∑ Fy = 0 = −2,000lb. + R1 − 12,0...
Beam Shear – Example 3 (pg. 68)
•

Determine the shear at
various points along the beam
Beam Shear – Example 3 (pg. 68)
•

Conclusions
– max. vertical shear =
12,800 lb.
•

disregard +/- notations

– shear pass...
Bending Moment
• Bending moment: tendency of a beam to bend
due to forces acting on it
• Magnitude (M) = sum of moments of...
Bending Moment
Bending Moment – Example 1
• Simple beam
– span = 20 feet
– 2 concentrated loads
– shear diagram from
earlier
• Construct ...
Bending Moment – Example 1
1) Compute moments at
critical locations
–

under 8,000 lb. load &
1,200 lb. load

( + ↵) M ( x...
Bending Moment – Example 2
• Simple beam
–
–
–
–

Span = 20 feet
1 concentrated load
1 uniformly distr. Load
Shear diagram...
Bending Moment – Example 2
1) Compute moments at
critical locations
–

When x = 11 ft. and under
6,000 lb. load

( + ↵) M ...
Negative Bending Moment
• Previously, simple beams subjected to
positive bending moments only
– moment diagrams on one sid...
Negative Bending Moment
• deflected shape has
inflection point
– bending moment = 0

• See example
Negative Bending Moment Example
– Simple beam with
overhanging end on right
side
• Span = 20’
• Overhang = 6’
• Uniformly ...
Negative Bending Moment Example
1) Determine the
reactions

(1) ( + ↑)∑ Fx = 0
( 2) ( + ↑)∑ Fy = 0 = R1 − (600lb ft. × 26'...
Negative Bending Moment Example
2) Determine the shear
at various points
along the beam and
draw the shear
diagram
V( x =1...
Negative Bending Moment Example
3) Determine where the
shear is at a
maximum and where
it crosses zero
–

max shear occurs...
Negative Bending Moment Example
4) Determine the
moments that the
critical shear points
found in step 3) and
draw the mome...
Negative Bending Moment Example
4) Find the location of the inflection
point (zero moment) and max.
bending moment

[(

) ...
Rules of Thumb/Review
• shear is dependent on the loads and reactions
– when a reaction occurs; the shear “jumps” up by th...
Rules of Thumb/Review
• moment is dependent upon the shear diagram
– the area under the shear diagram = change in the
mome...
Typical Loadings
• In beam design, only
need to know:
–
–
–
–

reactions
max. shear
max. bending moment
max. deflection
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Transcript of "Shear and moment diagram"

  1. 1. Shear and Moment Diagrams MD. INTISHAR RAHMAN Ambrose - Chapter 6
  2. 2. Beam Shear • Vertical shear: tendency for one part of a beam to move vertically with respect to an adjacent part – Section 3.4 (Ambrose) – Figure 3.3 =>
  3. 3. Beam Shear • Magnitude (V) = sum of vertical forces on either side of the section – can be determined at any section along the length of the beam • Upward forces (reactions) = positive • Downward forces (loads) = negative • Vertical Shear = reactions – loads (to the left of the section)
  4. 4. Beam Shear • Why? – necessary to know the maximum value of the shear – necessary to locate where the shear changes from positive to negative • where the shear passes through zero • Use of shear diagrams give a graphical representation of vertical shear throughout the length of a beam
  5. 5. Beam Shear – Example 1 (pg. 64) • Simple beam – Span = 20 feet – 2 concentrated loads • Construct shear diagram
  6. 6. Beam Shear – Example 1 (pg. 64) 1) Determine the reactions (1) ( + ↑ )∑ Fx = 0 ( 2) ( + ↑ )∑ Fy = 0 = R1 − 8000lb. − 1200lb. + R2 ( 3)( + ↵) ∑ M 1 = 0 = (8000lb. × 6' ) + (1200lb. ×16' ) − ( R2 × 20' ) Solving equation (3): 20' R2 = 48,000lb.− ft . +19,200 lb.− ft . 67,200lb.− ft . R2 = = 3,360lb. (↑) 20 ft . Solving equation (2): R1 = 8,000lb. +1,200lb. − 3,360lb. R1 = 5,840 lb. (↑) Figure 6.7a =>
  7. 7. Beam Shear – Example 1 (pg. 64) • Determine the shear at various points along the beam V( x =1) = 5,480 − 0 = +5,480lb. V( x =8) = 5,480 − 8,000 = −2,160lb. V( x =18) = 5,480 − 8,000 − 1,200 = −3,360lb.
  8. 8. Beam Shear – Example 1 (pg. 64) • Conclusions – max. vertical shear = 5,840 lb. • – max. vertical shear occurs at greater reaction and equals the greater reaction (for simple spans) shear changes sign under 8,000 lb. load • where max. bending occurs
  9. 9. Beam Shear – Example 2 (pg. 66) • Simple beam – Span = 20 feet – 1 concentrated load – 1 uniformly distr. load • Construct shear diagram, designate maximum shear, locate where shear passes through zero
  10. 10. Beam Shear – Example 2 (pg. 66) • Determine the reactions (1) ( + ↑)∑ Fx = 0 ( 2) ( + ↑)∑ Fy = 0 = R1 − (1,000lb ft. ×12' ) − 6,000lb. + R2 ( 3)( + ↵) ∑ M 1 = 0 = [(1,000lb ft. ×12' )(6' )] + (6000lb. ×16' ) − ( R2 × 24' ) Solving equation (3): 24' R2 = 72,000lb.− ft . + 96,000 lb.− ft . 168,000lb.− ft . R2 = = 7,000lb. (↑) 24 ft . Solving equation (2): R1 =12,000lb. + 6,000lb. − 7,000lb. R1 =11,000lb. (↑)
  11. 11. Beam Shear – Example 2 (pg. 66) • Determine the shear at various points along the beam V( x =1) = 11,000 − (1× 1,000) = 10,000lb. V( x = 2) = 11,000 − (2 × 1,000) = 9,000lb. V( x =12) = 11,000 − (12 × 1,000) = −1,000lb. V( x =16 − ) = 11,000 − (12 × 1,000) = −1,000lb. V( x =16 + ) = 11,000 − [ (12 × 1,000 ) + 6,000] = −7,000lb. V( x = 24 ) = 11,000 − [ (12 × 1,000) + 6,000] = − 7,000lb.
  12. 12. Beam Shear – Example 2 (pg. 66) • Conclusions – max. vertical shear = 11,000 lb. • – at left reaction shear passes through zero at some point between the left end and the end of the distributed load • x = exact location from R1 – at this location, V = 0 V = 0 = 11,000 − ( x ×1,000) x = 11 feet
  13. 13. Beam Shear – Example 3 (pg. 68) • Simple beam with overhanging ends – Span = 32 feet – 3 concentrated loads – 1 uniformly distr. load acting over the entire beam • Construct shear diagram, designate maximum shear, locate where shear passes through zero
  14. 14. Beam Shear – Example 3 (pg. 68) • Determine the reactions (1) ( + ↑)∑ Fx = 0 ( 2) ( + ↑)∑ Fy = 0 = −2,000lb. + R1 − 12,000lb. − 4,000lb. − (500lb ft. × 32' ) + R2 ( 3)( + ↵) ∑ M 1 = 0 = −(2,000lb. × 8' ) + (12,000lb. ×14' ) + (4,000lb. × 24' ) + [(500lb. ft . × 32' )((32 2 ) − 8)] − R2 (20' ) ( 4)( + ↵) ∑ M 2 = 0 = (4,000lb. × 4' ) − (12,000lb. × 6' ) − (2,000lb. × 28' ) − [(500lb. ft . × 32' )((32 2 ) − 4)] + R1 (20' ) Solving equation (3): 20' R2 = −16,000lb.− ft . +168,000lb.− ft . + 96,000lb.− ft . +128,000lb.− ft . 376,000lb.− ft . R2 = =18,800lb. (↑) ft . 20 Solving equation (4): 20' R1 = −16,000lb.− ft . + 72,000lb.− ft . + 56,000lb.− ft . +192,000lb.− ft . 304,000lb.− ft . R1 = =15,200lb. (↑) 20'
  15. 15. Beam Shear – Example 3 (pg. 68) • Determine the shear at various points along the beam
  16. 16. Beam Shear – Example 3 (pg. 68) • Conclusions – max. vertical shear = 12,800 lb. • disregard +/- notations – shear passes through zero at three points • R1, R2, and under the 12,000lb. load
  17. 17. Bending Moment • Bending moment: tendency of a beam to bend due to forces acting on it • Magnitude (M) = sum of moments of forces on either side of the section – can be determined at any section along the length of the beam • Bending Moment = moments of reactions – moments of loads • (to the left of the section)
  18. 18. Bending Moment
  19. 19. Bending Moment – Example 1 • Simple beam – span = 20 feet – 2 concentrated loads – shear diagram from earlier • Construct moment diagram
  20. 20. Bending Moment – Example 1 1) Compute moments at critical locations – under 8,000 lb. load & 1,200 lb. load ( + ↵) M ( x =6') = (5,840lb. × 6' ) − 0 = 35,040lb.− ft . ( + ↵) M ( x =16') = (5,840lb. ×16' ) − (8,000lb. ×10' ) = 13,440lb.− ft .
  21. 21. Bending Moment – Example 2 • Simple beam – – – – Span = 20 feet 1 concentrated load 1 uniformly distr. Load Shear diagram • Construct moment diagram
  22. 22. Bending Moment – Example 2 1) Compute moments at critical locations – When x = 11 ft. and under 6,000 lb. load ( + ↵) M ( x =11') = (11,000lb. ×11' ) − [(1,000lb ft ×11)(11 2 )] = 60,500lb.− ft . ( + ↵) M ( x =16') = (11,000lb. ×16' ) − [(1,000lb ft ×12)((12 2 ) + 4)] = 56,000lb.− ft .
  23. 23. Negative Bending Moment • Previously, simple beams subjected to positive bending moments only – moment diagrams on one side of the base line • concave upward (compression on top) • Overhanging ends create negative moments • concave downward (compression on bottom)
  24. 24. Negative Bending Moment • deflected shape has inflection point – bending moment = 0 • See example
  25. 25. Negative Bending Moment Example – Simple beam with overhanging end on right side • Span = 20’ • Overhang = 6’ • Uniformly distributed load acting over entire span – Construct the shear and moment diagram – Figure 6.12
  26. 26. Negative Bending Moment Example 1) Determine the reactions (1) ( + ↑)∑ Fx = 0 ( 2) ( + ↑)∑ Fy = 0 = R1 − (600lb ft. × 26' ) + R2 ( 3)( + ↵) ∑ M 1 = 0 = [(600lb ft. × 26' )(26 2 )] − ( R2 × 20' ) - Solving equation (3): 20' R2 = 202,800lb.− ft . 202,800lb.− ft . R2 = = 10,140lb. ( ↑ ) ft . 20 - Solving equation (4): R1 =15,600lb. −10,140lb. R1 = 5,460lb. (↑)
  27. 27. Negative Bending Moment Example 2) Determine the shear at various points along the beam and draw the shear diagram V( x =1) = 5,460 − (1× 600) = 4,860lb. V( x =10 ) = 5,460 − (10 × 600) = −540lb. V( x = 20 − ) = 5,460 − (20 × 600) = −6,540lb. V( x = 20 + ) = 5,460 + 10,140 − (20 × 600) = 3,600lb.
  28. 28. Negative Bending Moment Example 3) Determine where the shear is at a maximum and where it crosses zero – max shear occurs at the right reaction = 6,540 lb. V = 0 = 5,460 − ( x × 600) x = 9.1 feet
  29. 29. Negative Bending Moment Example 4) Determine the moments that the critical shear points found in step 3) and draw the moment diagram [( )] = 24,843 2 × 20') ( 20 )] = −10,800 2 )( M ( x =9.1) = (5,460lb. × 9.1' ) − 600lb ft × 9.1' 9.1 lb .− ft . M ( x = 20) = (5,460lb. × 20' ) − 600lb lb.− ft . [( ft
  30. 30. Negative Bending Moment Example 4) Find the location of the inflection point (zero moment) and max. bending moment [( ) ( 2 )] M = 0 = (5,460 × x) − 600lb ft × x x  x2  M = 0 = 5,460 x − 600   2   M = 0 = 5,460 x − 300 x 2 = −300 x 2 + 5,460 x − 5,460 ± (5,460) 2 − 4(−300)(0) x= 2(−300) − 5460 ± 5,460 x= = 0 feet;18.2 feet − 600 • • since x cannot =0, then we use x=18.2’ Max. bending moment =24,843 lb.-ft.
  31. 31. Rules of Thumb/Review • shear is dependent on the loads and reactions – when a reaction occurs; the shear “jumps” up by the amount of the reaction – when a load occurs; the shear “jumps” down by the amount of the load • point loads create straight lines on shear diagrams • uniformly distributed loads create sloping lines of shear diagrams
  32. 32. Rules of Thumb/Review • moment is dependent upon the shear diagram – the area under the shear diagram = change in the moment (i.e. Ashear diagram = ΔM) • straight lines on shear diagrams create sloping lines on moment diagrams • sloping lines on shear diagrams create curves on moment diagrams • positive shear = increasing slope • negative shear = decreasing slope
  33. 33. Typical Loadings • In beam design, only need to know: – – – – reactions max. shear max. bending moment max. deflection
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