Post on 06-May-2015
Shear and Moment Diagrams
Ambrose - Chapter 6
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 =>
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)
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
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
Solving equation (3):
Solving equation (2):
Figure 6.7a =>
)'20()'161200()'68000(03
1200800002
01
2..
1
2..
1
RM
RRF
F
lblb
lblby
x
..
..
2
....2
360,320
200,67
200,19000,48'20
lbft
ftlb
ftlbftlb
R
R
.
1
...1
840,5
360,3200,1000,8lb
lblblb
R
R
Beam Shear – Example 1 (pg. 64)
• Determine the shear at various points along the beam
.)18(
.)8(
.)1(
360,3200,1000,8480,5
160,2000,8480,5
480,50480,5
lbx
lbx
lbx
V
V
V
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
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
Beam Shear – Example 2 (pg. 66)
• Determine the reactions
Solving equation (3):
Solving equation (2):
)'24()'166000()]'6)('12000,1[(03
000,6)'12000,1(02
01
2..
1
2..
1
RM
RRF
F
lbftlb
lbftlby
x
..
..
2
....2
000,724
000,168
000,96000,72'24
lbft
ftlb
ftlbftlb
R
R
.
1
...1
000,11
000,7000,6000,12lb
lblblb
R
R
Beam Shear – Example 2 (pg. 66)
• Determine the shear at various points along the beam
.
)24(
.)16(
.)16(
.)12(
.)2(
.)1(
000,7000,6000,112000,11
000,7000,6000,112000,11
000,1)000,112(000,11
000,1)000,112(000,11
000,9)000,12(000,11
000,10)000,11(000,11
lbx
lbx
lbx
lbx
lbx
lbx
V
V
V
V
V
V
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
feetx
xV
11
)000,1(000,110
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
Beam Shear – Example 3 (pg. 68)• Determine the reactions
Solving equation (3):
Solving equation (4):
)'20(42
32)'32500()'28000,2()'6000,12()'4000,4(04
)'20(8232)'32500()'24000,4()'14000,12()'8000,2(03
)'32500(000,4000,12000,202
01
1.....
2
2.....
1
2...
1.
RM
RM
RRF
F
ftlblblblb
ftlblblblb
ftlblblblby
x
..
..
2
........2
800,1820
000,376
000,128000,96000,168000,16'20
lbft
ftlb
ftlbftlbftlbftlb
R
R
...
1
........1
200,15'20
000,304
000,192000,56000,72000,16'20
lbftlb
ftlbftlbftlbftlb
R
R
Beam Shear – Example 3 (pg. 68)
• Determine the shear at various points along the beam
.......
)32(
.......)28(
......)28(
......)22(
.....)22(
.....)8(
....)8(
000,4'32500000,12000,2800,18200,15
000,6'28500000,12000,2800,18200,15
800,12'28500000,12000,2200,15
800,9'22500000,12000,2200,15
200,2'22500000,2200,15
200,9'8500000,2200,15
000,6'8500000,20
lbftlblblblblbx
lbftlblblblblbx
lbftlblblblbx
lbftlblblblbx
lbftlblblbx
lbftlblblbx
lbftlblbx
V
V
V
V
V
V
V
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
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)
Bending Moment
Bending Moment – Example 1
• Simple beam – span = 20 feet– 2 concentrated loads– shear diagram from
earlier
• Construct moment diagram
Bending Moment – Example 1
1) Compute moments at critical locations
– under 8,000 lb. load & 1,200 lb. load
....
)'16(
...)'6(
440,13)'10000,8()'16840,5(
040,350)'6840,5(ftlblblb
x
ftlblbx
M
M
Bending Moment – Example 2
• Simple beam – Span = 20 feet– 1 concentrated load– 1 uniformly distr. Load– Shear diagram
• Construct moment diagram
Bending Moment – Example 2
1) Compute moments at critical locations
– When x = 11 ft. and under 6,000 lb. load
...
)'16(
...)'11(
000,56421212000,1)'16000,11(
500,6021111000,1)'11000,11(
ftlbftlblbx
ftlbftlblbx
M
M
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)
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 distributed load acting over entire span
– Construct the shear and moment diagram
– Figure 6.12
Negative Bending Moment - Example
)'20(226)'26600(03
)'26600(02
01
2.
1
2.
1
RM
RRF
F
ftlb
ftlby
x
1) Determine the reactions
- Solving equation (3):
- Solving equation (4):
..
..
2
..2
140,1020
800,202
800,202'20
lbft
ftlb
ftlb
R
R
.
1
..1
460,5
140,10600,15lb
lblb
R
R
Negative Bending Moment - Example
2) Determine the shear at various points along the beam and draw the shear diagram
.)20(
.)20(
.)10(
.)1(
600,3)60020(140,10460,5
540,6)60020(460,5
540)60010(460,5
860,4)6001(460,5
lbx
lbx
lbx
lbx
V
V
V
V
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.
feetx
xV
1.9
)600(460,50
Negative Bending Moment - Example
4) Determine the moments that the critical shear points found in step 3) and draw the moment diagram
...
)20(
...)1.9(
800,10220'20600)'20460,5(
843,2421.9'1.9600)'1.9460,5(
ftlbftlblbx
ftlbftlblbx
M
M
Negative Bending Moment - Example
4) Find the location of the inflection point (zero moment) and max. bending moment
• since x cannot =0, then we use x=18.2’• Max. bending moment =24,843 lb.-ft.
feetfeetx
x
xxxxM
xxM
xxxM ftlb
2.18;0600
460,55460
)300(2
)0)(300(4)460,5(460,5
460,5300300460,50
2600460,50
2600)460,5(0
2
22
2
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
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
Typical Loadings
• In beam design, only need to know:– reactions– max. shear– max. bending moment– max. deflection