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