T - Shear Force and Bending Moment Diagr-1

INTERNAL INTERNAL FORCESFORCES

Shear Force and Bending Moment Shear Force and Bending Moment Equations and DiagramsEquations and Diagrams

LEARNING OBJECTIVESLEARNING OBJECTIVES

Be able to draw shear force and Be able to draw shear force and bending moment diagrams for bending moment diagrams for beamsbeams

PRE-REQUISITE KNOWLEDGEPRE-REQUISITE KNOWLEDGE

Units of measurementUnits of measurement Trigonometry conceptsTrigonometry concepts Rectangular component conceptsRectangular component concepts Equilibrium of forces in 2-DEquilibrium of forces in 2-D Internal force conceptsInternal force concepts

STRESSES AND STRAINSSTRESSES AND STRAINS

Stress is defined as load divided by the Stress is defined as load divided by the cross-sectional area on which it acts.cross-sectional area on which it acts.

Strain is defined as the change in length divided by the original length.

Stress = σ = Q/A

Strain = ε = ΔL/L


Stress = σ = Q/A

Q

Q

A = πr2

L

ΔL

Strain = ε = ΔL/L


Stress = σ = Q/A

Q

Q

A = πr2

L

ΔL

Strain = ε = ΔL/L


Stress = σ = Q/A

Q

Q

A = πr2

L

ΔL

Strain = ε = ΔL/L

σ

ε

Δσ

Δε

Young’s Modulus = Δσ/Δε

If Q increases gradually

SHEAR FORCE AND BENDING MOMENT DIAGRAMS

Shear force diagram or SFD is a graphical representation of the vertical shear force (V) along a structural member

Bending moment diagram or BMD is a graphical representation of the bending moment (M) along a structural member

SIGN CONVENTION OF SHEAR FORCE AND BENDING MOMENT

The internal shear force is considered positive if it causes clockwise rotation of the structural member under consideration as shown below.

Positive Negative

SIGN CONVENTION OF SHEAR FORCE AND BENDING MOMENT

The internal bending moment is considered positive if it causes compression of the top fiber of the structural member under consideration as shown below.

Positive Moment Negative Moment

Tension at Bottom Tension at top

STEPS FOR THE INTERNAL SHEAR FORCE DIAGRAM

1. Draw FBD of the entire structure and calculate the support reactions

2. Calculate the internal shear force along the member under consideration from left to right. The shear force is equal to the integral of the externally applied load (for distributed load, the internal shear force is equal to the area under the load).

3. Draw SFD based on the calculated internal shear force from left to right.

STEPS OF THE INTERNAL BENDING MOMENT DIAGRAM1. Draw FBD of the entire structure and calculate the

support reactions2. Calculate the bending moment along the member

under consideration from left to right. The bending moment is equal to the integral of the internal shear force equation or the area under the shear force diagram.

3. The maximum bending moment takes place at the points where the shear force is equal to zero.

4. Draw BMD based on the calculated bending moment from left to right

EXAMPLE 1 - EXAMPLE 1 - SIMPLY SUPPORTED BEAMSIMPLY SUPPORTED BEAM

200 lb

5 ft 5 ft

x

y

Determine the internal shear force (V) and bending moment (M) at any section located at x distance from the left support.


200 lb

5 ft 5 ft

x

y

Step 1 – Draw FBD and determine the reactions at the support

200 lb

RAY RBY

A B


5 ft 5 fty

Step 1 – Draw FBD and determine the reactions at the support

ΣMA = 0.0; RBY(10) – (200)(5) = 0.0; RBY = 100 lb

ΣFY = 0.0; RAY + RBY – 200 = 0.0; RAY = 100 lb

200 lb

RAYRBY

xA B


5 ft 5 fty

200 lb

RAYRBY

xA B

Step 2 – draw FBD for the section in question.For this problem, 2 sections should be analyzed one from A to the left of the 200 lb force and the other from A to the right of the 200 lb force.

EXAMPLE 1 - EXAMPLE 1 - SIMPLY SUPPORTED BEAMSIMPLY SUPPORTED BEAM y

xRAY = 100 lb

x

FBD for 0.0 < x < 5.0

MV

Shear V = 100 lb (constant and independent of xMoment M = 100 x

y

xRAY = 100 lb

x

FBD for 5.0 < x < 10.0

MV

Shear V = 100-200 = -100 lb independent of x Moment M = 100x – 200(x-5)

200 lb

5 ft


Variation of the shear (Variation of the shear (VV) and bending moment () and bending moment (MM) ) along the beamalong the beam

0.0 2.0 4.0 6.0 8.0 10.0

SHEAR

100

0.0

-100

MOMENT

500

0.0

(Lb-ft)(lb)

Shear

Moment


xRAY = 100 lb

x

FBD for 0.0 < x < 5.0

MV

What does the moment do at any section?

Compression

TensionShear

σt

σc


Compression

TensionShear

Compression

TensionShear

h 2h/3

EXAMPLE 2EXAMPLE 2

If P = 600 lb, a = 5 ft and b = 7 ft, draw the If P = 600 lb, a = 5 ft and b = 7 ft, draw the SFD and BMDSFD and BMD

SOLUTION 2SOLUTION 2Solve for support reactions

600 lb

5 ft 7 ftRAY

RAX

RBY

MA = 0; RBY(12) – 600(5) = 0; RBY = 250 lb

FY = 0; RAY – 600 + 250 = 0; RAY = 350 lb

Fx = 0; RAX = 0

SOLUTION 2 -Contd.SOLUTION 2 -Contd.Draw SFD and BMD

350 lb 250 lb

600 lb

5 ft 7 ft

SFD

BMD

350 lb

– 250 lb

0 lb

0 + 5(350) = 1750 lb-ft

1750 – 7(250) = 0 lb-ft

EXAMPLE 3EXAMPLE 3

Draw SFD and BMD for the beamDraw SFD and BMD for the beam

200 lb/ft

5 ft

1000 lb

EXAMPLE 3EXAMPLE 3200 lb/ft

5 ft

1000 lb

RAY

RAX

RBY

+

+

SOLUTION 3SOLUTION 3Determine V and M as a function of x for 0.0 < x < 20

RAY

RAX

RBY

xxVC MC

200 lb/ft

•C


RAY

RAX

RBY

xxVC MC

200 lb/ft

•C

+


RAY

RAX

RBY

x VD MD

200 lb/ft

•D


RAY

RAX

RBY

x

VD MD

200 lb/ft

•D

+


RAY

RAX

RBY

x VD MD

200 lb/ft

•E

1000lb5ft


RAY

RAX

RBY

x VD MD

200 lb/ft

•E

1000lb5ft

+

SOLUTION 3 -Contd.SOLUTION 3 -Contd.

Applied External Load

Internal Shear Force

Internal Bending Moment

-2260 lb

1740 lb

7569 lb-ft

200 lb-ft

1000 lb

5200 lb

EXAMPLE 4 – EXAMPLE 4 – Cantilever BeamCantilever Beam

4 feet

6 inch

75 lb/ft/ft

1. Calculate the reaction at the support

2. Plot the bending moment and shear force diagrams along the beam

3. Calculate the maximum internal stresses

EXAMPLE 4 – EXAMPLE 4 – Cantilever StructureCantilever Structure

4 feet

6 inch

75 lb/ft/ft

300 lb/ft

300 lb/ft600 lb-ft/ft

2 ft 2 ft


75 lb/ft/ft

300 lb/ft

600 lb-ft/ft xV

M

V = 300 – 75x, linear relative to x

M = -600 +300x – 75x2/2 nonlinear relative to x


4 feet

6 inch

75 lb/ft/ft

300 lb/ft

600 lb-ft/ft


4 feet

6 inch

75 lb/ft/ft

300 lb/ft

600 lb-ft/ft

Assume that along the beam cross-section, the shear stresses are uniformly distributed and the bending stresses are linearly distributed and that the neutral axis remains in the middle of the beam, calculate the maximum shear and bending stresses


4 feet

6 inch

75 lb/ft/ft

300 lb/ft

600 lb-ft/ft

The 300 lb/ft shear force creates shear stress distributed along the beam cross-section, while the -600 lb-ft/ft bending moment creates bending stresses distributed against the beam cross-section.


The average shear stress can be calculated by dividing the shear force by the cross-sectional area as follows:

For V = 300 lb/ft, The average Shear Stress =

τaverage = 300/0.5 = 600 lb/ft/ft

Distribution of the shear stress along the cross section


The Maximum bending moment = -600 lb-ft/ft, this creates tensile and compressive bending stresses that are linearly distributed along the beam cross-section. The maximum stresses can be calculated as follows:

σc = σt ; (σc)[(h/2)/2)(2)[2(h/2)/3][1/122] = M lb-ft/ft

σc = σt ; (σc)(h2/6) = M lb-ft/ft

σ t max

σ c max

Neutral Axis

h/2

h/2


The Maximum bending moment = -600 lb-ft/ft, this creates tensile and compressive bending stresses that are linearly distributed along the beam cross-section. The maximum stresses can be calculated as follows:

σc = σt ; (σc){(3/2)(2)(6)/3}{1/144} = 600 lb-ft/ft

σc = σt = (6)(600)/36 = 14400 lb/ft2 =100 psi

σ t max

σ c max

Neutral Axis

3 inch

3 inch


5 feet

9 inch

110 lb/ft/ft

RV

M

800 lb/ft 1 ft

1 point – Calculate the reaction RV at the support

1 point – Calculate the moment M at the support

1 point – Draw the shear diagram (values on the axis)

1 point – Draw the moment diagram (values on the axis)


800 lb/ft550 lb/ft

RV = 1350 lb/ft

M = 4575 lb-ft/ft

9 inch

ΣFY = 0; RV – 550 – 800 = 0; RV = 1350 lb/ft

ΣM = 0; M – 550(2.5) – 800(4) = 0;

M = 4575 lb-ft/ft


110 lb/ft/ft

1350 lb/ft

-4575 lb-ft/ft0 < X < 4 ft

V

M

ΣFY = 0; 1350 – 110X – V = 0; V = 1350 – 110X

ΣM@X = 0; 1350 X – 4575 - 110X (X/2) + M = 0;

M = 4575 + 55X2 – 1350X


110 lb/ft/ft

1350 lb/ft

-4575 lb-ft/ft4 < X < 5 ft

V

M

ΣFY = 0; 1350 – 110X – 800 - V = 0; V = 1350 – 110X - 800

800 lb4 ft

ΣM@X = 0; -4575 + 1350X – 110X (X/2) – 800 (X-4) + M = 0;

M = 1375 – 550X + 55X2

EXAMPLE 5 EXAMPLE 5

The cantilever beam shown below is subjected to two triangular distributed load. Draw the moment and shear diagram, determine the locations of maximum shear and maximum moment and calculate the maximum tensile and compressive stresses (the beam cross-section is 1 ft by 1 ft).

EXAMPLE 5 - Solution EXAMPLE 5 - Solution

Step 1 – Calculate the reactions at the support

FYA

MA

3 kN 1 m3 kN1 m

mkN81M

0(3)(5)-(3)(1)-M0Aat M

kN6F033F0F

A

A

YAYAY

EXAMPLE 5 EXAMPLE 5

X

VD; MD

D

2-2X/3 kN/m2X/3 kN/m

Total rectangle; 2X-2X2/3 kN located at X/2 from A

Total triangle; X2/3 located at X/3 from A

2X3

X6V

0V)3

2X(2X

3

X60;F

2

D

D

22

Y

9

XX6X18

3

XX

9

2X6X18M

0M2

X

3

2X2X

3

2X

3

X6X180;M@D

32

32

3

D

D

22

0>X>3m

EXAMPLE 5 EXAMPLE 5

E

VE; ME

3 kN

3

3)-(X3V

0V2

3)(X

3

2360;F

2

E

E

2

Y

EXAMPLE 5 EXAMPLE 5

E

VE; ME

3 kN

9

3315M

0M3

3

2

3X3

3

21)-3(X6X-180;@M

3

E

E

XX

XXE

SHEAR & MOMENT DIAGRAMSSHEAR & MOMENT DIAGRAMS

STRESSESSTRESSES

• Maximum moment = 18 kN-m• Maximum shear = 6 kN• Maximum shear stress = 6/1= 6 kPascal• Assume the neutral axis is located in the middle

of the beam, • The maximum tensile stress = the maximum

compressive stress

• σt = σc = Mmax(6)/(12) = 108 kPascal

EXAMPLE 7 EXAMPLE 7

Shear

Replace the distributed load by equivalent forces and determine their locations:

Bfromfeet5atlocatedkips055(10)

Bfromft 2at and Afromfeet2atlocatedkips515(6)/2

orAeither

15 kip50 kip

15 kip

EXAMPLE 7 EXAMPLE 7

Determine the reactions at supports A and B:

kip40FF BYAY

Shear

15 kip50 kip

15 kip

Because of symmetry:

FAY = 40 kip FBY = 40 kip

EXAMPLE 7 EXAMPLE 7

Shear

36

5X15

3

X

12

5X15M0@XM

12

5XV0F

32

X

2

XXY

1. Make a cut at 0 < X < 6 ft

/65X

X

Vx MX

(5X/6)(X/2) = 5X2/12 2. Calculate the magnitude of the

distributed force at the cut 3. Replace the distributed load by

an equivalent force

4. Calculate the shear force and moment at the cut

EXAMPLE 7 EXAMPLE 7 5.Make a cut at 6 < X < 16 ft & replace the distributed load

2

6)-5(X)6(40)4(1515M0@XM

)6X(5(25)65(X4015V0F2

X

XXY

XX

Vx MX

X

15 kip @ 2’ from A 5(X-6) @(X-6)/2 from A

6.Calculate the shear and moment at the cut

EXAMPLE 7 EXAMPLE 7

Shear

7.Make a cut at 16 < X < 22ft

Vx MXX

15 kip @ 2’ from A 50 kip @5’ from A [5-(5/6)(X-16)][X-16] @ (X-16) /2 from B

(5/6)(X-16)2/2 @ [(X-16)/3]’from B

8.Replace the distributed load by equivalent forces

40 kip

3

162

2

)16(

6

5

2

)16(16

6

55)16(40

)11(50)6(40)4(1515M0@XM

16XX226

5]16XX22

6

5-[540-504015V0F

22

X

XXY

XXXXX

XXX

EXAMPLE 7EXAMPLE 7

Shear

Vx MXX

15 kip @ 2’ from A 50 kip @5’ from A

40 kip

[5-(5/6)(X-16)][X-16] @ (X-16) /2 from B

(5/6)(X-16)2/2 @ [(X-16)/3]’from B

Calculate the shear and moment at the cut

EXAMPLE 7 EXAMPLE 7

EXAMPLE 7 EXAMPLE 7

Shear

EXAMPLE 7 EXAMPLE 7

Shear

Assume the cross section of the beam is 1 ft by 1 ft as shown, calculate the maximum tensile stress.

h = 1 ft

b = 1 ft

EXAMPLE 7EXAMPLE 7The maximum moment at the support is -45 kip-ft causing tension at the top

σt

σC

σt

σC

Assume the neutral axis is located in the middleFt = FC = σt(b)(h/2)/2= σt(bh)/4 located at h/6 from top or bottom M = σt(bh/4)(4h/6) = σt(bh2)/6 = 2σt(bh2/12) σt = (1/2)M(12/bh2)(h/h) = M(h/2)(12/bh3) = MC/I

EXAMPLE 8 EXAMPLE 8

Draw the shear and moment diagrams along the beam

EXAMPLE 8 EXAMPLE 8

Replace the distributed load by equivalent load and calculate the reactions at supports A and C using symmetry.

4.5 kN @ 2m from C4.5 kN @ 2m from A

FCY = 4.5 kNFAY = 4.5 kN

EXAMPLE 8 EXAMPLE 8

1. Make a cut at 0 < X < 3 m

VX MX

4.5 kN

3X/3=X

2. Calculate the value of the distributed load at the cut

EXAMPLE 8 EXAMPLE 8

C VX MX

4.5 kN

3X/3=X

3. Calculate the internal shear and moment∑FY=0; 4.5 – X2/2 - VX = 0; VX = 4.5 – X2/2 ∑M@C=0; 4.5X – (X2/2)(X/3) - MX = 0MX = 4.5X – X3/6

EXAMPLE 8 EXAMPLE 8

VX MX4.5 kN

3-3(X-3)/3

4. Make a cut at 3 < X < 6 m

5. Calculate the value of the distributed load at the cut

D

EXAMPLE 8 EXAMPLE 8

VX MX4.5 kN

3-3(X-3)/3

6. Break up the trapezoidal area to rectangle and triangle and calculate the height of the triangle

3(X-3)/3

EXAMPLE 8 EXAMPLE 8

6. Replace the distributed load by equivalent loads

VX MX4.5 kN

3-3(X-3)/3[3-3(X-3)/3)](X-3)

3(X-3)/34.5

3(X-3)2/6

EXAMPLE 8 EXAMPLE 8

VX MX4.5 kN

[3-3(X-3)/3)](X-3)4.5

3(X-3)2/6

6. Calculate VX

∑VX = 0; 4.5 - 4.5 - 3(X-3)2/6 - [3-3(X-3)/3][(X-3)] – VX =0

VX = -3(X-3)2/6 + 9X- X2 -18 –VX = 0VX = -1.5X2 +12X - 22.5

EXAMPLE 8 EXAMPLE 8

6. Calculate MXMX = 4.5X - 4.5(X-2) – [3(X-3)2/6][2/3][X-3] - [3-3(X-3)/3]

[(X-3)][(X-3)/2]MX = 9 –[3(X-3)2(2/3)(X-3)]/6 -3(X-3)2/2 + (3(X-3)3/6)

VX MX4.5 kN

[3-3(X-3)/3)](X-3)4.5

3(X-3)2/6

EXAMPLE 8 EXAMPLE 8

EXAMPLE 8 EXAMPLE 8

EXAMPLE 8 EXAMPLE 8


EXAMPLE 8 EXAMPLE 8

1. Replace the distributed load by equivalent load

2. Calculate the reactions at support A

WL/4 @ L/6 from A WL/4 @ 4L/6 from A

A

FAX = 0FAY = WL/2MA = 5WL2/24

EXAMPLE 8 EXAMPLE 8

3. Make a cut for 0 < X < (L/2)4. Divide the remaining distributed load to rectangular and triangular areas.5. Replace the distributed load by equivalent loads

w-2Xw/LX(w-2Xw/L) @ X/2 from A

X2w/L) @ X/3 from A

A

XFAY = WL/2MA = 5WL2/24

VX; MX

2Xw/L

EXAMPLE 8 EXAMPLE 8

6. Calculate VX and MX

VX = wL/2 – X2w/L – X(w-2Xw/L)MX = 5wL2/24 - XwL/2 + (X/3)(X2w/L) + (X/2)(X)(w-2Xw/L)

w-2Xw/L

2Xw/L

X(w-2Xw/L) @ X/2 from A

X2w/L) @ X/3 from A

A


VX; MX

EXAMPLE 8 EXAMPLE 8

3. Make a cut for (L/2) < X < (L)4. Divide the remaining distributed loads to rectangular and triangular areas.5. Replace the distributed load by equivalent loads

w-2(X-L/2)w/ L

2(X-L/2)w/L(X-L/2)(w-2Xw/L) @ X/2 from A

(X-L/2)2w/L) @ (X-L/2/3 from A

A


VX; MX

WL/4 @ L/6 from A

EXAMPLE 8 EXAMPLE 8

6. Calculate VX and MX

VX = wL/2 – wL/4 – (X-L/2)2w/L) - (X-L/2)(w-2Xw/L) MX = 5wL2/24 – XwL/2 + (wL/4)(X-L/6) + (X-4L/6)(X-L/2)2w/L)(X-4L/6) + (X-L/2)(w-2Xw/L)(L/4+X/2)

w-2(X-L/2)w/ L

2(X-L/2)w/L(X-L/2)(w-2Xw/L) @ (L/4+X/2) from A

(X-L/2)2w/L) @ (X-4L/6) from A

AX

FAY = WL/2MA = 5WL2/24

VX; MX

WL/4 @ L/6 from A

EXAMPLE 9 EXAMPLE 9

EXAMPLE 9 EXAMPLE 9


EXAMPLE 9 EXAMPLE 9

Replace the distributed load by equivalent loads

20 kN @ 5m from A4 kN @ 12m from A7.5 kN @ 5m from A

6 kN @14m from A

EXAMPLE 9 EXAMPLE 9

Calculate the reactions at the support A

20 kN @ 5m from A4 kN @ 12m from A7.5 kN @ 10m from A

6 kN @14m from A

FAX

MA

FAY

FAX = 0.; FAY = 20 +7.5 + 4 + 6 = 37.5 kNMA = (20)(5) + (7.5)(10) + (4)(12) + (6)(14) + 40 = 347 kN-m

EXAMPLE 10 EXAMPLE 10







EXAMPLE 9 EXAMPLE 9

QUESTIONQUESTION

Which statement is correct?Which statement is correct?

a)a) The shear force diagram represents the The shear force diagram represents the area under the bending moment diagram.area under the bending moment diagram.

b)b) The bending moment diagram represents The bending moment diagram represents the area under the shear force diagramthe area under the shear force diagram

c)c) Both of the aboveBoth of the above

d)d) None of the aboveNone of the above

What does the values on the BMD What does the values on the BMD indicate ?indicate ?

A) External momentA) External moment

B) Internal momentB) Internal moment

C) Sum of external and internal C) Sum of external and internal momentsmoments

QUESTIONQUESTION

QuestionsQuestionsFor the beam shown, which of the following correctly For the beam shown, which of the following correctly represents the shape of SFD?represents the shape of SFD?

A)A)

B)B)

C)C)

D)D)

QUESTIONSQUESTIONSFor the SFD as shown below, which of the following For the SFD as shown below, which of the following correctly represents the shape of BMD?correctly represents the shape of BMD?

A)A)

B)B)

C)C)

D)D)

250 N

– 250 N

250 N 250 N

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T - Shear Force and Bending Moment Diagr-1