Chapter 07 Internal Forces - Statics 4300:201 07 PP Internal Forces...&KDSWHU ²,QWHUQDO )RUFHV 1H[W...

Chapter 7 – Internal Forces


Chapter Objectives• To show how to use the method of sections for determining the internal loadings in a

member.

• To generalize this procedure by formulating equations that can be plotted so thatthey describe the internal shear and moment throughout a member.


In our past discussions on trusses we have considered the internal forces in a straighttwo-force member.• These forces produced only “tension” or “compression” in the member.

• The internal forces in any other type of member will usually produce “shear” and“bending” as well.

This chapter is devoted to the analysis of the internal forces in “beams.”• Beams are long, straight prismatic members designed to support loads that are

applied perpendicular to the axis of the member and at various points along themember.


7.1 Internal Loadings Developed in Structural MembersConsider the case of a straight two-force member.

Cut the member at C.• To maintain equilibrium an internal axial

force must exist.


Next consider the case of a multi-forcemember.• The internal forces in beam AB are not

limited to axial tension or compressionas in the case of straight two-forcemembers.

• The internal forces also include “shear”and “bending.”


By the “method of sections” we can take a cut at anypoint along the length of the member to find theinternal resisting effects – axial force (P), shearforce (V), and bending moment (M).• If the entire beam is in equilibrium, then any

portion of the beam is in equilibrium.

• Equilibrium of the isolated portion of the beam isachieved by the internal resisting effects, where

P – axial force

V – shear force

M – bending moment


Various Types of Loading and SupportA “beam” is a structural member designed to support loads applied at various pointsalong the member.• In most cases, the loads are applied perpendicular to the axis of the beam and will

cause only shear and bending.

• However, axial forces may be present in the beam when the applied loads are notperpendicular to the axis of the beam.


Beams are usually long, straight, and symmetrical (prismatic) in cross section (such aswide flange sections, commonly called I-beams or W-sections).• Designing a beam consists essentially in selecting the cross section that will provide

the most effective resistance to bending, shear, and deflection produced by theapplied loads.

The design of a beam includes the following steps.1. Determine the shear forces and bending moments produced by the applied loads.

2. Select the best suited cross section.


A beam may be subjected to various types of applied loads.

Concentrated loads Distributed loads* Both

*Distributed loads may be uniform, trapezoidal, or triangular.


A beam may be supported in a number of ways.• The distance between the supports, L, is referred to as the “span.”

• The following beams are “statically determinate.”

Simply supported Overhanging Cantilever


The following beams are “statically indeterminate.”

Continuous Propped cantilever Fixed-Fixed


Free-Body DiagramThe following steps are used to begin the analysis or design of a beam.• The entire beam is taken as a free-body diagram and the reactions are determined at

the supports.

• Then, to determine the internal forces at any point along the length of the beam, wecut the beam and draw the free-body diagram of that portion of the beam.

• Using the three equations of equilibrium we may determine the axial force (P), shearforce (V), and bending moment (M) at any point along the length of the beam.

∑ Fx = 0 yields the axial force (P)∑ Fy = 0 yields the shear force (V)∑ Mcut = 0 yields the bending moment (M)


Sign ConventionsThe following sign conventions are used for the internal effects.

Axial force

Shear force

Bending moment


Methods of analysisThree methods of analysis will be demonstrated.• Determining values for shear and bending moment at distinct points on a beam.

• Writing the equations for shear and bending moment.

• Drawing shear and bending moment diagrams by understanding the relationshipsbetween load, shear, and bending moment.


Example – Values of Shear and Moment at Specific Locations on a Beam

Given: The beam loaded as shown.

Find: V and M @ x = 0+’, 2’, 4-’, 4+’,6’, 8-’, 8+’, 10’, and 12-’


x = 0+’∑ Fy = 0 = 10 – V V = 10 k

∑ Mcut = 0 = - 10 (0) + M M = 0 k-ft

x = 2’∑ Fy = 0 = 10 – V V = 10 k

∑ Mcut = 0 = - 10 (2) + M M = 20.0 k-ft


x = 4-’∑ Fy = 0 = 10 – V V = 10 k

∑ Mcut = 0 = - 10 (4) + M M = 40.0 k-ft

x = 4+’∑ Fy = 0 = 10 – 12 - V V = - 2 k

∑ Mcut = 0 = - 10 (4) + 12 (0) + MM = 40.0 k-ft


x = 6’∑ Fy = 0 = 10 – 12 - V V = - 2 k

∑ Mcut = 0 = - 10 (6) + 12 (2) + MM = 60 – 24 = 36.0 k-ft

x = 8-’∑ Fy = 0 = 10 – 12 - V V = - 2 k

∑ Mcut = 0 = - 10 (8) + 12 (4) + MM = 80 – 48 = 32.0 k-ft


x = 8+’∑ Fy = 0 = 10 – 12 – 6 - V

V = - 8 k

∑ Mcut = 0 = - 10 (8) + 12 (4) + 6 (0) + MM = 80 – 48 – 0 = 32.0 k-ft

x = 10’∑ Fy = 0 = 10 – 12 – 6 - V

V = - 8 k

∑ Mcut = 0 = - 10 (10) + 12 (6) + 6 (2) + MM = 100 – 72 – 12 = 16.0 k-ft


x = 12-’∑ Fy = 0 = 10 – 12 – 6 - V

V = - 8 k

∑ Mcut = 0 = - 10 (12) + 12 (8) + 6 (4) + MM = 120 – 96 – 24 = 0 k-ft


To simplify the calculations, use a free body diagram from the right side.

x = 10’∑ Fy = 0 = V + 8 V = - 8 k

∑ Mcut = 0 = - M + 8 (2) M = 16.0 k-ft

x = 12-’∑ Fy = 0 = V + 8 V = - 8 k

∑ Mcut = 0 = - M + 8 (0) M = 0 k-ft

Note: The moment is zero at the supports of simply supported beams.


Example – Values of Shear and Moment at Specific Locations on a Beam


Find: V and M @ x = 0+’, 3’, 6-’, 6+’,7.5’, and 9-’


x = 0+’∑ Fy = 0 = 19 – V V = 19 k

∑ Mcut = 0 = - 19 (0) + M M = 0 k-ft

x = 3’∑ Fy = 0 = 19 – ½ (6) 3 – ½ (4.5) 3 – V

V = 19 – 9 – 6.75 = 3.25 k

∑ Mcut = 0 = - 19 (3) + ½ (6) 3 (2/3) 3 + ½ (4.5) 3 (1/3) 3 + MM = 57 – 18 – 6.75 = 32.25 k-ft


x = 6-’∑ Fy = 0 = 19 – ½ (6) 6 - ½ (3) 6 – V

V = 19 – 18 – 9 = - 8.0 k

∑ Mcut = 0 = - 19 (6) + ½ (6) 6 (2/3) 6+ ½ (3) 6 (1/3) 6 + M

M = 114 – 72 – 18 = 24.0 k-ft

x = 6+’∑ Fy = 0 = V + 8 V = - 8.0 k

∑ Mcut = 0 = - M + 8 (3) M = 24.0 k-ft


x = 7.5’∑ Fy = 0 = V + 8 V = - 8.0 k

∑ Mcut = 0 = - M + 8 (1.5) M = 12.0 k-ft

x = 9-’∑ Fy = 0 = V + 8 V = - 8.0 k

∑ Mcut = 0 = - M + 8 (0) M = 0 k-ft


7.2 Shear and Moment Equations and DiagramsThe plot of values of shear force and bending moment against a distance “x” are graphscalled “shear and bending moment diagrams.”• Rather than pick points along a beam and approximate these diagrams by determining

values of shear and bending moment at a limited number of points, we can writeequations to define functions of shear and moment at any “x” along the beam.


The equations for shear and bending moment may be developed by the following steps.• First, determine the number of sets of equations needed to define shear force and

bending moment across the entire length of the beam.- The shear and bending moment equations will be discontinuous at points where a

distributed load begins or ends, and where concentrated forces or concentratedcouples are applied.

• Next, for each set of equations, draw the necessary free-body diagram.


• Then, for each set of equations, write the equilibrium equations and solve for shearforce (V) and bending moment (M) in terms of position on the beam (i.e. “x”, where“x” is measured from the left end of the beam).- Use ∑Fy = 0 to develop the equations for shear force.

- Use ∑Mcut = 0 to develop the equations for the bending moment


Example – Shear and Moment Equations

Given: Beam loaded as shown.

Find: Write the V and M equations and draw theV and M diagrams.

Two sets of equations are required.


First, solve for the reactions at the supports.∑ MB = 0 = - Ay L + P (L/2)

Ay = P/2

∑ MA = 0 = By L - P (L/2)By = P/2

Note: The beam is symmetrical and symmetricallyloaded; thus, the reactions are symmetrical as well.


0 < x < L/2∑ Fy = 0 = P/2 – V

V = P/2

∑ Mcut = 0 = - (P/2) x + MM = Px/2

L/2 < x < L∑ Fy = 0 = P/2 – P – V

V = - P/2

∑ Mcut = 0 = - (P/2) x + P (x – L/2) + MM = Px/2 – Px + PL/2M = P/2 (L – x)


0 < x < L/2V = P/2M = Px/2

L/2 < x < LV = - P/2M = P/2 (L – x)

The shear and moment diagrams forthe beam are shown at the right.




Find: Write the V and M equations and draw theV and M diagrams.

One set of equations is required.


0 < x < L∑ Fy = 0 = wL/2 – wx - V

V = w (L/2 – x)

∑ Mcut = 0 = - (wL/2) x + wx (x/2) + MM = - wx2/2 + wLx/2M = (wx/2) (L – x)


0 < x < LV = w (L/2 – x)

M = (wx/2) (L – x)

The shear and moment diagrams for thebeam are shown at the right.




Find: Write the V and M equations and draw the V andM diagrams.

Three sets of equations are required.


First, solve for the reactions at the supports.• First, take moments about the roller support at the

right end of the beam to find the vertical reaction atthe pin support at the left end of the beam.

∑ MR = 0 = - 10 Ay + 10 (8) – 46 + ½ (3) 6 [2 + (2/3) 6]+ ½ (6) 6 [2 + (1/3) 6]

10 Ay = 80 – 46 + 9(6) + 18(4) = 160.0

Ay = 16.0 kips ↑

Note: The distributed load (a trapezoid) is treated as twotriangles.


• Then, take moments about the pin support at theleft end of the beam to find the vertical reaction atthe roller support at the right end of the beam.

∑ ML = 0 = 10 By - 10 (2) – 46 - ½ (3) 6 [2 + (1/3) 6]- ½ (6) 6 [2 + (2/3)6]

10 By = 20 + 46 + 9(4) + 18(6) = 210.0

By = 21.0 kips ↑


0’ < x < 2’∑ Fy = 0 = 16 - V

V = 16

∑ Mcut = 0 = - 16x + MM = 16x


2’ < x < 8’Since the intensity of the distributed load varies, an equation is needed to define theintensity of the distributed load (p) as a function of position “x” along the beam.

Using the general equation of a line: p = m x + c• Determine the slope m.

m = (6 – 3)/(8 – 2) = ½ So, p = ½ x + c

• Next, determine the “y-intercept” value “c”.Known points on the line include:

when x = 2, p = 3 and when x = 8, p = 6.

Substituting the second condition into the equation p = ½ x + c:6 = ½ (8) + c c = 6 – 4 = 2 Thus, p = ½ x + 2


Now write the equations for shear and moment.

∑ Fy = 0 = 16 – 10 – ½ (3) (x – 2) – ½ (½ x + 2) (x – 2) - V

V = 16 – 10 – (3/2) (x – 2) – ½ (½ x + 2) (x – 2)= 6 – 3x/2 + 3 – ½ (x2/2 + x – 4)= 6 – 3x/2 + 3 - x2/4 - x/2 + 2

V = - x2/4 – 2x + 11 V = 0 k @ x = 3.75’


Now write the equations for shear and moment.∑ Mcut = 0 = - 16x + 10 (x – 2) + ½ (3) (x – 2) (2/3) (x – 2)

+ ½ (½ x + 2) (x – 2) (1/3) (x – 2) + M - 46

M = 16x - 10 (x – 2) - (x – 2)2

– (1/6) (½ x + 2) (x – 2)2 + 46

= 16x – 10x + 20 - (x2 – 4x + 4)– (1/6) (x3/2 – 6x + 8) + 46

= 6x + 20 – x2 + 4x – 4 – x3/12 + x – 1.33 + 46

M= – x3/12 – x2 + 11x + 60.67

M = 83.5 k-ft @ x = 3.75’


8’ < x < 10’∑ Fy = 0 = V + 21

V = - 21 k

∑ Mcut = 0 = - M + 21 (10 – x)M = 21 (10 – x)


0’ < x < 2’V = 16 M = 16x

2’ < x < 8’V = - x2/4 – 2x + 11M= – x3/12 – x2 + 11x + 60.67@ x = 3.75’, V = 0, M = 83.5 kip-ft

8’ < x < 10’V = - 21 k M = 21 (10 – x)

The shear and moment diagrams for the beamare shown at the right.


7.3 Relations between Distributed Load, Shear, and MomentThe methods outlined so far for drawing shear and bending moment diagrams becomeincreasingly cumbersome the more complex the loading.• The construction of shear and bending moment diagrams, however, can become

greatly simplified by understanding the relationships that exist between thedistributed load, the shear force, and the bending moment.


Relation between the Distributed Load and ShearFrom the free body diagram:

∑ Fy = 0 = V – (V + ∆V) – w ∆x,where w = w (x) = constant for small ∆x

∆V = – w ∆x

∆V/∆x = - w(x), then letting ∆x → 0,by definition of a derivative,

dV/dx = - w(x)


First equation: dV/dx = - w(x)

If we integrate this expression between two points,then

dV = - w(x) dx

∫ dV = - ∫12 w(x) dx

V2 – V1 = - w(x) dx


Interpretation of these first two expressions:

dV/dx = - w(x) The value of the slope on the shear diagram is equal to theheight of the load diagram at that point times minus one.

V2 – V1 = - ∫12 w(x) dx The change in shear between two points is equal to the area

under the load diagram times minus one.

Concentrated ForcesThese equations are not valid under a concentrated load.• The shear diagram is discontinuous at the point of a concentrated load.


Relation between the Shear and Bending MomentFrom the free body diagram:

∑ MR = 0 = - M – V ∆x + w (∆x)2/2 + M + ∆M

∆M = V ∆x – ½ w (∆x)2

∆M = V ∆xNote: When ∆x → 0, (∆x)2 ≈ 0

∆M/∆x = V, then letting ∆x → 0,by definition of a derivative,

dM/dx = V


Third equation: dM/dx = V

If we integrate this expression between two points,then

dM = V dx = V(x) dx

Note: Let V = V(x) since the shear may vary.

∫ dM = ∫ V(x) dx

M2 – M1 = ∫12 V(x) dx


Interpretation of these two expressions:

dM/dx = V(x) The value of the slope on the bending moment diagram isequal to the height of the shear diagram at that point.

Note: When shear is zero, the slope on the moment diagram is zero corresponding toa point of maximum bending moment.

M2 – M1 = ∫12 V(x) dx The change in bending moment between two points is equal

to the area under the shear diagram.

Concentrated CouplesThese equations are not valid under a concentrated couple.• The bending moment diagram is discontinuous at the point of a concentrated couple.


Example – Shear and Moment Diagrams


Find: Draw the shear and moment diagrams.


Draw the shear diagramBetween the left end and right end of the beam:• The area under the load diagram = + wL

∆V = - Area = - wL

V2 = V1 + ∆V = wL/2 – wL = - wL/2


Draw the moment diagramBetween the left end of the beam and mid-span:• The area under the shear diagram

= ½ (wL/2)(L/2) = wL2/8

∆M = Area = wL2/8

M2 = M1 + ∆M = 0 + wL2/8 = wL2/8

• At mid-span (i.e. x = L/2), V = 0, so dM/dx = 0.


Between mid-span and the right end of the beam:• The area under the shear diagram

= - ½ (wL/2)(L/2) = - wL2/8

∆M = Area = - wL2/8

M2 = M1 + ∆M = wL2/8 + (- wL2/8) = 0

In general, when V = 0 at a point, the slope on themoment diagram at that point is zero (i.e. ahorizontal tangent).






Draw the shear diagram


Draw the moment diagram






Draw the shear diagram


Draw the moment diagram






Solve for the reactions at the supports.• Take moments about point A to find the vertical reaction

at point D.

∑MA = 0 = - ½ (4) 12 [(2/3) 12] – 20 (15) – 120 + 18 Dy

18 Dy = 192 + 300 + 120 = 612.0Dy = 34.0 k ↑

• Take moments about point D to find the vertical reactionat point A.

∑MD = 0 = ½ (4) 12 [6 + (1/3) 12] + 20 (3) – 120 - 18 Ay

18 Ay = 240 + 60 - 120 = 180.0Ay = 10.0 k ↑


Draw the shear diagramBetween x = 0’ and x = 12’• The area under the load diagram

= + ½ (12) 4 = + 24

∆V = - Area = - 24

V2 = V1 + ∆V = 10 – 24 = - 14


Draw the moment diagramIn order to draw the first part of the moment diagram, the location for the point ofmaximum moment (i.e. V = 0 kips) needs to be determined so that the area under theshear diagram may be calculated.• Write an equation for shear between x = 0’ and x = 12’.

∑ Fy = 0 = 10 – ½ (x) (x/3) – VV = 10 – x2/6

Let V = 0, thus 0 = 10 – x2/6x2 = 60.0x = 7.75’


• The change in moment from x = 0’ to x = 7.75’ isequal to the area under the shear diagram.

Δ M = (2/3) 10 (7.75’) = 51.7 kip-ft

From x = 7.75’ to x = 12’, the change in moment isequal to the area under the shear diagram.• The area under the portion shear diagram between

x = 7.75’ and x = 12’ cannot be calculated usingstandard formulas since there is no horizontaltangent at x = 7.75’ or at x = 12’.


• Determine the value of moment at x = 12’ and plot this value onthe moment diagram.- Take a cut at x = 12’, draw the free body diagram, and write the

equilibrium equation to determine the value of moment.

∑ Mcut = 0 = MB – 10 (12) + ½ (12) 4 (1/3)(12)

MB = 120 – 96 = 24.0 kip-ft

- Then plot this value on the moment diagram.


• The change in moment from x = 12’ to x = 15’ is equalto the area under the shear diagram.

Δ M = - 14 (3) = - 42

M2 = M1 + ∆M = 24 + (- 42) = - 18


• The concentrated couple at point C causes themoment diagram to jump up by 120 kip-ft.- Since the applied moment is acting clockwise, the

internal resisting moment increases, thus thejump “up” on the moment diagram.

• The change in moment from x = 15’ to x = 18’ is equalto the area under the shear diagram.

Δ M = - 34 (3) = - 102

M2 = M1 + ∆M = 102 + (- 102) = 0






Draw the shear diagramBetween x = 0’ and x = 2’• The area under the load diagram = 0

∆V = 0

Between x = 2’ and x = 8’• The area under the load diagram

= + ½ (3 + 6) 6 = + 27∆V = - Area = - 27V2 = V1 + ∆V = 6 – 27 = - 21

Between x = 8’ and x = 10’• The area under the load diagram = 0

∆V = 0


Draw the moment diagramBetween x = 0’ and x = 2’• The area under the shear diagram

= 16 (20 = + 32∆M = + 32M2 = M1 + ∆M = 0 + 32 = 32


In order to draw the part of the moment diagram beyond x = 2’, the location for thepoint of maximum moment (i.e. V = 0 kips) needs to be determined.• Write an equation for shear for 2’ < x < 8’.

∑ Fy = 0 = 16 - 10 – ½ (3) (x - 2) - ½ (½ x + 2) (x - 2) - VV = 6 – 3/2 (x – 2) – ½ (½ x + 2) (x – 2)

= 6 – 3x/2 + 3 – ½ (x2/2 + x – 4)= 6 - 3x/2 + 3 – x2/4 - x/2 + 2

V = - x2/4 – 2x + 11


The maximum moment occurs when V = 0.• Let V = 0; thus, 0 = - x2/4 – 2x + 11

• Using the quadratic equation, find “x” when V = 0.

x = - (- 2) ± [(- 2 )2 – 4 (- 1/4 ) (11)]1/2

2 (- 1/4)

x = - 11.75’ (Not reasonable: not within thelimits of 2’ < x < 8’.)

x = 3.75’


• Find M @ x = 3.75’

∑ Mcut = 0 = - 16 (3.75) + 10 (1.75) – 46+ ½ (3) 1.75 (2/3) 1.75+ ½ (3.75/2 + 2) (1.75) (1.75/3) + M

M = 16 (3.75) – 10 (1.75) + 46– ½ (3) 1.75 (2/3) 1.75- ½ (3.875) 1.75 (1.75/3)

M = 60 – 17.5 + 46 – 3.06 – 1.98

M = 83.5 k-ft


• Plot M @ x = 3.75’

M = 83.5 k-ft


From x = 3.75’ to x = 8’, the change in moment is equal to thearea under the shear diagram.• The area under the portion shear diagram between x = 3.75’

and x = 8’ cannot be calculated using standard formulas sincethere is no horizontal tangent at x = 3.75’ or at x = 8’.

• Determine the value of moment at x = 8’ and plot this value onthe moment diagram.- Take a cut at x = 8’, draw the free body diagram, and write

the equilibrium equation to determine the value of moment.

∑ Mcut = 0 = M – 21 (2) M = 42 kip-ft

- Then plot this value on the moment diagram.


From x = 8’ to x = 10’, the change in momentis equal to the area under the sheardiagram.• The area under the shear diagram

= (- 21) 2 = - 42

∆M = - 42

M2 = M1 + ∆M = 42 + (- 42) = 0

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