Lecture 5 1

Lecture 51

Outline:

-Shear stress and shear flow

-Built-up beams

-Shear stress distribution on the cross-section

1. Shear stress

Imagine that boards are stacked on top of each other as shown in Figure 1. In the first case there is no

bond between the boards so that during deformation they can slip over each other. In the second case the

boards are bonded together (by means of glue or nails) so that they doe not slip over each other and act

together as one unit.

(a) No bond between the boards (b) Boards are bonded together

Fig. 1 Beams stacked on top of each other

What is preventing slip in between the boards is the shear stress carried by the bond (glue or nails) as

shown in Figure 2.

(a) Slip between the unbonded boards (b) Shear stress between the bonded boards

Fig. 2 Slip and shear stress between the boards

As illustrated for the simply supported beam in Fig 1.a, when the boards slip over each other, at the

interface of the two boards the ends of the board on the top gets a head of the ends of the board on the

bottom as shown in Fig. 2(b). When the boards are bonded in order to prevent slip, the shear stress acting

at the bottom surface of the top board is pulling the ends of the beam back, on the other hand the stresses

1 Mechanics of Solids Dr Emre Erkmen

Lecture 5 2

acting at the top surface of the bottom beam is pushing the ends of the beam forward as shown in Fig.2 so

that they stay together as in Fig1. b.

2.1. How to calculate the shear stresses

Consider a small segment x∆ isolated from the loaded beam as shown in the Fig.3.

Fig. 3 Loaded beam

Vertical loads will cause internal shear force and bending moment. In Fig.4 notice that the internal forces

and the applied load are shown according to the positive sign convention on both sides of the beam

segment. If at x the internal shear force is V and bending moment is M, then at x x+ ∆ , they are V V+ ∆

and M M+ ∆ because in general, for different coordinates along the axis of the beam there will be

different shear force and bending moments. Because x∆ is small, M∆ and V∆ are also small.

Fig. 4 Isolated segment of x∆

We can write a moment equilibrium at x (counter-clockwise positive), i.e.

( ) ( ) 02

xM M M V V x q x

∆+ ∆ − − + ∆ ∆ − ∆ =

Because x∆ is small, the terms 2x∆ and x V∆ ∆ are very small so that they are negligible. As a result from

the equation above we can obtain a relation between the increment in the moment and the shear force as

M V x∆ = ∆

In the limit case 0x∆ → this relation becomes

d

d

MV

x=

The above equation states that if there is no internal shear force (due to applied loading) there is no

change in the bending moment.

Lecture 5 3

Longitudinal shear stress

The above equation states that if there is no internal shear force (due to applied loading) there is no

change in the bending moment.

By using the bending formula that we derived in lecture 4, i.e. My

Iσ = − we can express the normal

stresses acting on the left and right cross-sections as shown in Fig. 5.

M+ M

x

M

V V+ V

NA

y

tσ t tσ σ+ ∆

b b b

Fig. 5 Normal stresses acting on both sides of segment of x∆

Notice that at coordinate is x the stress at the top fibre is tσ and at the bottom fibre is bσ due to internal

bending moment M, which become t tσ σ+ ∆ at the top fibre and b bσ σ+ ∆ at the bottom fibre is at

coordinate x x+ ∆ due to change in the bending moment.

If we cut the segment horizontally (shown with the dotted line in Fig.5) and consider the longitudinal

equilibrium for the free-body shown in Fig. 6 a, since the normal stress distribution is different on the left

and right cross-sections there should be an equivalent force acting on the bottom surface in tangential

direction sF ,

tσ t tσ σ+ ∆q

Fs

Fig. 6 Free-body diagram of a piece cut from the beam

Lecture 5 4

Consider the forces acting in the axial direction as shown on the Free-body diagram in Fig. 7. The cross-

sectional area of the piece cut from the segment x∆ is shown as A’.

dA

( )M M y

I

+ ∆= −

σ My

I= −

σ σ+ ∆

Fs

A’

∆x

y

bNeutral axis

Fig. 7 Free-body diagram shown in perspective (only the forces in axial direction are shown)

From longitudinal equilibrium, i.e. 0xF =∑ , we can write

( )' '

d d 0s

A A

F A Aσ σ σ+ + ∆ − =∫ ∫

By using the bending formula, stresses at a distance y from the neutral axis can be expressed in terms of

the moments and the above equations becomes

' ' '

d d ds

A A A

M M M MF y A y A y A

I I I

+ ∆ ∆= − =

∫ ∫ ∫

Notice that the width is shown as b in Fig. 7, and as in Lecture 4 the axis is passing through the neutral

axis. Also note that the cut of with be is not necessarily at the neutral axis. In Fig. 7, indeed it is a cut

above the neutral axis (also check Fig.5).

Remember from lecture 1 that the shear stresses τ can be calculated from the shear force sF as

sF

b xτ =

∆

From the above two equations the average shear stress at the cut surface can be expressed as

'

dA

My A

xbIτ

∆=

∆∫ when 0x∆ → '

dA

Vy A

bIτ = ∫

When obtaining bending stresses we imposed kinematic conditions because it was a statically

indeterminate problem. When obtaining shear stresses we did not impose any further kinematic

conditions and we used only equilibrium. Thus obtaining shear stresses is a statically determinate

problem.

Lecture 5 5

New definition related to the cross-section

Since the shear force V, the moment of inertia I and the width b is independent of the coordinate y of a

point on the cross-section that is under axial stress, in the shear stress formula we can take them out of the

integration.

'

dA

V Vy A Q

bI bIτ = =∫

The integral term (which will be shown as Q ) is the first moment of the cut area with respect to the

neutral (centroidal) axis of the whole section. The first moment of area can be written as

'Q yA=

in which y is the distance of the centroid of the cut section with area A’ to the neutral axis of the whole

section as shown in Fig. 8..

A’

y

Centroid of Area A’

Fig. 8 Distance from the neutral axis to the centroid of the cut section

Shear flow q is defined as the shear force per unit length, i.e

sFq

x=∆ when 0x∆ →

Vq Q

I=

Summary

Shear stress:

V is the internal resultant shear force, determined from the method of section and

equilibrium.

I is the moment of inertia of the entire cross-sectional area about the neutral axis.

b is the width of the member’s cross sectional area, measured at the point where

shear stress is to be determined.

Q is the first moment of the area around the neutral axis.

VQ

bIτ = Shear flow:

VQq

I=

Lecture 5 6

Summary of the procedure for finding longitudinal shear stress or shear flow

1. Determine the location of the neutral axis and calculate the moment of inertia I of the entire cross sectional area about the neutral axis

2. Pass an imaginary horizontal section through the point where the shear stress (or flow) is to be determined.

3. Measure the width of the area at this section 4. Calculate the first moment Q for the area laying either above or below the section

Example

A beam is made of three planks, nailed together.

Knowing that the spacing between nails is 25 mm and

that the vertical shear in the beam is V = 500 N,

determine the shear force in each nail.

Solution

Based on the plan view shown below it can be discussed that if the planks were glued together, on the

glue a shear force of F would be acting for a span of L cut from the beam. That force should now be

shared by nails.

F

Fn

L

q q

Fig. Plan view

We can determine the shear flow q on the lower surface of the upper plank by using the formula

VQq

I=

The shear force for the problem is given as V=500kN

For the given I shape cross-section the moment of inertia can be calculated as

Lecture 5 7

( )( ) ( )( ) ( ) ( )3 3 21 112 12

6 4

0.020m 0.100m 2[ 0.100m 0.020m 0.020m 0.100m 0.060m ]

16.20 10 m

I

−

= + + ×

= ×

Calculating the first moment of area is the part of the solution that one must be most careful about since it

depends on the cut.

We are interested in the shear flow between the two planks, so we cut from there and look under the top

plank which forming the top flange. In this case the area above the cut is the area of the top flange.

'Q A y=

( )( ) 6 30.020m 0.100m 0.060m 120 10 mQ −= × = ×

The shear flow can be calculated as

6 3

-6 4

(500N)(120 10 m ) N3704m16.20 10 m

q−×

= =×

For a length of L=25mm the shear flow will be equivalent to a shear force nF qL= which is the area that

each nail covers, i.e.

(0.025m) (0.025m)(3704 92.6NnF q N m= = =

What about the nail connecting the bottom flange with the web?

Lecture 5 8

( )( ) 6 30.020m 0.100m 0.060m 120 10 mQ −= × − = − ×

6

-6

500 ( 120 10 ) N3704m16.20 10

q−× − ×

= = −×

The change of sign means the shear flow is in the opposite direction

(0.025m) (0.025m)( 3704 92.6NnF q N m= = − = − Since the section is symmetric about N.A the shear

force in each nail is same as the top nails however the direction is different,

Example

Find the shear flow in the welds holding a 150x10 plate on the top flange of the T-beams shown below to

carry 100N shear force.

150

welds

10

11.36

78.64zC

Izz = 2.084 x 106 mm4

Solution

The welds are providing the necessary shear force to hold the plate and the top flange together.

That means we should take a cut in between the plate and the top flange and look under the plate.

Thus the area that we should consider for the first moment of area Q is as shown in the figure below.

N.A

16.36

150

10

11.36

3 3150 10 16.36 24.54 10Q Ay mm= = × × = ×

By applying the flow formula

6

100 24,5401.178 /

2.084 10

VQq N mm

I

×= = =

×

This means each weld should carry q/2

/ 2 0.59N/mmq =

Lecture 5 9

Shear stress distribution on the cross-sectional surface

Can we find the shear stress distribution across the section (similar to the bending stresses in Lecture 4)?

C∆x

V

Fig. 9 Shear stresses acting on the cross-section

By using the expression developed for the longitudinal shear stresses (in x direction) acting at the cut

surface we can find the shear stress distribution across-the section (in y direction).

For this purpose we should consider the moment equilibrium cut from the segment whose front surface is

on the cross-sectional surface and top surface is at the intersection of the top and bottom components.

As shown in the side view of this cube in Fig. 10 the shear stress acting at the intersection surface is

xτ and the shear stress acting at the front surface is yτ .

Fig. 10 Side view of a cube cut from the segment

From the moment equilibrium with respect to A we can write

0AM = →∑ ( ) ( ) 0x ydxdz dy dydz dxτ τ× − × =

From which we obtain x yτ τ= . Thus, longitudinal shear stress always has its associated transverse shear

stress acting on the cross-section and two stresses are equal.

∆x

b

F s

F s

C

ττ

dz

dy

dx

yτdy

xτ

yτ

xτ

Lecture 5 10

Shear stress distribution for rectangular section

Take a in a beam of solid rectangular cross section with width b and height h transmitting a vertical shear

V as shown in Fig. 11.

Fig. 11 Rectangular section carrying shear force

The distribution of the shear stress throughout the cross section can be determined by computing the

shear stress at an arbitrary height y from the neutral axis as shown in Fig.12

VQ

bIτ =

Fig. 12 An arbitrary cut at y to calculate shear stresses

The first moment of area according to this arbitrary cut can be written as

2

21 1' '

2 2 2 2 4

h h hQ A y y y y b y b

= = + − − = −

The shear stress according to the arbitrary cut can be written as

2

2

3

6

4

VQ V hy

Ib bhτ

= = −

Thus, the shear stress is a quadratic function of y based on which the distribution across the section can be

shown as in Fig. 13.

Neutral axis

y

Lecture 5 11

Fig. 13 Shear stress distribution on the rectangular cross-section

The minimum value is at the top and the bottom of the section , where y = h/2 min 0τ→ =

The maximum value is at the neutral axis, where y = 0 2

max 3

6 31.5

4 2

V h V V

bh bh Aτ→ = = =

Remember the average shear stress is average

V

Aτ =

The maximum shear stress is 50% greater than the average shear stress!

Example

A cantilever beam of length L=2m supports a load P=8kN. The beam is made of wood with cross-

sectional dimensions 120mm x 200mm. Calculated the shear stresses due to the load P at points located 25mm, 50mm, 75mm and 100mm from the top surface of the beam. From these results, plot a graph

showing the distribution of shear stresses from top to bottom of the beam

For rectangular sections we know that the shear stress distribution can be calculated as

2

2

3

6

4

V hy

bhτ

= −

Lecture 5 12

Based on the above formula for shear stress distribution we can write

2

2

6

8000 200

160 10 4yτ

= − ×

Shear stress distribution for an I-section

Fig. 14 I-section subjected to shear force

Similar to rectangular sections the shear stress is a quadratic function of the y coordinate, however

because of the sudden change in the width b, a jump occurs on the shear stress distribution as shown in

Fig. 15.

Fig. 15 Shear stress distribution on the I-section

Lecture 5 13

Example

A wide-flange I beam is subjected to shear force of V=80kN. Plot the shear stress distribution acting over

the beam’s cross-sectional area. Also, determine the shear force resisted only by the web.

Solution

The shear-stress distribution will be parabolic and due to symmetry only the shear stresses at points B’,

shown in the figure below and the centroid of the whole section will be computed.

We must first determine the moment of inertia of the cross-section

( ) ( ) ( ) ( ) ( )( )3 3 21 112 12

6 4

0.015m 0.200m 2[ 0.300m 0.02m 0.300m 0.200m 0.110m ]

155.6 10 m

I

−

= + + ×

= ×

At point B’, width is b=0.3m and A’ is the dark shaded area shown in the figure above. The first moment

area can be calculated as

( )3 3

0.110m 0.300m 0.020m

0.66 10 m

Q yA

−

′= = ×

= ×

The shear stress at point B’ can be calculated as

3

6

80 0.66 101.13

156 10 0.3

VQ

Ibτ

−

−

× ×= = =

× ×MPa

At point B, width is b=0.015m. A’ is the same dark shaded area shown in the figure above and because of

that the first moment area is the same as 3 30.66 10 mQ yA −′= = × .

Lecture 5 14

The shear stress at point B can be calculated as

3

6

80 0.66 1022.6

156 10 0.015

VQ

Ibτ

−

−

× ×= = =

× ×MPa

At the centroid of the I-section the width is b=0.015m. The first moment of area can be calculated as

( ) ( )3 3

0.110m 0.300m 0.020m 0.050m 0.100m 0.015m

0.74 10 m

Q yA

−

′= = × + ×

= ×

The shear stress at the centroid can be calculated as

3

6

80 0.74 1025.2

156 10 0.015

VQ

Ibτ

−

−

× ×= = =

× ×MPa

The shear stress distribution is as shown in the figure below.

The shear force resisted only by the web ( ) ( ) ( )222.6 15 200 25.2 22.6 15 200 73

3webV kN= × × + − × × =

91% of the shear force is carried by the web. Usually in thin-walled beams web carry the most of the

vertical shear force acting on the cross-section.

Built-up thin-walled beams

What if the shear flow (or stress) that we are concerned with is not the one on horizontal surface as in Fig

16 a but on the vertical surface as in Fig. 16b.

(a) Vertical nails are needed (b) Horizontal nails are needed

Fig. 16 Built-up sections

Lecture 5 15

In cases where we need to connect to pieces whose contact surface is vertical we can still determine the

shear flow (or stress) according to a vertical cut as shown in Fig. 17.

(a) I-section subjected to shear force (b) Free-body of a piece cut from the flange

Fig. 17 Shear flow calculations in horizontal direction

In order to keep the dark shaded part shown in Fig. 17 a to the rest of the flange there has to be shear flow

acting as shown in Fig. 17 b. Similar to the previous calculations this shear flow can be calculated from

the longitudinal equilibrium of the free-body shown in Fig. 17 b.

At the front surface normal stresses due to bending moment add up to an equivalent force of F F+ ∆ . At

the back surface the normal stresses add up to an equivalent force of F. The shear flow on the side occurs

in order to balance that difference F∆ in longitudinal direction. If 0x∆ → then F dF∆ →

Example

Nails having a total shear strength of 40 kN are used in a beam that can be constructed either as in the

case I or case II. If the nails are spaced at 9cm, determine the largest vertical shear that can be supported

at each case so that the fastener will not fail.

Lecture 5 16

Since the geometry is same in both cases the moment of inertia is the same, which can be obtained as

( ) ( ) ( ) ( )3 3 41 112 123cm 5cm 2[ 1cm 4cm ] 20.58cmI = − =

Case I

An horizontal cut that is the contact surface between the web and the flange should be made in order to

determine the shear flow

Thus whole flange should be considered in calculating the first moment of area

( ) 32.25cm 3cm 0.50cm 3.375cmQ yA′= = × × =

The shear strength of a nail is 40N which covers a span of 9cm. In this case the shear flow that can be

carried by the nails can be calculated as

40 3.375

9 20.58

VQ Vq

I

×= = =

From the above equation the shear force capacity can be calculated as 27.1V N=

Case II

A vertical cut that is the contact surface between the web and half the flange should be made in order to

determine the shear flow.

Thus only half of the flange is considered

( ) 32.25cm 1cm 0.50cm 1.125cmQ yA′= = × × =

The shear strength of a nail is 40N which covers a span of 9cm. In this case the shear flow that can be

carried by the nails can be calculated as

40 1.125

9 20.58

VQ Vq

I

×= = =

From the above equation the shear force capacity can be calculated as 81.3V N=

Lecture 5 17

Example

The timber beam is fabricated of four 30x150 mm boards glued together to form a box section 150mm

wide and 210mm deep. Maximum allowable shearing stresses in the timber and in the glue are 1.2N/mm2

and 0.8N/mm2 respectively. Determine the maximum possible value of w.

30mm

90mm 30mm30mm

30mm

150mm

As the first step maximum shear force that is acting along the beam should be calculated.

For this, the shear force diagram has to be drawn to pick the maximum shear force.

The support reactions are shown on the free-body diagram below.

From the moment equilibrium

From the equilibrium in vertical direction

The horizontal reaction at A is zero.

Based on the calculated support reactions the shear force diagram can be drawn in terms of the applied

force w.

0AM = →∑ B3w 1.5+R 6 1.5w 0− × × − =BR w→ =

0YF = →∑ A BR +R 3w=AR 2w→ =

A BC

3m 3m 1m

w kN/m1.5w kNm

Lecture 5 18

Shear force diagram

The maximum shear force is 2w

The moment of inertia around the neutral axis of the box section can be calculated as

Case 1 If the shear failure is in timber

If the failure is assumed to occur in timber then the maximum shear stress occurs when y=0, because te

first moment of area Q is maximum.

3 3

max 150 105 52.5 90 75 37.5 573.75 10Q mm= × × − × × = ×

The shear stress in this case can be calculated as

max maxmax

V Q

Ibτ =

3 3

6

2 10 573.75 10

90.45 10 60

w× × ×=

× × allowτ= 21.2 /N mm= 5.675 /w kN m→ =

Case 2 If the shear failure is in glue

If the failure is assumed to occur in the glue the shear stress should be calculated at the contact surface.

3 3150 30 90 405 10glueQ mm= × × = ×

The shear stress in this case can be calculated as

max glue

glue

V Q

Ibτ =

3 3

6

2 10 405 10

90.45 10 60

w× × ×=

× × glueτ= 20.8 /N mm= 5.3629 /w kN m→ =

Case 2 governs because it allows a less applied load w.

3 31150 210 90 150

12I = × − ×

6 490.45 10 mm= ×