Transverse Shear Stress Page A 1

To develop τ = ƒ(V)1 Write fundamental equations

3 Write statics equation for top slice ΣFx = 0

2 Write differential equations

4. Find Fs = F2 – F1

5 Write τ = Fs / As 6. Define ∫ y dA = Q and use V = dM/dx

τ=FS

AS

σ=(M )( y)

I

V=dMdx

∑ F x=0=F1+FS−F2

dF=σ (dA )

∫ dF=∫ (M )( y )I dA

F1=∫ (M )( y )I dA

F2=∫ (M+dM )( y)I dA

F1−F2=∫ (M )( y )I dA−∫ (M+dM )( y)

I dA

FS=∫ (dM ) yI

dA

τ=∫ (dM ) y

IdA

(t )(dx)

τ=dMdx ∫ ydA

¿

τ=V ∫ ydA¿

τ=VQ¿

Shear

Shear stress in Glue

Worst at neutral axis (largest Q)Or least thickness (smallest t)

0,0

1

2

3

(5)(20)

35(5)(30)

15

32.5

(5)(40)

22.510

25 (5)(20)

Transverse Shear Stress Page A 2

1 Prepare cross-section 1a. put x,y datum (0,0) 1b. number sections

2 Make table to find Y-bar and Ix

Rows=section #s Columns from equations

3. Calculate Q for areas above Y-bar (our use bottom section)

4 Calculate τ caused by V

Y=Σ (A ∙ y )ΣA

I total=Σ I+Σ ( A d2 )

Q=Σ(A p ∙ y )using topQ= (5mm ) ( 40mm ) (22.5mm )+ (5mm ) (20mm ) (10mm )=5500mm3

using bottomQ=(5mm ) (20mm ) (32.5mm )+(5mm ) (30mm ) (15mm )=5500mm3

τ=VQ¿

τ=VQ¿

τ=(675 N )(5500mm3)

( 265830mm4 )(5mm)

τ=2.79 Nmm2

8ft 6ft 4ft

80 K

A B

15 K 12 K

2ft

V (K)

x (ft)

M (K ft)

x (ft)

34.5

-44.5

8ft8ft

(8ft)(34.5K+2.5K)/2 =148Kft

8K

148K ft-(8ft)(12.5K+44.5K)/2=-80Kft

4ft

72.5K34.5K

2.5

-12.5

2812

Transverse Shear Stress Design Page B 1

Cross-sections Pg 700

Guidelines Pg723

Guideline Pg723

Materials Prop Pg 717

1 Find A and B

2 Draw V and M diagrams 3 Guidelines for bending

4 Find S for wide flange beam

5 Calculate τmax using web shear formula pg 437

6 Compare to shear guidelines, safe?

∑M A=0=−(15 K ) (8 ft )− (80K ) (10 ft )+B (16 ft )−(12K )(20 ft)

B=72.5 K∑ F y=0=A−15 K−80 K+72.5 K−12 K A=34.5K

σ d=S y

1.5 Sy=50ksi

σ= MS

50 K¿2

1.5=148Kft ¿¿

S=53.3¿3

w14 x 43 S=62.5¿3t=0.305∈¿h = 13.7in

τ max≈Vt h

τ max=44.5 K

¿¿

τ d=0.4 S y

Sy=50ksiτ d=0.4 (50ksi )=20ksiSAFE

0,0

1

2

3

Nail

Transverse Shear Flow (q) Page C 1

q= forcelength τ=

F(L )( t)

=V QI t q=V Q

I

Sections of z shape are held together with nails that can support a shear force of 100 N. Find the nail spacing for a transverse shear load of 500 N.

If glue is used, find the shear stress in the glue.

τ=VQ¿

τ=(500 N ) (5mm ) (15mm )(17.5mm)

(72917mm4 )(5mm)

τ=1.8 Nmm2

Q=Σ(A p ∙ y )using section1Q=(5mm ) (15mm ) (17.5mm )

q= FL

=100NL

=VQI

100NL =

(500 N ) (5mm )(15mm)(17.5mm)72917mm4

L=11.1mm

Logic for Shear flow

Given=Forcenail

Find=VQI

=Forcemm

ForcenailForcemm

=mmnail

Transverse Shear Flow (q) Page C 2