Investmech: Force, moment, torque

and stress distributions

Dr. Michiel Heyns Pr.Eng.T: +27 12 664-7604

C: +27 82 445-0510

mheyns@investmech.com

Topics

• Shear force distribution

• Bending moment distribution

o Bernoulli-Euler beam modelling

• Normal stress distribution

• Bending stress distribution

• Torque shear stress distribution

• Shear force stress distribution

2017-09-19 2

Shear force diagram

2017-09-19 3

𝐹𝑅𝐴 𝑅𝐵a b

From Newton’s law for forces:

𝑅𝐴 − 𝐹 + 𝑅𝐵 = 0From Newton’s law for moments around A:

𝑅𝐵𝐿 − 𝐹𝑎 = 0

𝑅𝐵 =𝐹𝑎

𝐿Substitute into the first equation above:

𝑅𝐴 = 𝐹 − 𝑅𝐵

= 𝐹 −𝐹𝑎

𝐿

= 𝐹 1 −𝑎

𝐿Shear force diagram:

• 𝑅𝐴 for x ∈ [0; 𝑎]• −𝑅𝐵 for x ∈ [𝑎; 𝐿]

𝐹 1 −𝑎

𝐿

−𝐹𝑎

𝐿

𝐿 = 𝑎 + 𝑏

x

y

x

y

Add distributed load

2017-09-19 4

𝐹𝑅𝐵

a b

𝐿 = 𝑎 + 𝑏

𝑤

From Newton’s law for forces:

𝑅𝐴 − 𝐹 + 𝑅𝐵 − 𝑤𝑎 = 0From Newton’s law for moments around A:

𝑅𝐵𝐿 − 𝐹𝑎 − 𝑤𝑎 ∙𝑎

2= 0

𝑅𝐵 =𝐹𝑎 +

𝑤𝑎2

2𝐿

Substitute into the first equation above:

𝑅𝐴 = 𝐹 − 𝑅𝐵 + 𝑤𝑎

= 𝐹 −𝐹𝑎 +

𝑤𝑎2

2𝐿

+ 𝑤𝑎

= 𝐹 1 −𝑎

𝐿+ 𝑤

𝑎2

𝐿+ 𝑎

Shear force distribution over 𝑥 ∈ [0; 𝑎]:𝑉 𝑥 = 𝑅𝐴 − 𝑤𝑥= 𝑅𝐴 − 𝑤𝑥

Shear force distribution over 𝑥 ∈ [𝑎, 𝐿]:𝑉 𝑥 = 𝑅𝐴 − 𝑤𝑎 − 𝐹

𝑉 𝑥 = 𝑅𝐴 − 𝑤𝑥

𝑉 𝑥 = 𝑅𝐴 − 𝑤𝑎 − 𝐹

x

y

Bending moment diagram

2017-09-19 5

Euler-Bernoulli beam theory:

𝐸 𝑥 𝐼 𝑥𝑑2𝑤

𝑑𝑥2= 𝑀(𝑥)

For y-axis pointing towards reader

Sign of M depends on direction of

z-axis

If z-axis points upwards, the equation

becomes (y-axis pointing away from

reader):

𝐸 𝑥 𝐼 𝑥𝑑2𝑤

𝑑𝑥2= 𝑀𝑦(𝑥)

+𝑀

+𝑉

+𝑀𝐵

+𝑉𝐵

+𝜃𝐴

+𝑤𝐴

x

z

x+𝑀𝐴

Apply to cantilever beam

2017-09-19 6

𝐹

𝐿

𝑀𝐴

𝐹𝐴From statics:

𝑀𝐴 = 𝐹𝐿𝐹𝐴 = 𝐹

𝑥

The bending moment at position 𝑥:

𝑀𝑦(𝑥) = −𝐹𝐴𝑥 +𝑀𝐴

= −𝐹𝑥 + 𝐹𝐿Euler-Bernoulli for this coordinate system:

𝐸 𝑥 𝐼 𝑥𝑑2𝑤

𝑑𝑥2= 𝑀𝑦

𝑑2𝑤

𝑑𝑥2=

𝑀

𝐸 𝑥 𝐼 𝑥

=𝐹

𝐸𝐼(−𝑥 + 𝐿)

Integrate once to calculate beam slope:

𝑑𝑤

𝑑𝑥=

𝐹

𝐸𝐼−1

2𝑥2 + 𝐿𝑥 + 𝐴

The rotation is zero at 𝑥 = 0 (built in):

𝜃𝐴 =𝑑𝑤

𝑑𝑥𝑥 = 0

= 𝐴 = 0

𝜃 𝑥 = 𝐿 =1

2

𝐹𝐿2

𝐸𝐼: tip will lift up, + slope

Integrate to calculate deflection:

𝑤 =𝐹

𝐸𝐼−1

6𝑥3 +

1

2𝐿𝑥2 + 𝐵

The deflection is zero at 𝑥 = 0. Therefore,

𝐵 = 0Deflection at the end is:

𝑤 =1

3∙𝐹𝐿3

𝐸𝐼Clockwise moment at x is + for

chosen coordinate system

x

z

Simply supported beam

2017-09-19 7

𝐿𝑎𝑥

A B

𝐹

𝑅𝐴 𝑅𝐵

Apply the static equations

• Sum of forces:

𝐹𝑧 = 0 = 𝑅𝐴 + 𝑅𝐵 − 𝐹

𝑅𝐴 + 𝑅𝐵 = 𝐹• Sum of moments around A:

𝑀𝑦 = 0

𝐹𝑎 − 𝐿𝑅𝐵 = 0

𝑅𝐵 =𝐹𝑎

𝐿Therefore:

𝑅𝐴 = 𝐹 −𝐹𝑎

𝐿

x

z

Bending moment equations

From A to F (x=0 to a):

𝑀𝑦 = 𝑅𝐴𝑥

= 𝐹 −𝐹𝑎

𝐿𝑥

From F to B (x=a to L):

𝑀𝑦 = 𝑅𝐴𝑥 − 𝐹 𝑥 − 𝐿

= 𝐹 1 −𝑎

𝐿𝑥 − 𝐹 𝑥 − 𝑎

= −𝐹𝑎

𝐿𝑥 + 𝐹𝑎

= 𝐹𝑎 1 −𝑥

𝐿

For F = 1 N; a = 1.2 m & L = 2 m, the

bending moment diagram is:

Shear force diagram and shear force to

bending moment by integration

2017-09-19 8

𝐿𝑎𝑥

A B

𝐹

𝑅𝐴 𝑅𝐵

x

z

𝐹 −𝐹𝑎

𝐿

−𝐹𝑎

𝐿From A to F:

𝑀𝑦 = න𝑥=0

𝑥

𝐹 −𝐹𝑎

𝐿𝑑𝑥

= 𝐹 −𝐹𝑎

𝐿𝑥

From F to B:

𝑀𝑦 = න𝑥=𝑎

𝑥

−𝐹𝑎

𝐿𝑑𝑥 + 𝐹 −

𝐹𝑎

𝐿𝑎

= −𝐹𝑎

𝐿𝑥 +

𝐹𝑎2

𝐿+ 𝐹𝑎 −

𝐹𝑎2

𝐿

= 𝐹𝑎 1 −𝑥

𝐿

Bending stress distribution

2017-09-19 9

x

y

𝑀𝑥

𝑀𝑦

𝑊

𝐻

General equation:

𝜎𝑏,𝑖 =𝑀𝑥𝑦𝑖𝐼𝑥𝑥

−𝑀𝑦𝑥𝑖

𝐼𝑦𝑦Where:

𝑥𝑖: x-coordinate of the point where stress is required

𝑦𝑖: y-coordinate of the point where stress is required

𝐼𝑥𝑥: Second moment of area around the x-axis [m4]

𝐼𝑦𝑦: Second moment of area around the y-axis [m4]

Applied to Point A:

𝜎𝑏,𝐴 =𝑀𝑥𝑦𝐴𝐼𝑥𝑥

−𝑀𝑦𝑥𝐴

𝐼𝑦𝑦

𝑥𝐴 =𝑊

2, yA =

H

2Applied to Point B:

𝜎𝑏,𝐵 =𝑀𝑥𝑦𝐵𝐼𝑥𝑥

−𝑀𝑦𝑥𝐵

𝐼𝑦𝑦

𝑥𝐴 = −𝑊

2, yA = −

H

2

A

B

Bending stress distribution

2017-09-19 10

x

y

𝑀𝑥

𝑀𝑦

𝑊

𝐻

A

B

Say 𝐻 = 100;𝑊 = 50 𝑚𝑚, then:

𝐼𝑥𝑥 =1

12𝑊𝐻3 = 4.1667 × 10−6 𝑚4

𝐼𝑦𝑦 =1

12𝐻𝑊3 = 1.25 × 10−5 𝑚4

Bending stresses at A and B for 𝑀𝑥 = 1000 𝑁𝑚 &𝑀𝑦 = 100 𝑁𝑚

𝜎𝑏,𝐴 = 11.8 𝑀𝑃𝑎; 𝜎𝑏,𝐵 = −11.8 𝑀𝑃𝑎

The bending moment distribution over the cross section for

Investmech algorithm: stressbendingmoment.m

Bending stress distribution at x=0

2017-09-19 11

General equation:

𝜎𝑏,𝑖 =𝑀𝑥𝑦𝑖𝐼𝑥𝑥

−𝑀𝑦𝑥𝑖

𝐼𝑦𝑦At 𝑥 = 0, the bending stress distribution reduces to:

𝜎𝑏,𝑖 =𝑀𝑥𝑦𝑖𝐼𝑥𝑥

For a positive bending moment 𝑀𝑥𝑥:• Tensile stress for positive 𝑦• Compressive stress for negative 𝑦

Investmech algorithm: stressbendingunsymmetric.m

Bending stress: Circular tube

2017-09-19 12

x

y

𝐷𝑜

𝐷𝑖

General equation:

𝜎𝑏,𝑖 =𝑀𝑥𝑦𝑖𝐼𝑥𝑥

−𝑀𝑦𝑥𝑖

𝐼𝑦𝑦Where:

𝐼𝑥𝑥 =𝜋

64𝐷𝑜4 − 𝐷𝑖

4

𝐼𝑦𝑦 = 𝐼𝑥𝑥𝐼𝑧𝑧 = 𝐼𝑥𝑥 + 𝐼𝑦𝑦= 2𝐼𝑥𝑥

Investmech algorithm: stresscircular.m

Bending stress: Solid circular bar

2017-09-19 13

x

y

𝐷𝑜

Bending stress in unsymmetrical

sections

2017-09-19 14

z

y

The equation for unsymmetrical bending under 𝑀𝑦:

𝜎 =𝐹

𝐴+𝑀𝑦𝐼𝑧𝑧 +𝑀𝑧𝐼𝑦𝑧

𝐼𝑧𝑧𝐼𝑦𝑦 − 𝐼𝑦𝑧2 𝑧 −

𝑀𝑧𝐼𝑧𝑧 +𝑀𝑦𝐼𝑦𝑧

𝐼𝑧𝑧𝐼𝑦𝑦 − 𝐼𝑦𝑧2 𝑦

Where:

𝐼𝑦𝑦 =1

12𝑡𝑤ℎ

3 + 𝑤𝑡𝑓ℎ

2

2

× 2

𝐼𝑧𝑧 =1

12ℎ𝑡𝑤

3 + 2 ×1

3𝑡𝑓𝑤

3

𝐼𝑦𝑧 = 𝑡𝑓𝑤ℎ

2∙𝑤

2+ 𝑡𝑓𝑤

−ℎ

2∙−𝑤

2

=1

2𝑡𝑓𝑤

2ℎ

See stress distribution on next page

ℎ

𝑤

𝑤

𝑡𝑤

𝑡𝑓

𝑡𝑓

2017-09-19 15

Normal stress distribution

• Is uniform

𝜎𝑛 =𝐹

𝐴

2017-09-19 16

Shear stress distribution-solid

rectangular bar

2017-09-19 17

The first moment of area around the

x- and y-axes are:

𝑄𝑥 = න𝑦𝑑𝐴

𝑄𝑦 = න𝑥𝑑𝐴

From which the shear stress is:

𝜏𝑦 =𝑉𝑦𝑄𝑥

𝐼𝑥𝑥𝑊

𝜏𝑥 =𝑉𝑥𝑄𝑦

𝐼𝑦𝑦𝐻

𝜏 = 𝜏𝑥 + 𝜏𝑦

Average shear stress:

𝜏𝑎𝑣𝑒 =𝑉

𝐴

x

y

𝑉𝑦

𝑊

𝐻

B

𝑦

𝑑𝐴 = 𝑊𝑑𝑦Shear flow equation

𝑄𝑥(𝑦) = න0

𝑎

𝑎𝑊𝑑𝑎

=1

2𝑊𝑎2

0

𝑎

=1

2𝑊𝑎2

With maximum value at 𝑎 =𝐻

2:

𝑄𝑥,𝑚𝑎𝑥 =1

2𝑊𝐻2

4

𝜏𝑚𝑎𝑥 =𝑉𝑦𝑊𝐻2

8112

𝑊2𝐻3

=3

2∙𝑉𝑦

𝑊𝐻

= 1.5 ×𝑉𝑦

𝐴

𝑎

𝑑𝑎

Shear stress: Circular section

2017-09-19 18

x

y

𝑅 =𝐷

2

𝑦

𝑑𝑦

The first moment of area around the x-axis (the minus

sign is because the integral from R to y is in the opposite

direction as shown in the figure):

𝑄𝑥 𝑦 = න𝑅

𝑦

−𝑦𝑑𝐴

= න𝑅

𝑦

−𝑦 ∙ 2𝑤𝑑𝑦

= 2න𝑅

𝑦

−𝑦 𝑅2 − 𝑦2 0.5 𝑑𝑦

= 2 ∙1

3(𝑅2−𝑦2)1.5 𝑅

𝑦

=2

3𝑅2 − 𝑦2 1.5

The shear stress is maximum when 𝑦 = 0:

𝜏 =𝑉𝑄

𝐼𝑥𝑥(2𝑅)

𝐼𝑥𝑥 =𝜋

4𝑅4

𝜏𝑚𝑎𝑥 =𝑉 ∙

23𝑅3

𝜋4𝑅4 ∙ 2𝑅

=4

3

𝑉

𝜋𝑅2=4

3𝜏𝑎𝑣𝑒

2𝑤

𝑤 = 𝑅2 − 𝑦2 0.5

Average shear stress:

𝜏𝑎𝑣𝑒 =𝑉

𝐴

=𝑉

𝜋𝑅2

Consult http://integral-table.com/ for integrals

Stress distribution

2017-09-19 19

For radius 𝑅 = 100 𝑚𝑚, the stress distribution is the following function times shear force 𝑉:

x

y

𝑅 =𝐷

2

𝑦

𝑑𝑦2𝑤

Shear stress distribution in boxed

section

2017-09-19 20

𝑎

𝑑𝑎

𝑤𝑡𝑤𝑡𝑤

𝑡𝑓

𝑡𝑓

𝑊

𝐻ℎ

Area:

𝐴 = 𝑊𝐻 −𝑤ℎSecond moment of area:

𝐼𝑥𝑥 =1

12(𝑊𝐻3 − 𝑤ℎ3)

First moment of area:

𝑄𝑥 = න𝑦𝑑𝐴

= න0

𝑡𝑓 𝐻

2− 𝑎 𝑊𝑑𝑎

𝑇𝑜𝑝 𝑓𝑙𝑎𝑛𝑔𝑒

+න𝑡𝑓

𝐻−𝑡𝑓 𝐻

2− 𝑎 2𝑡𝑤𝑑𝑎

𝑊𝑒𝑏𝑠

+න𝐻−𝑡𝑓

𝐻 𝐻

2− 𝑎 𝑊𝑑𝑎

𝐵𝑜𝑡𝑡𝑜𝑚 𝑓𝑙𝑎𝑛𝑔𝑒

The equation for the shear stress:

𝑡 =𝑉𝑦𝑄𝑥

𝐼𝑥𝑥𝑏Where 𝑏 is the thickness, ranging from 𝑊 to 2𝑡𝑓,

depending on 𝑎

x

y

𝑦

Shear stress distribution for Vy = 1 N for

H = 0.2; W = 0.1; tw = 0.01; tf = 0.01

2017-09-19 21

𝑡𝑤

𝑡𝑤

𝑡𝑓

𝑡𝑓

𝑊

𝐻

x

𝑉𝑦

y

Investmech algorithm used: stressshearboxed.m

Why is Weld A better than B

Weld A

Weld

B

𝜏𝑎𝑣𝑒 = 178.6 𝑃𝑎𝑡𝑚𝑎𝑥 = 315.5 𝑃𝑎𝜏𝑚𝑎𝑥

𝜏𝑎𝑣𝑒= 1.77

Shear stress in unsymmetric sections

2017-09-19 22

The shear stress for unsymmetric sections is:

𝜏 =𝐼𝑧𝑧𝑄𝑦 − 𝐼𝑦𝑧𝑄𝑧

𝑏 𝐼𝑧𝑧𝐼𝑦𝑦 − 𝐼𝑦𝑧2

𝑉𝑧 +𝐼𝑧𝑧𝑄𝑧 − 𝐼𝑦𝑧𝑄𝑦

𝑏 𝐼𝑧𝑧𝐼𝑦𝑦 − 𝐼𝑦𝑧2

𝑉𝑦

𝑄𝑦 = න𝑧𝑑𝐴

𝑄𝑧 = න𝑦𝑑𝐴

y

z

Torque

2017-09-19 23

x

y

𝑅 =𝐷

2

𝑟

Shear stress due to torque:

𝑡𝑇 =𝑇𝑟

𝐼𝑧𝑧

Investmech algorithm: stresscircular.m

Circular tube

2017-09-19 24

x

y

𝑅 =𝐷

2

𝑅𝑖

Investmech algorithm: stresscircular.m

Weld group subjected to torsion and

shear

2017-09-19 25

Problem statement

The attachment is subjected to the force 𝑃 as shown.

Compile equations from which the weld throat area, 𝑎, can

be calculated.

Solution

The second moments of area for the weld group is:

𝐼𝑥𝑥 = 2𝑎𝑏𝑑

2

2

=1

2𝑎𝑏𝑑2

𝐼𝑦𝑦 = 2 ∙1

12𝑎𝑏3

=1

6𝑎𝑏3

𝐼𝑧𝑧 = 𝐼𝑥𝑥 + 𝐼𝑦𝑦

= 𝑎1

2𝑏𝑑2 +

1

6𝑏3

The area of the weld group is:

𝐴 = 2𝑎𝑏

𝑃𝐿

𝑏

𝑑 x

y

• Shear stress due to shear forceo Assume uniform distribution

𝜏𝑦𝑉 = −𝑃

𝐴= −

𝑃

2𝑎𝑏• Shear stress due to torque

𝜏𝑦𝑇 = −𝑃𝐿𝑥

𝐼𝑧𝑧

𝜏𝑥𝑇 =𝑃𝐿𝑦

𝐼𝑧𝑧• From direction of stresses, critical point is Point A

𝑥𝐴 =𝑏

2; 𝑦𝐴 =

𝑑

2

𝜏𝑦,𝐴 = −𝑃𝐿𝑏

2𝑎12𝑏𝑑2 +

16𝑏3

−𝑃

2𝑎𝑏

𝜏𝑥𝑇,𝐴 =𝑃𝐿𝑑

2𝑎12𝑏𝑑2 +

16𝑏3

o Calculate residual stress on the surface and compare with factored resistance

𝜏 = 𝜏𝑦,𝐴2 + 𝜏𝑥,𝐴

2 ≤ 0.672𝑥𝑢

2017-09-19 26

𝑃𝐿

𝑏

𝑑 x

y

A

Shear stress due to shear force and

torque to be considered

y

r

𝜃

𝜃

𝑥; 𝑦 = 𝑟(cos 𝜃; sin 𝜃)

• Substitute further:

𝜏𝑦,𝐴 = −𝑃𝐿𝑏

2𝑎12𝑏𝑑2 +

16𝑏3

−𝑃

2𝑎𝑏

=𝑃

𝑎−

𝐿𝑏

212𝑏𝑑2 +

16𝑏3

−1

2𝑏

𝜏𝑥,𝐴 =𝑃𝐿𝑑

2𝑎12𝑏𝑑2 +

16𝑏3

=𝑃

𝑎

𝐿𝑑

212𝑏𝑑2 +

16𝑏3

2017-09-19 27𝑏

𝑑 x

y

A

Using the idea

• Say, 𝑑 = 100 𝑚𝑚, 𝑏 = 50 𝑚𝑚, 𝐿 = 400 𝑚𝑚,𝑃 = 15 000 𝑁• Weld metal is E70XX, with ultimate tensile strength 490 𝑀𝑃𝑎

o Factored resistance: 𝜏𝑅 = 0.672 ∙ 490 = 220𝑀𝑃𝑎o Calculate stress and find throat size. From graph it is 𝑎 ≥ 6 𝑚𝑚

o 𝑒 ≥ 8.5 𝑚𝑚

2017-09-19 28

Using the equation

𝜏𝑦,𝐴 =𝑃

𝑎−

𝐿𝑏

212𝑏𝑑2 +

16𝑏3

−1

2𝑏

= −1

𝑎∙ 7.0385 × 105

𝜏𝑥,𝐴 =𝑃

𝑎

𝐿𝑑

212𝑏𝑑2 +

16𝑏3

=1

𝑎∙ 1.1077 × 106

𝜏 =1

𝑎7.0385 × 105 2 + 1.1077 × 106 2

=1.3214

𝑎𝑀𝑃𝑎 ≤ 220

𝑎 = 0.006 𝑚𝑒 ≥ 8.4 𝑚𝑚

2017-09-19 29

Test with forces

• The factored shear resistance of 8 mm weld is:

𝜏′𝑅 = 0.672 ∙8

2∙ 490 = 1.24 𝑘𝑁/𝑚𝑚

• The weld lengths are 50 mm, giving:

𝑉𝑅 = 62.2 𝑘𝑁 per weld

• The torque that can be resisted is:

𝑇𝑅 = 2𝑑

2𝑉𝑅 = 2 ∙ 0.05 × 𝑉𝑅

= 6.2 𝑘𝑁𝑚• The applied torque is:

𝑇 = 𝑃𝐿 = 0.4 × 15 = 6 𝑘𝑁𝑚• This is less than the resistance

• The shear force resistance is 112.4 kN, more than the shear force or 15 kN

• This is not ideal, because at Point A the weld is subject to the highest shear stress due to torque and shear force combined o However, above approximation using forces give quick check

2017-09-19 30

𝑃𝐿

𝑏

𝑑 x

y

A𝑉𝑅

𝑉𝑅

Welded end-connection

2017-09-19 31

𝐻

𝑒𝑓𝑒

𝐵

Problem statement

An I-beam with height, 𝐻, width 𝐵, flange thickness, 𝑡𝑓 and web thickness, 𝑡𝑤 is joined at the end by

the all around weld of size 𝑒. A force, 𝐹 is applied at the other end.

Give the equations from which the following can be calculated in the weld group:

1. Bending stress

2. Normal stress

3. Shear stress

4. Combined stress

Assume for all calculation that the weld throat is positioned at the outside lines of the cross-section.

This is a conservative assumption and will reduce calculations substantially.

𝐿𝐹𝑦

x

y

y

z𝐹𝑥

𝑀𝑧 = torque

The weld throat area

2017-09-19 32

𝐻

𝐵

x

y

Area:

𝐴 = 2𝑎𝑤 𝐻 − 2𝑡𝑓 + 2𝑎𝑤 𝐵 − 𝑡𝑤 + 2𝑎𝑤𝐵 + 4𝑎𝑤𝑡𝑓Second moments of area:

𝐼𝑥𝑥 =1

12𝑎𝑤 2 𝐻 − 2𝑡𝑓

3+ 4tf

3

+H − tf2

2

4awtf

+2aw B − twH

2− tf

2

+2awBH

2

2

𝐼𝑦𝑦 =1

12𝑎𝑤 4𝐵3 − 2𝑡𝑤

3 + 2𝑎𝑤 𝐻 − 2𝑡𝑓𝑡𝑤2

2

+4𝑎𝑤𝑡𝑓𝐵

2

2

𝐼𝑧𝑧 = 𝐼𝑥𝑥 + 𝐼𝑦𝑦Using these equations, the normal and shear stress can be

calculated at any point in the weld group

For the torque, calculate x- and y-components to add or subtract as

needed

Stress at any point

General equations:

𝜎𝑖 =𝑀𝑥𝑦𝑖𝐼𝑥𝑥

−𝑀𝑦𝑥𝑖

𝐼𝑦𝑦+𝐹𝑧𝐴

𝜏𝑦 = 𝑘𝑦𝑉𝑦

𝐴+𝑇𝑥

𝐼𝑧𝑧

𝜏𝑥 = 𝑘𝑥𝑉𝑥𝐴−𝑇𝑦

𝐼𝑧𝑧

𝛼 = atan𝑦

𝑥, 𝑥 ≠ 0

𝜏𝑤 = 𝜏𝑥2 + 𝑡𝑦

2 + 𝜎2

Where:o 𝐹𝑧: Normal force [N]

o 𝑉𝑥, 𝑉𝑦: Shear force in the x- and y-directions [N]

o 𝑇 = 𝑀𝑧: Moment around the z-axis = torque [Nm]

o 𝑥𝑖: x-coordinate of the point where stress is required

o 𝑦𝑖: y-coordinate of the point where stress is required

o 𝐼𝑥𝑥: Second moment of area around the x-axis [m4]

o 𝐼𝑦𝑦: Second moment of area around the y-axis [m4]

o 𝐼𝑧𝑧: Second moment of area around the z-axis [m4]

2017-09-19 33

x

y

r

𝜃

𝜃

𝑥; 𝑦 = 𝑟(cos 𝜃; sin 𝜃)

• Do an example with numbers in class

2017-09-19 34

Welded splice

2017-09-19 35

Problem statement:

An I-beam with height, 𝐻, width 𝐵, flange thickness, 𝑡𝑓 and web thickness, 𝑡𝑤 is joined by the welded

splice as shown in the figure above. The dimension of the strengthening on the flange is 𝐿𝑓𝑠, 𝑊𝑓𝑠 and 𝑡𝑓𝑠respectively. The double plates on the web each has dimensions: is 𝐿𝑤𝑠, 𝑊𝑤𝑠 and 𝑡𝑤𝑠.Type equation here. The gap is assumed to be 𝑔. Assume the weld size on the flange and web to be

𝑒𝑓 and 𝑒𝑤 respectively.

Calculate the weld size required on the flange and web, assume them of same size, to resist the shear

force and bending moment over the cross section.

𝐿𝑓𝑠

𝐿𝑤𝑠

𝑊𝑤𝑠

𝑊𝑓𝑠

𝐻

𝐵

𝑡𝑓

𝑡𝑤

𝑡𝑓𝑠

𝑡𝑤𝑠𝑡𝑤𝑠

𝑒𝑓𝑒𝑓𝑒𝑓

𝑒𝑓

𝑒𝑓𝑒𝑤

𝑒𝑓𝑒𝑤

The inserted metals in the gap

2017-09-19 36

The second moment of area around the x-axis:

𝐼𝑥𝑥 =1

12𝑡𝑤𝑠𝑊𝑤𝑠

3 × 2 + 2 𝑦𝑓2 𝑊𝑓𝑠𝑡𝑓𝑠 +

1

12𝑊𝑓𝑠𝑡𝑓𝑠

3

𝑦𝑓 =𝐻

2+𝑡𝑓𝑠

2The area:

𝐴 = 2 𝑊𝑓𝑠𝑡𝑓𝑠 +𝑊𝑤𝑠𝑡𝑤𝑠The bending stress distribution:

𝜎𝑏 =𝑀𝑥𝑦

𝐼𝑥𝑥−𝑀𝑦𝑥

𝐼𝑦𝑦The bending force in the top and bottom section for 𝑀𝑥:

𝐹𝑡𝑜𝑝 =𝑀𝑥

𝐼𝑥𝑥න𝐻2

𝐻2+𝑡𝑓𝑠

𝑦𝑊𝑓𝑠𝑑𝑦

=𝑀𝑥

𝐼𝑥𝑥∙𝑊𝑓𝑠

2

𝐻 + 2𝑡𝑓𝑠

2

2

−𝐻

2

2

=𝑀𝑥

𝐼𝑥𝑥∙𝑊𝑓𝑠

8(𝐻2+4𝐻𝑡𝑓𝑠 + 4𝑡𝑓𝑠

2 ) − 𝐻2

=𝑀𝑥

𝐼𝑥𝑥∙𝑊𝑓𝑠

2𝐻𝑡𝑓𝑠 + 𝑡𝑓𝑠

2

This force must be resisted by the weld on the flanges

x

y

𝑦𝑓𝐻

2

𝑡𝑓𝑠

𝑡𝑓𝑠

𝐻𝑊𝑠

𝑡𝑤𝑠 𝑡𝑤𝑠

𝑊𝑓𝑠

% bending moment resisted by the

flange plates

2017-09-19 37

x

y

𝑦𝑓

If assumed that the forces on the flanges act at the centroid of the top and bottom

plates, 𝑦𝑓, the percentage of the bending moment carried by the top and bottom

plates is:

𝑀 = 2𝑦𝑓𝑀𝑥

𝐼𝑥𝑥∙𝑊𝑓𝑠

84𝐻𝑡𝑓𝑠 + 4𝑡𝑓𝑠

2

𝑀

𝑀𝑥=𝑦𝑓

𝐼𝑥𝑥∙𝑊𝑓𝑠

44𝐻𝑡𝑓𝑠 + 4𝑡𝑓𝑠

2

=𝑦𝑓

𝐼𝑥𝑥∙ 𝑊𝑓𝑠 𝐻𝑡𝑓𝑠 + 𝑡𝑓𝑠

2

𝑦𝑓 =𝐻

2+𝑡𝑓𝑠

2

254 x 254 x 73 parallel flange H-section

2017-09-19 38

Designation h b tw tf r1 m A Ix Zx rx

mmxmmxkg/m mm mm mm mm mm kg/m 103 mm2 106 mm4 103mm3 mm

254x254x73 254.2 254 8.6 14.2 12.7 73.1 9.29 114 896 111

Iy Zy ry J Cw Zplx Zply h/tf hw

106 mm4 103mm3 mm 103mm4 109 mm6 103mm3 103mm3 mm

38.8 306 64.6 578 559 990 463 17.9 200

Source: https://www.macsteel.co.za/files/saisc_structural_steel_section_properties.xls

From the tables above, it is clear the H-beam has the following dimensions:

𝐻 = 254.2 𝑚𝑚𝐵 = 254.2 𝑚𝑚𝑡𝑓 = 14.2 𝑚𝑚

𝑡𝑤 = 8.6 𝑚𝑚Assume the splice is made with the following dimensions:

𝑊𝑓𝑠 = 200 𝑚𝑚

𝑡𝑓𝑠 = 12 𝑚𝑚

𝐻𝑤𝑠 = 120 𝑚𝑚𝑡𝑤𝑠 = 10 𝑚𝑚

Calculate the length of plate required on the flanges to resist a bending moment 𝑀𝑥 = 179.2 𝑘𝑁𝑚. That is,

calculate the weld length required to resist the force in the outer flanges for a weld size of 5 mm.

Flange weld calculation

2017-09-19 39

Cross-section parameters

H 0.2542 m B 0.2542 m Ixx 1E-04 m4

tf 0.0142 m tw 0.0086 m A 0.009 m2

Flange plate

W fs 0.2 m tfs 0.012 m Afs 0.002 m2

Web plate

Wws 0.12 m tws 0.01 m Aws 0.001 m2

Cross-section parameters for the plates

yf 0.1331 m Ixx 8.79725E-05 m4

g 0.005 m

Bending moment: Mx 1.79E+05 Nm

Force in top plate: Fz 738 698 N

739 kN

% Moment by flange plates: 97%

Welding on flanges

xu 4.90E+08 Pa fweld 1.73E+03 N/mm

efs 0.005 m Lweld 4.26E+02 mm

afs 0.00354 m Lfs/2 213 mm

Lfs 426 mm Gap added

The plates on the flanges

resist 97 % of the bending

moment

To be continued in 2018

Shear flow and shear stress in

membraned sections

• Not in your scope

2017-09-19 40

References

• Cross section properties

https://www.macsteel.co.za/files/saisc_structural_s

teel_section_properties.xls

2017-09-19 41