objective: to design simple connections and determine weather the material fails or safe, taking...

Objective:Objective:

To design simple connections and

determine weather the material fails or

safe, taking into consideration the

computed stresses and the natural

strength.

Allowable Stress

x

Stress

Strain

allowable

failure

ultimate

Working area

Factor of safety = F.S. = Ultimate stress

Allowable stress

all

ult

all

ult

SF

SF

..

..

Normal and Shear Stress

Bearing Stress

Example:Example:

The two members are pinned to gather at B. Top view of the pin connections at A and B are also given in the figure. If the pin have allowable shear stress of allow = 90 MPa and the allowable tensile stress of rod CB is (t)allow = 115 MPa, determine to the nearest mm the smallest diameter of pins A and B and the diameter of rod CB necessary to support the load.

Solution:Solution:

mmdd

mkPa

kN

T

VA

mmdd

mkPa

kN

T

VA

BB

Allow

BB

AA

Allow

AA

7.9 4

1011.741090

67.6

3.6 4

1056.341090

84.2

226

3

226

3

dA = 7 mm Ans….

dB = 10 mm Ans….

Diameter of the Pins.Diameter of the Pins.

Diameter of Rod.Diameter of Rod.

mmdd

mkPa

kNPA BC

BC

allowtBC 59.8

410358

10115

67.26 226

3

dBC = 9 mm Ans…..

We will Choose

Example:Example:

The control arm is subjected to the loading.\

Determine to the nearest 5 mm the required diameter of the steel pin

at C if the allowable shear stress for the steel is allow = 55 MPa.

Note: in the figure that the pin is subjected to double shear.

Solution:Solution:

Required Area:Required Area:

mmd

mmd

mmkN

kNVA

allow

8.18

45.2762

1045.276/ 1055

205.15

22

2623

Use a pin having a diameter of

d = 20 mm Ans….

Example:Example:

(a) (b)

SolutionSolutionDiameter of Rod:Diameter of Rod:

.... 6.20 0206.0

103333.04

103333.0/ 1060

1020

222

2326

3

Ansmmmd

md

A

mmN

NPA

allow

Thickness of disk:Thickness of disk:

.... 55.41055.4

02.02

105714.0

10571.01035

1020

323

2326

3

Ansmmmm

mt

mmN

NVA

allow

Since the sectioned area A = 2(0.02 m)(t), the requiredthickness of the disk is

Example:Example:

SolutionSolution

Normal Stress:Normal Stress:

kNP

m

PmN

A

Pallow

8.51

03.0

31055 2

26

Bearing Stress:Bearing Stress:

kNP

m

PmN

PA

allow

0.55

10199.2

31075 ;

2326