Tutorial4

Tutorial 4

MECH 101

Liang Tengfei

tfliang@ust.hk

Office phone : 2358-8811

Mobile : 6497-0191

Office hour : 14:00-15:00 Fri



1

Outline

 Example about normal stress and normal strain

 Example about shear stress, shear strain, and Hooke’s

law in shear

Example 1

The two bar truss ABC has pin supports at points A and C, which are 2.0m

apart. Members AB and BC are steel bars, pin connected at joint B. The

length of BC is 3m. A sign weighing 5.4kn is suspended from bar BC at D

and E, which are located 0.8m and 0.4m, respectively, from the ends of the

bar. Determine the required cross-sectional area of AB and the required

diameter of pin at C if the allowable stresses in tension and shear are 125

MPa and 45 Mpa.

Draw Free Body Diagram

Supposing counterclockwise moments are positive,



 Mc  0, R AH (2.0m)  (2.7kN )(0.8m)  (2.7kN )(2.6m)  0

RAH  4.590kN

F horiz  0, RCH  RAH  4.590kN

M B  0,  Rcv (3.0m)  (2.7kN )(2.2m)  (2.7kN )(0.4m)  0

RCV  2.340kN



Return to free body diagram of the entire truss,



F vert  0, RAV  RCV  2.7kN  2.7kN  0

RAV  3.060kN .



Reaction force at A and C,



RA  ( RAV ) 2  ( RAH ) 2  5.516kN

Rc  ( RCV ) 2  ( RCH ) 2  5.152kN

Tensile force in bar AB

Shear force acting on the pin at C



FAB  RA  5.516kN

Vc  Rc  5.252kN



Required area of the bar





FAB

AAB   44.1mm 2

 allow

Required diameter of the pin

Vc

Apin   57.2mm 2

2 allow

d pin  4 Apin /   8.54mm

practice









 c  5.21MPa

Shear strain : change in the shape of the element











Hook’s law in shear   G

Example 2

A flexible connection consisting of rubber pads( t=9mm) bonded to steel

plates is shown in the figure. The pads are 160mm long and 80mm wide.

(a)Find the average shear strain γaver in the rubber if the force P=16kN and

the shear modulus for the rubber is G=1250kPa.

(B) Find the relative horizontal displacement δ between the interior plate and

the outer plates.

 aver  V

ab

 aver

Shear strain :   V

abGe

Ge







In most practical situations the shear strain γ is a small angle, and in such cases

tan γ can be replaced by γ.

Shear force V=P/2,

Shear Strain:

 aver  abV





 aver

   V

abGe  0.5

Ge

Horizontal displacement δ,



hV

d  h   4.5mm

abGe

Shear stress and Bearing stress

Shear stress : tangential to the surface

V aL P d 2

Average shear stress:  aver  Where V  A

A 2 2 4

F

Average bearing stress:  b  b Where Fb  aL  P Ab  dL

Ab

V

m n

p

L a m n



p q  d 2

A

V 4

One shear Two shear

surface surfaces









V

m n F

V F V

L F

  L F

  2

p q

A 1 d2 p q A 1 d2

V 4 V

4

Example 3

A steel strut S serving as a brace for a boat hoist transmits a compressive forces

P=54KN to the deck of a pier. The strut has a hollow square cross section with wall

thickness t=12mm, and angle θ is 40 °. A pin through the strut transmits the

compressive force from the strut to two gussets G that welded to the base plate B.

Four anchor bolts fasten the base plate to deck. The diameter of the pin is dpin=18mm,

the thickness of the gussets is tG=15mm, the thickness of the base plate is tB=8mm, the

diameter of the anchor bolts is dbolt=12mm.

Determine the following stress: (a)the bearing stress between the strut and the pin;(b) the

shear stress in the pin;(c) the bearing stress between the pin and the gussets,(d) the bearing

stress between the anchor bolts and the base plate,(e) the shear stress in the anchor bolts.

Solution

 (a)the bearing stress between the strut and the pin

= The force in the strut / the total bearing area of the strut



P 54kN

 b1    125MPa

2td pin 2(12mm)(18mm)

Bearing

surface

 (b) the shear stress in the pin( the pin tends

to shear the two planes between the gussets and

the strut)



Shearing

surface

 (c) the bearing stress between the pin and the gussets

The pin bear against the gussets at two locations, so the

bearing area is twice the thickness of the gussets times the pin

diameter;

Bearing

surface







(d) the bearing stress between the anchor bolts and the base plate

(The vertical component of force is transmitted to the pier by

direct bearing between the base plate and the pier; the

horizontal component, is transmitted through the anchor bolts.)



P cos 40 (54kN )(cos 40)

 bolt    108MPa

4t B dbolt 4(8mm)(12mm)



Bearing Bearing

surface force

 (e) the shear stress in the anchor bolts



P cos 40 (54kN )(cos 40)

 bolt    91.4 MPa

4 dbolt / 4

2

4 (12mm) / 4

2









Shearing

surface



Shearing force

Practice





 aver  31.8MPa;

 b  41.7 MPa

Practice







 max  7330 psi;

 b  12800 psi