Tutorial 4 MECH 101

Liang Tengfeitfliang@ust.hk

Office phone : 2358-8811Mobile : 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 1The 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

0, 2.7 2.7 0

3.060 .vert AV CV

AV

F R R kN kN

R kN

2 2

2 2

( ) ( ) 5.516

( ) ( ) 5.152

A AV AH

CV CH

R R R kN

Rc R R kN

Return to free body diagram of the entire truss,

Reaction force at A and C,

Supposing counterclockwise moments are positive,

0, (2.0 ) (2.7 )(0.8 ) (2.7 )(2.6 ) 0

4.590

0, 4.590

0, (3.0 ) (2.7 )(2.2 ) (2.7 )(0.4 ) 0

2.340

AH

AH

horiz CH AH

B cv

CV

Mc R m kN m kN m

R kN

F R R kN

M R m kN m kN m

R kN

2

2

5.516

5.252

44.1

57.22

4 / 8.54

AB A

c c

ABAB

allow

cpin

allow

pin pin

F R kN

V R kN

FA mm

VA mm

d A mm

Tensile force in bar ABShear force acting on the pin at C

Required area of the bar

Required diameter of the pin

practice

5.21c MPa

Hook’s law in shear G

Shear strain : change in the shape of the element

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.

Vaver ab

e

aver VabGGe

Shear strain :

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:

Horizontal displacement δ,

0.5e

Vaver ab

aver VabGGe

4.5hV

d h mmabGe

Shear stress and Bearing stress

A

Vaver

b

bb A

F PaLFb

Shear stress : tangential to the surface

Average shear stress:

Average bearing stress: Where dLAb

22

PaLV

4

2dA

Where

n

V

m

q

V

p

aL nmp

4

2dA

One shear surface

Two shear surfaces

2

21

4

FV

A d

n

V

m

q

V

p

L

q

V

p

L F F

214

V F

A d

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

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

to shear the two planes between the gussets and

the strut)

1

54125

2 2(12 )(18 )bpin

P kNMPa

td mm mm

Bearing surface

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;

(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.)

cos 40 (54 )(cos 40 )108

4 4(8 )(12 )boltB bolt

P kNMPa

t d mm mm

Bearing force

Bearing surface

Bearing surface

(e) the shear stress in the anchor bolts

2 2

cos 40 (54 )(cos 40 )91.4

4 / 4 4 (12 ) / 4boltbolt

P kNMPa

d mm

Shearing force

Shearing surface

Practice

31.8 ;

41.7aver

b

MPa

MPa

Practice

max 7330 ;

12800b

psi

psi