**Internal Code :****Subject Code :****University :****Subject Name :**Mechanical Engineering

The first thing to be considered is the sum of moments around point A. It has been assumed that the sum of moments around point A is zero. The point A is chosen in proving that this is doable at the beams either ends so long it is supported by pin.

It is generally a necessity of idealizing the behavior of support in facilitating an analysis. The taken approach has similarity to the frictionless, massless pulley. Although there is no existence of a pulley, they have usefulness in enabling the learning with respect to certain issues. Therefore, mass and friction are frequently ignored in the behavior consideration of support or connection. It is a key aspect in realizing that all of the supports graphical representations are actual physical connections idealization. Efforts must be made in searching out and comparing the reality with the numerical and/or graphical model. It is sometimes easier in forgetting the idealization assumed to be different strikingly than the reality.

The diagram shows the moments and/or forces that have been active or available at each support type. It is expected that these moments and forces if calculated properly, can be bringing about equilibrium in each element of the structure.

The pinned joined shown in the diagram resembles a door hinge. This has the allowance of rotation about one axis, although preventing every new relative motion. In this problem, at the joint no relative motion is possible. There have to be reaction forces 3 components that act to prevent motion.

Roller support

To the lateral forces, it is not possible for the roller support in providing the resistance. It shall be remaining in place so long the structure would be supporting it and probably a perfect vertical load. When any kind of lateral loads pushing the structure, in response to the force it will not roll away. The lateral load, in this problem, has been a shove. The roller's support in the diagram have the freedom of rotating and translating along the surface where the rollers rest. The surface is horizontal in this problem. The reaction that has resulted is a single force away from, and perpendicular to the surface. The end of this beam is where the roller support is located. This would be allowing the bridge structure to be contracting and expanding with changes in temperature. However, the supported can be fractured by the expansion forces.

The rotating shaft is supported by the use of bearings. The bearings in various ways have been constraining the shaft. The bearing in the given diagram resembles a pin joint. This would be allowing rotation about one axis, although preventing rotation about the remaining two axes, and does prevention of the shafts all relative displacement. At this kind of bearing, no relative motion has the possibility. The reaction force must be having 3 components. In one axis, the relative rotation is allowable that has been parallel to the shaft, although the other two are prevented. Of the reaction moment, there has to be two components.

We have assumed that sum of the moments of A and B is zero (MB)

Therefore,

MB 0 200 (2) A (4)

Or A (4) 400

Or A 100

A B 200 0 MB

B 100 -200 0

B 100

Thus, the reaction forces at the pin support at A are 100 kN and the roller support at B is also 100 kN.

The stress in the strut BC (Fns) is determined in the following way.

Fns fceAcsfce 0.85sfc

s 1.0

fc cylinder compressive strength of concrete i.e. d 0.05 m

Acs Struts cross-sectional area 0.2 m2

Therefore,

Fns 0.85x1.0x0.05x0.2 .0085

Shear stress in the bolt F/A

Where,

F force

A Cross sectional area

F tp 0.2 m

A 0.2 m2

Therefore,

Shear stress in the bolt 1

The contact pressure is the bearing stress between separate bodies.

The bearing stress can be calculated in the following way.

b Pb/AbPb compressive load i.e. tb 0.15 m

Ab characteristics area perpendicular to PbThe platform at the pin connection can be determined by the following equation.

The properties that are essential to a bolt pattern are its reaction module and elastic center.

Out of the plane axis is the principal elastic axis where the external axis that acts on it cease to be tilting for supporting the plane. The elastic center in the bolt pattern has been the intersection of the bolt pattern plane and principal elastic axis. The elastic centers coordinates is the bolts coordinates stiffness-weighted averages. Engineers can be determining the Y and X coordinates of the elastic with the use of

The center of gravity (CG) or the mass centers coordinates are the mass points coordinate the mass-weighted averages.

Both the rocking moment and the pull-out force does generation of tension in bolts. On the elastic center, a pull out force stretches uniformly all bolts by P/K and share of each bolt of reaction has proportionality to its stiffness. Thus, the bolts that are stiffer bears a bigger share of the reaction in pulling out force on elastic center

Off the elastic center, pull out force is replicable by an identical magnitude force at a moment and the elastic center. The forece, P, is what bolts are subjected to along with the induced momentM eP. The resultant of this has been that the reaction for the bolts will be bigger than P/N. In preventing this, the eccentricity is reduced i.e. P is put under the elastic center or to bring CG in the direction of the elastic center.

The rotation of an external moment is with the entire bolt pattern by M/K. The reaction moment and the external moment are in equilibrium.

Reaction force Shear stress average/Area

Shear Stress average F/(2 r2)

Or Shear Stress average 4F (2 d2)(1)

Shear Stress average shear load/area

For G,

Shear load 68 kNArea 86.4 mm

Therefore,

Shear Stress average 0.78

Coming back to equation (1)

F Shear stress average/4(2 d2)

3.14157

d 25

d2 625

Thus,

F 0.78/4(2 x 3.14157 x 625)

F 0.78/2(3926.96)

F 0.78/7853.92

F 0.0000993

Therefore, the reaction force for Ax, Ay is 0.0000993 lbs and the average sheer stress is 0.78 N/mm2.

For R,

Shear load 68 kNArea 86.4 mm

Therefore,

Shear Stress average 0.78

Coming back to equation (1)

F Shear stress average/4(2 d2)

3.14157

d 20

d2 400

Thus,

F 0.78/4(2 x 3.14157 x 400)

F 0.78/2(2513.25)

F 0.78/5026.50

F 0.0001551

Therefore, the reaction force for R is 0.0001551 lbs and the average sheer stress is 0.78 N/mm2.

The deformation of this problem is a case of axial deformation taking place in the body because of the axial loading. Following the axial deformation, the structures axis did not change. The planes cross section area also will be remaining unchanged. Following the axial deformation, the structures cross section cease to rotate about the identical axis. The axial deformation is dependent on the cross sectional position that has been normal to longitudinal axis.

The deformation or deflection in the bar is provided underneath.

Here,

A is the cross sectional area

E is the elasticity modulus

P is the axial load

is the deformation

L is the lengths uniform member

However, this formula is suitable for the given problem. This formula is usable in the event of satisfying the following conditions.

There is axial load

The bars cross section area is uniform, and

Stress must be contained in the proportional limit

However, the bars in question are non-uniform bars.

For the bars that are non-uniform, the deformations formula is provided beneath

Here,

g has been the acceleration because of the gravity

Bars total mass is M

Now,

For ABC,

M 2000 N

g 3000 N

L 200 50 250 mm

E 210 80 GPa 290 GPaA 152 402 225 1600 1825 mm2

Therefore,

ABCs deformation is

2000 x 3000 x 250/2 x 290 x 1825 1500000000/1058500 1417.09

For DEF,

M 2000 N

g 3000 N

L 200 50 250 mm

E 210 80 GPa 290 GPaA 152 402 225 1600 1825 mm2

Therefore,

DEFs deformation is

2000 x 3000 x 250/2 x 290 x 1825 1500000000/1058500 1417.09

Question 4

Along the length of the beam (the x axis), at some distance it experiences the internal bending moment (M) that is obtainable with the use of the bending moment diagram provided. The generic formula of the normal stress or the bending stress is as follows.

bend My/I

Here,

M the internal moment of bending about the neutral axis of the section

y the perpendicular distance to a sectional point from the neutral axis

I the inertia moment of the area in the section about the neutral axis

It is natural that if a particular section of the beam is given, the maximization f the bending stress will be by the distance from the neutral axis (y). Therefore, the occurrence of the maximum bending stress can be either at the bottom or the top of the beam section which depends of the distance that is larger.

bend, max Mc/I

Here,

c from the neutral axis the perpendicular distance to the sections farthest point.

For AB,

bend, max Mc/I

M 600 kNmmc 150 mm

I 802 x 50 320000

Therefore,

bend, max 600 x 150/320000

bend, max 90000/320000 0.28

For BC,

bend, max Mc/I

M 120 kNmmc 120 mm

I 1202 x 40 576000

Therefore,

bend, max 120 x 120/576000

bend, max 14400/320000 0.045

For CD,

bend, max Mc/I

M 240 kNmmc 150 mm

I 602 x 30 108000

Therefore,

bend, max 240 x 150/108000

bend, max 36000/108000 0.33

In this problem, the shaft is comprised of 3 different segments (1) AB (2) BC and (3) CD. Each of the segments has different diameter and have torques with applications at a number of cross sections. The bars each region between the changes and between the applied loads in cross section has been in pure torsion. Therefore, the derived formula can be applied here. Thus, from the internal torque, each regions angle of twist can be calculated with help of the following formula.

T/J t/r

Where,

t bend, max

For AB

t bend, max

t 0.28

r 80

Therefore, the angle of twist for AB 0.28/80 0.0035

For BC

t bend, max

t 0.045

r 120

Therefore, the angle of twist for BC 0.045/120 0.000375

For CD

t bend, max

t 0.33

r 60

Therefore, the angle of twist for CD 0.33/60 0.0055

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