Mechanical Engineering (1)

A Course

For

1st Year Bio-Medical Engineering

lecture 2

Structural Analysis

By

Dr. Sief A.Y. Khorshid

Mechanical Design and Production Dept.

Cairo University – Faculty of Engineering

2.1. Structural bars

2.2. Structural Plates and sheets

2.3.Structural supports

2.4.Structural Connections

2.5.Structural Static Analysis

2. Structural Elements

Structural elements are divided into four groups namely:

  • Structural Bars.
  • Structural plates and sheets.
  • Structural supports
  • Structural Connections

2.1. Structural bars

A structural bar is a body formed by the movement of a constant cross section along its centroidal line. The centroidal axis of the bar is defined as the line or curve containing the loci of the bar cross sectional c.g.

Types of bars

Bars are defined by several names according to the type of loading acting on each. Engineers to indicate the type of bar loading generally use the following names;

Tie: A bar subjected to axial tensile force acting along its axis.

Strut: A bar subjected to axial compressive force acting along its axis.

Beam: A bar subjected to transverse load acting normal to its axis or

bending moments acting normal to its axis.

Shaft : A circular bar bar subjected mainly to torque.

Pipe : A hollow circular bar mainly subjected to internal pressure.

2.2. Structural Plates and sheets

Structural bars may be arranged in an order ro bear loads. Plates and sheets may be used to fill in between these bars. Plates and sheets are subjected to loads along its surface and transverse to its surface. The following drawing shows the use of plates and sheets with bars.

2.3.Structural supports

Structural supports are used to describe the relation between bar ends and the supporting body. In the following table , displacements are given as free or restricted. Resulting loads such as forces and moments are given either in quantitive value or null value.

No / Support
Name / Description & Sketch / Displacements / Loads
ux / uy / z / Rx / Ry / Mz
Rigid / / 0 / 0 / 0 / 1 / 1 / 1
Pivoted / / 0 / 0 / 1 / 1 / 1 / 0
Roller / / 1 / 0 / 1 / 0 / 1 / 0
No / Support
Name / Description & Sketch / Displacements / Loads
ux / uy / z / Rx / Ry / Mz
Rigid Slider / / 1 / 0 / 0 / 0 / 1 / 1
Pivoted Slider / / 1 / 0 / 1 / 0 / 1 / 0

2.4.Structural Connections

Bars are connected together through one of two methods namely:

  • Rigid connection
  • Hinged connection

The following figure shows the geometric presentation and the equilibrium conditions controlling the acting loads.

The following relations are applied at point A;

For rigid connection
/ For Hinged connection

2.5.Structural Static Analysis

Structural static analysis aims at the determination of loads acting on body composed of an assembly of structural elements. Acting loads means specifically the acting and reacting forces and moments operating on the elements of this body. For a body made of bars, connected at their ends and supporting loads, the following tips are considered useful during static analysis.

  • Load determination is always possible if equilibrium conditions are enough to obtain the unknown loads
  • A bar connected to other bodies through its two end hinges will only allow force to be transmitted along the line passing by the these two hinges.
  • Bar subjected to more than two loads acting at different points should allow t all these loads are meeting at one point where a resultant force and moment are equal to zero.
  • A system of bars in equilibrium is composed of bars where each bar satisfy equilibrium separately.

Example (3)

A man is climbing on a wooden ladder resting on frictionless ground as shown. The weight of the man is 85 kilograms while the ladder weight is 40 kilograms., obtain the following;

  1. The reactions at points A and C if the man is standing at the middle height of the ladder.
  2. The load acting on the ladder sides, middle tie and the top hinge if the man is sitting on the top of the ladder.
  3. Determine the reactions at points A and C when the man at top of the ladder is applying a horizontal pull force of 200 N . Assume that the contact between ladder end C and the ground is frictional . Give the loas acting on the ladder side BC .

Loads

wt1 = 85  981 = 833.85 N

wt2 = 40  9.81 = 392.4 N

Equilibrium During Climbing

wt1 +wt2 = Ay +Cy

Moment about C

Ay  1500 – 833.85  1500/4- 392.4  1500/2

Ay = 404.66 N

Cy = 833.85 + 392.4 –404.66 = 821.59 N

Results

Ay = 404.66 N, Cy = 821.59 N

Equilibrium During sitting on top of ladder

wt1 +wt2 = Ay +Cy

Moment about C

Ay  1500 – (833.85 + 392.4)  1500/2

Ay = Cy = 613.125 N

Equilibrium of Ladder side A-B

Bx + Dx = 0

Dx = - Bx

By – 833.85/2 –392.4/2 +613.125 = 0

By = 0

Moment about point A

Bx  3000 + Dx  1500 + (833.85/2)  (1500/2) + (392.4/2)  (1500/4) = 0

2Bx + Dx = -515.025 N

2Bx – Bx = - 515.025 N

Bx = -515.025 N & Dx = 515.025 N

Results

Ay = 613.125 N, Bx = -515.025 N, By = 0 Dx = 515.025 N

Pulling Situation

For the whole system

Fp + Cx =0

Cx = -Fp = -200 N

wt1 + wt2 = Ay + Cy = 833.75 + 392.4

moment about C

Ay  1500 + Fp 3300 – ( wt1+wt2)  (1500/2) =0

Ay = (833.85 + 392.4)  (1500/2) – 200  3300 = 173.125 N

Cy = 833.85 + 392.4 – 173.125 = 1053.125 N

For element BC

Cx + Ex + Bx + Fp = 0

-200 + Ex + Bx + 200 = 0 , Ex + Bx = 0

(-wt1/2) - wt2 + Cy + By = 0

By = (833.85/2) + 392.4 - 1053.125 = - 243.8 N

Moment about point E

Cx  1500 + Cy  (750/2) + (833.85/2)  750 + 200  (3300-1500) + Bx  3000 + By  (750/2) = 0

Dividing by 1500

-200 + 1053.125/4 + 833.85/4 –200  ( 1800/1500) + 243.8/4 = 2 Bx

Bx = 46.35 N

Ex = -46.35 N

Results

Ay = 173.125 N, Bx = 46.35 N, By = -243.8 N ,

Cx = -200 N, Cy = 1053.125 N, Ex = -46.35 N .

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