strength of materials most steel item used in ship building are divided into 4 general categories....

Strength of Materials

Most steel item used in ship building are divided into 4 general categories.

–Beams–Plates–Columns–shafts


Each of the 4 groups have their individual function, but also must support the other 3 groups

The coordinated functions of all groups result in a vessel hull forming a long square beam or girder

Strength for the hull girder is determined by calculation based on a Trochoidal Wave


Trochoidal Wave• Wave length from crest to crest is equal to

the length of the vessel• Wave height is equal to 1/20th of the length

of the vessel

The use of a wave of this size will determine the maximum load on the hull girder, and scantlings required for construction


Structural Terminology:

• Load- The total force acting on a structure, usually expressed in pounds or tons

• Stress- The force per unit area, usually expressed in pounds or tons per square inch

• Strain- The distortion resulting from stress


• Tensile Stress- Occurs between 2 parts of a body when each draws the other toward itself– Ex. A tie rod subjected to a steady pull

Formula:

Ts=Pull = P

Area A


Compressive Stress- This is just the reverse of Tensile Stress– Ex. A tie rod subjected to a steady push

The same formula applies


Shearing Stress- The tendency of one body to slide over another part is known as shear

The magnitude of this tendency to slide at any point is termed Shearing Stress

Formula:

Ss= Pull = P

Area of the rivet A


Ultimate tensile strength of mild steel (primary steel used in ship construction) is 28 to 32 tons per square inch

Ultimate shearing Strength of mild steel is 22 tons per square inch

Steel flattens when compressed at about 18 tons per square inch


Diagram explanation:

Working Range: linear line between the 0 point and point “A”


Point “A” is also refered to as the “Limit of Elasticity”

• The point at which metal will no longer return to its original shape


If the material is loaded (even slightly) beyond point “A”, it will yield at point “B”, the “Yield Point”.

A further slight increase in load will cause the metal to stretch uniformly until it reaches point “C”, the “Yield Overcome”.


At point “C”, the metal becomes harder, so the yield is overcome.

If a slight additional weight is then added, the metal will stretch out of all proportion to the load applied until it reaches point “D”, the “Ultimate Strength”.

The metal will fail at point “E”


Curves such as the one illustrated for mild steel are available for most metals used in ship construction

The generation of these curves is known as “Testing to Destruction” as the sample must be destroyed to determine the failure point


Safety Factors

Factors of safety (F.S.) are build into ships to allow for exceptional loads not considered by the standard trochoidal wave model. Vessel may be subjected to extreme weather conditions or, due to corrosion and wastage, the original hull steel may be weakened. The standard F.S. is considered 4.


Safety Factors are based on the Ultimate Strength of the sample, or point “D”. As the working range is about ½ of that, the actual, practical factor of safety is only 2.


When referring to ships, there are 2 very broad phases of strength:

Local Strength

Hull-girder Strength


Local Strength

The strength of individual parts of a ship

Hull-girder Strength

The strength of the ship as a whole


Class Societies

Class societies are the regulator that actually determine the scantlings of such construction components as beams, stiffeners and shell plating. Most are insurance driven and are based on catastrophic loss in the past.


Bending Moment-

A moment of a force about any line is the product of the force times the perpendicular distance to that line.


Example A

100 lbs is located 4’ from the end of a board imbedded in a wall

B.M. = Weight x Distance

B.M. = 100lbs x 4’

B.M. = 400 ft-lbs


Example B



B.M. = 100lbs x 8’

B.M. = 800 ft-lbs


Example C



B.M. = 200lbs x 4’

B.M. = 800 ft-lbs


Beams-Horizontal strength members loaded verticallyThe beam in the next slide is a homogeneous material and rectangular, it is symmetrical

The ends of the beam are considered free as they are not imbedded in the wall


A load applied to the center of the beam will cause a deflection. The upper surface will shorten, the lower surface will lengthen, and the middle will remain neutral.

The upper surface must be under compression.

The lower surface must be under tension


As compression and tension are opposite forces, there must be a layer in the beam where the forces are neutral.

The zero point is located along the center of gravity (centroid) of the beam.


If a beam of similar size is made up of 5 individual layers that are free to slip over one another, the top 2 layers would not be under compression as they are allowed to slip, the bottom 2 layers are not under tension. The middle will not be neutral.

This beam will carry only 1/5th the load of a solid or laminated beam.


The depth of a solid beam controls its resistance or strength. Another way of considering this is the farther from the neutral axis the compression and tension edges are, the stronger the beam.


Example:Formula:

Relative Strength=(D2/D1)2

where D1 is the depth of the smaller of the 2 beams considered.

Using a 2” x 4” and a 2” x 8” we can see(8/4)2= 4 or the 2” x 8” beam is 4 times stronger than the 2” x 4” beam


In the beam with 5 individual members, each 1” thick, compared to a similar sized solid beam 5” thick, the same formula applies

(D2/D1)2=Relative Strength

(1”/5”)2= 1/25th as strong for each pieceThe five pieces considered together will be 1/5 as strong


Because the maximum force (compression and tension) on a beam is concentrated at the upper and lower edge, the material of the rectangular beam can be re-distributed to the upper and lower edges, it will gain a great deal of strength without gaining weight,


There are 5 factors that determine the size of a beam:

1) Type and amount of load on the beam

2) Distance between supports3) Type and efficiency of end connections4) Number of supports5) The material the beam is constructed of.


Type and amount of load on a beam

A) Concentrated load

B.M.= WL/4

W= weight

L= distance between supports

Example: 2 tons x 10 ft/ 4 = 5 ft-tons


Type and amount of load on a beam

B) Uniform loan

B.M.= WL/8

W= weight

L= distance between supports

Example: 2 tons x 10 ft/8 = 2.5 ft-tons


Distance between supports ( often referred to as span)

A) Deflection of a rectangular free-end beam varies as the cube of the span

Example: A 10’ span has a deflection of 1”

What will be the deflection on a beam with a 20’ span?

(S2/S1)3= (20’/10’)3= 8 inches


Distance between supports

B) Strength of a rectangular free-end beam varies inversely as the span

Example: A 10’ span will support 10 tons

How much weight will a 20’ span support?

Relative strength= span A/span B

10’/20’=.5

10 tons x .5 = 5 tons


Type and efficiency of end connections

A) A fixed-ended rectangular beam will support twice as much concentrated load as a free-ended beam

B) The deflection of a fixed-ended rectangular beam is 1/4th that of a free-ended beam


Effects of number of supports

The greater the number of supports in a given distance, the shorter the span. A shorter span means a smaller bending moment


Material the beam is constructed of

Most ship construction is mild-steel.

Other materials used include:

Stainless Steel

High tensile steel

Aluminum

Strength of Material

Columns

A strut placed such that it is loaded vertically (also referred to as a stanchion or pillar)

Columns are usually symmetrical (round)


Shafts

A shaft subjected to a twisting moment is said to be in torsion

The twisting moment is referred to as torque

Torque (lb-ft)= horsepower x 5252/RPM

The higher the RPM, the less torque


Example:

A 5,000 horsepower 80 RPM engine will require about the same size shaft as a 20,000 horsepower 320 RPM engine

Torque= 5,000 x 5252/80=328250 lb-ft

Torque= 20,000 x 5252/320=328250 lb-ft

Strength of Material

Continuity of strength

Vessels must be constructed such that stresses may be gradually and continuously dissipated.

No part should be oversized or undersized

A discontinuity or change in shape will cause a concentration of stresses and may result in a failure

strength of materials most steel item used in ship building are divided into 4 general categories....

Documents

yield point

point b

point d

materialsat point c

limit of elasticitythe

original hull steel

ultimate strength

square inchstrain