MECHANICS OF SOLIDS

MECHANICS OF SOLIDSCHAPTER# 3STRESSES IN BEAMS

Beam:A beam is a horizontal structural member used to support loads. Beams are zused to support the roofs and floors in buildings. Beam may be defined as a structural member whose length is relatively large in comparison to its thickness and depth and which is loaded that produce significant bending effect as oppose to twisting or axial effects.Classification of beams:Beams are generally classified according to their geometry and the manner in which they are supported.

Beams used in Buildings and Bridges: Girders:Girder is the main horizontal support of the structure which supports smaller beams. Usually these are most important beams, which are, frequently used t wide spacing.

Joists:Usually less important beams, which are closely spaced, frequently with truss type webs.

Stringers: Longitudinal bridge beams spanning b/w floor beams.

Purlins:Roof beams spanning b/w trusses.

Girts:Horizontal wall beams serving principally to resist bending due to wind on the side of an industrial building.

Lintels:Members supporting a wall over a window or door openings.

Introduction To Beams:The parallel portions on an I-beam or H-beam are referred to as the flanges. The portion that connects the flanges is referred to as the web.

An important variable in beam design is the moment of inertia of the cross section, denoted by I. Inertia is a measure of a bodys ability to resist rotation. As I increases, bending and deflection will decrease.Moment of inertia is a measure of the stiffness of the beam w.r.t the cross section and the ability of the beam to resist bending.

Its units are, E.G. .

The area moment of inertia is the secod moment of the area around a given axis. For example given, the axis and the shaded area shown, one calculates;

Moment of inertia about x-axis

Moment of inertia about y-axis

Moment of inertia about z-axis Moment of inertia as an important value to determine The state of stress in a cross section Resistance to buckling The amount of deflection in a beam Radius of Gyration Example:If a designer is given a certain set of constraints on a structural problem (i.e. loads, spans, end conditions), a required value of the moment of inertia can be determined. Ten, any other structural element which has at least that specific moment of inertia will be able to be utilized in the design.If the inverse of the above problem in design i.e. if a specific element is given in a design. Then the load bearing capacity of the element could be determined.I can be different for any common area using calculus. However, moment of inertia equation for common cross section (e.g rectangular, circular, triangular ) are readily available in math and engineering.For a rectangular cross section,

Here b is the dimension parallel to the bending axis. h is the dimension perpendicular to the bending axis.

Where is known as 2nd moment of Area about x-axis which is also centroidal axis. Parallel axis theorem:The parallel axis theorem is a relation b/w the moment of inertia about an axis passing through the centroid and the moment of inertia about any parallel axis.

Where A= Area of Cross Sectiond=Distance of axis from centroidal axisin general the parallel axis theorem for any two parallel axes as long as one [asses through the centroid.

Parallel axis theorem proof:

Problem no.1:Calculate the moment of inertia about the X-axis and Y-axis for a yardstick that is 20mm high and 5mm thick.Solution:For X-axis

For Y-axis

These calculations are very simple for a solid, symmetric cross section.Now consider slightly more complex symmetric cross sections, e.g. hollow box beams. Calculating the moment of inertia takes a little more effortConsider a hollow box beam;

the same equation for moment of inertia, ,can be used.X-axis

Treat the outer dimensions as a positive area and the inner dimension as a negative area, at the centroids of the both are about the same axis i.e. X-axis. Problem no 2;

Calculate the moment of inertia about the X-axis for the positive area and the negative area using. The outer dimensions will be denoted with O and the inner dimensions will be denoted with I.

Solution: for ,

Solution:for ,

Thus,

Moment Of Inertia Of An I-Beam:The moment of inertia of I-beam can be calculated in a similar manner.Identify the positive and negative areas. And calculate the moment of inertia similar to the box beam (note the negative area dimensions and that it is multiplied by 2).

. Moment Of Inertia Of H-Beam:The moment of inertia of H-Beam can be calculated in the similar manner. However the H-Beam is divided into three positive areas.

Problem no 3:

calculate the deflection in the I-Beam. The I-Beam is composed of three steel plates welded together.

Solution:First, calculate the moment of inertia for an I-Beam as we know i.e. divide the cross section of the beam into positive and negative areas.

As, .

Now calculating deflection .

,

Problem no 4:

Calculate the mass and volume and mass of the beam if the density of the steel is 490 .Solution:Volume = (Area) (Length)

P.T.O.

Now Calculating Mass Of The Beam

Converting into cubic feet

Mass = Density Volume

Moment of inertia of built-up sections:Let us take three rectangular cross section and arrange in different orders.Its moment of inertia can be calculated by two ways;1)

2)

Moment of inertia of built-up section:Let us take three rectangular cross section and arrange in same order.1) first method:

2nd method:

Moment of inertia of built-up section:Let us take three rectangular cross section and arrange in same order.1) 1st method:

2) 2nd method:

Calclution of moment of inertia:1) 1st method:

2) 2nd method:

From above we conclude that three rectangular cross section which are arrange in different order, has same area but different moment of inertia.

Stresses in beams:Forces and couples acting on the beam cause bending (flexural stresses) and shearing stresses on any cross section of the beam and deflection perpendicular to the longitudinal axis of the beam. If couples are applied to the ends of the beam and no forces act on it, the bending is said to be pure bending. If forces produce the bending, the bending is called ordinary bending.

Centroid:It is the middle point of a geometrical figure. For two dimensions the centroid is center of area, and for three dimension the centroid is the center of volume.For discrete Areas:For continuous Areas

P.T.O

Neutral axis :The fibers above the neutral axis are in compression and the fibers below the axis are in tension. This hypothetical line is call the neutral axis and upon which the stress value is zero. radius of Curvature R:

Change in length per original.Length L is called strain

For a linearly elastic material,

Bending equation

Problem. No. 5:A 250mm depth 150mm width rectangular beam is subjected to bending moment of 750kNm. Determine;i. The maximum stress in the beam ii. If the value of E is 200GPa, find the radius of curvatureiii. The value of stress at a distance of 65mm from the top surface of beam.Solution:Bending Stress:Moment of inertiaDistance for neutral axisBending stress value

b = 150mmd = 250mmM = 750kNmE = 200GPa = ?R = ?

Bending stress Value at Y= 0.065Radius of curvature

Problem. No. 6:An I-section girder, 200mm wide by 300mm deep, with flange and web of thickness 20mm is used as a simply supported beam over a span of 7m. The girder carries a distributed load of 5 kN/m and the concentration load of 20kN at mid-span. Determine:(a) The second moment of area of the cross-section of the girder,(b) The maximum stress set-up.

Problem. No. 7:

A symmetrical section 200mm deep has a moment of inertia of about its neutral axis. determine the longest span over which, when simply supported, the beam would carry a uniformly distributed load of 4kN/m run without the stress due to bending exceeding

Data;

Section Modulus:The section modulus of a beam is the ratio of a Moment of inertia to the distance of the extreme compressive fiber from the neutral axis. Section modulus of the cross sectional shape is of significant importance in designing beams. It is a direct measure of the strength of the beam. A beam with larger section modulus will be stronger.

Rectangular section:

S = section modulus I = Moment of inertia about NAY = distance from NAD = depth of beamB = thickness of beam

Hollow Rectangular section:

Twin Beam;Consider the case of two rectangular- sectioned beams lying one on top of the other and supported on simple supports. If some form of vertical loading is applied the beams and with negligible friction b/w the mating surfaces of the beams each beam will bend independently of the other and as a result the lower surface of the top beam will slide relative to the upper surface of the lower beam.If the beams are replaced by a single solid bar of depth equal to the combined depth of the initial two beams. Then there must be stresses, set up at the central fibers within the beam to prevent the sliding as bending takes place. Stress and forces at N.A. of Beam: The portion of a beam of length dx and an element AB distance y from the N.A.

Under any loading system the B.M. across the beam will change from M at B to (M + dM). at A. now as a result of bending.Bending moment at B = MBending moment at A = M + dM

Longitudinal stress at A

Longitudinal force at A

Longitudinal stress at B

Longitudinal force at B

Net force at NA

Therefore total out of balance shear force from all sections above height y

Therefore total out of balance shear force from all sections above height y. thus if the shear stress is , then

= Q= first moment of area of shaded portion about the N.A.

= = where A is the area of shaded portion and y is the distance of its centroid from the N.A.

= rate of change of the B.M.= shear force= VTherefore equation 1 become

Where

= Shear Stress V = Shear forceI = moment of inertiaQ = 1st moment of areab = thickness

Neutral axis, Shear Plane and Shear flow:Shear flow q is the horizontal shear force per unit distance along the longitudinal axis of the beam

Problem. No. 8:Three board each of size 38 90mm rectangular cross section and nailed together to form a beam that is subjected to vertical shear of 1 kN. Knowing that the spacing between each nail is 60 mm. determine the shearing force in each nail.

Moment of inertia1st moment of areaShear flowShear force at 60mm of each nail

Problem. No. 9:A beam is made of three planks, nailed together. Knowing that the spacing between nails is 25mm and that the vertical shear in the beam is V = 500N. determine the shear force in each nail.

Moment of inertia1st moment of areaShear flow

Shear force at 25mm of each nail

Problem. No. 10:A square box beam is constructed from four planks. Knowing that the space b/w nails is 1.5in and the beam is subjected to a vertical shear of magnitude V = 600lb, determine the shearing force in each nail.

Moment of inertia:

1st Moment of area:

Shear Flow:

Edge force per unit length:

Shear force at 25mm of each nail

Chapter 4:CIRCULAR CHAFT: Torsion:Torsion is the twisting of an object due to an applied torque.When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted.

The parallel lines with the axis of the shaft distort (bend) into helixes. Line perpendicular to the axis of the shaft remains perpendicular. Therefore, plane section remains undistorted.

Plain section warps (Non circular sections):Cross section of noncircular shaft are distorted when subjected to torsion

Angle of Twist:The quantity is the rate of change of the angle of twist w.r.t. the distance x measured along the axis of the bar. It is denoted by and this refers as the angle of twist per unit length or the rate of twist.

Shear strain in Torsion:

Shear stress in Torsion:If the material is elastic, and has modulus of rigidity G then the circumferential shearing stress is given as

J or polar moment of inertia:Polar moment of inertia or polar 2nd moment of area is an object's ability to resist torsion.

Where

is the moment of inertia about x axis

Is the moment of inertia about y axis

or J is the polar moment of inertia about z axis The larger the Polar Moment of Inertia the less the beam will twist.

Solid Shaft:If the shaft has a solid circular cross section, the polar moment of inertia can be determined using an area element in the form of a differential ring.

Hollow Shaft: If the shaft has a solid circular cross section, the polar moment of inertia can be determined using an area element in the form of a differential ring.

Twisting moment:

Let the cross section of the shaft be considered as divided into elements of radius r and thickness dr as shown in following figure each subjected to a shear stress

This force will produce moment about the central axis of the shaft providing the contribution to the torque.

Torsion Formula for Solid and Hollow Circular Shaft.

= Shear Stressr = radius of circular shaftT = Torque or Twisting momentJ = Polar moment of inertiaG = Modulus of rigidity = Angle of twistL = length of circular shaft

Problem. No. 11:

A shaft is made of a steel alloy having an allowable shear stress of .If the diameter of the shaft is 1.5 in. (a) Determine the maximum torque T that can be transmitted. (b) What would be the maximum torque if a 1-in.-diameter hole is bored through the shaft? (c) Sketch the shear-stress distribution along a radial line in each case.

Data:

Shaft material = Alloy steel

Allowable stress = Diameter of shaft = 15 inDetermine: Maximum torque for solid shaft = T = ?

Maximum torque for hollow shaft = = ?Shear Stress distribution for Solid shaft = ?Shear Stress distribution for Hollow shaft =?

T = torque

= Shear stress J = Polar Moment of Inertia

Problem 12: The copper pipe has an outer diameter of 40mm and an inner diameter of 37mm. If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe.

Problem 12: What torque, applied to a hollow circular shaft of 25 cm outside diameter and 17.5 cm inside diameter will produce a maximum shearing stress of 75 MN/m2 in the material.

Problem 13: Ship's propeller shaft has external and internal diameters of 25 cm and 15 cm. What power can be transmitted at 110 rev/minute with a maximum shearing stress of 75 MN/m2, and what will then be the twist in degrees of a 10 m length of the shaft? G = 80 GN/m2.

Chapter 5:SPRINGS / STRAIN ENERGY:

Spring:Springs are energy-absorbing units whose function it is to store energy and to release it slowly or rapidly depending on the particular application.

Classification of springs:

Helical Spring :It is made of wire coiled in the form of helix. Type of spring formed by winding strips of metal around a cylinder. The helical spring is the most common spring. Material of spring: Carbon steels Alloy steels Corrosion resisting steels Phosphor bronze Spring brass Beryllium copper Nickel alloy steels Titanium alloy steels

Standards of specifications UNS: Unified Numbering System ASTM American Society for Testing and Materials AISI: Stands for American Iron and Steel Institute AMS: SAE / Aerospace Materials Standards SAE: Society Of Automotive Engineers Federal, MIL. Sp.: Federal and Military specifications JIS: Japanese Standards Association BS: British standards Compression Helical Spring:A helical spring is a spiral wound wire with a constant coil diameter and uniform pitch. The most common form of helical spring is the compression spring but tension springs are also widely used. . The strength of the steel used is one of the most important criteria to consider in designing springs. Specifications:

Problem 14: A close coil helical spring is to carry a load of 500 N. Its mean diameter is to be 10 times that of a wire diameter. Calculate the dia if the maximum shear stress in the material of the spring is to be 80 Mpa.

Problem 15: A close-coiled helical spring is required to absorb 2.25 x l03 joules of energy. Determine the diameter of the wire, the mean diameter of the spring and the number of coils necessary if (a) The maximum stress is not to exceed 400 MN/m2; (b) The maximum compression of the spring is limited to 250 mm; (c) The mean diameter of the spring can be assumed to be eight times that of the wire. For the spring material G = 70 GN/mZ.

Problem 16: A leaf spring of steel 1 m long is to support a central load of 5.8 kN of a Carriage. If the maximum deflection of spring is not to exceed 45 mm and maximum stress should not exceed 300 Mpa, calculate thickness of each plate and number of plates. Take E = 200 Gpa.

BY IRFAN SALAH & THOR KHAN BS(civil 3RD)B BUITEMS Page 29