Lecture 2 W2 Stress _nrr.pdf

STRESS Bolts on this structure are subjected to stress.

FACULTY OF CHEMICAL ENGINEERING, FKK, UiTM SHAH ALAM, 2015

Stress     

Force It is the intensity of force. Stress = Area Force per unit area. Normal stress: When the force acts normal to the area Tensile stress: If the normal force ‘pulls’ on the area. Compressive stress: If the normal force ‘pushes’ on the are. F σ avg =

A + for tensile normal stress − for compressive normal stress

Normal stress under axial loading  

Body must be able to withstand safely the intensity of internal force, if not it may rupture/deform. Axial force: Force shown here on the eyebar are collinear with the centroidal axis, and produce tensile loading of the bar.

If cut at a-a plane, the FBD is shown on the right. The distribution of internal force has a resultant force F, normal to the exposed area, equal in magnitude with P.

Units of Stress Commonly used units for stress is psi.  SI unit is N/m2, also known as Pa.  Stress magnitudes normally encountered in engineering applications are MN/m2 or MPa.  Since metal can sustain stress of several thousand pound per square inch, unit of ksi (kip per square inch) always been used (1 ksi =1000 psi). 

FACULTY OF CHEMICAL ENGINEERING, FKK, UiTM SHAH ALAM, 2019

Example Problem 5: Stress 

The flat steel bar has axial loads applied at point A, B, C and D as shown. If the bar has cross sectional area of 3 inch2, determine the normal stress in the bar. a) on a cross section 20 inch, to the right of point A. b) on a cross section 20 inch, to the right of point B. c) on cross section 20 inch, to the right of point C

Shearing stress in connections  

Loads applied are generally transmitted to the individual member through connections e.g. rivet, bolts, pin, nails. The resultant shear force V must be equal to applied load P

V A V = shear force A = bolt area

τ avg =

Single shear

Double Shear. Shear force on each cross section equal to half of the applied load

Example Problem 6- Shear Stress 

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A brass tube with an outside diameter of 2 inch and wall thickness of 0.375 inch is connected to a steel tube with an inside diameter of 2 inch and wall thickness of 0.25 inch by using 0.75 inch diameter pin. Determine:a)The shearing stress in the pin when the joint is carrying an axial load P of 10 kip b) The length of joint required if the pin is replace by a glued joint and the shearing stress in the glue must be limited to 250 psi

Bearing stress (Compressive normal stress)  

It is normal stress resulting from contact between two different bodies. It is the force pushing against a structure divided by the area. Bearing stress occur on surfaces of contact:

σb =

1) Between head of bolt and top plate 2) Between the nut and the bottom plate 3) Between the shanks of bolts (and pins) that are pressed against the side of the hole which they pass.

F A

F = force transmitted across the contact surface A = projected area

SI unit for stress =>

N / Pa 2 m For simplicity

Stresses on inclined plane in axially loaded member For inclined :-

An = A

cos θ

N P cos θ P P = = cos 2 θ = (1 + cos 2θ ) 2A An A A cos θ − P sin θ − P −P V = = sin θ cos θ = sin 2θ τn = A An A 2A cos θ

σn =

σ avg =

τ avg =

F A

V A

Angle θ is measured counterclockwise from positive xaxis to positive n-axis. Why it is cos and the other is sine? N and V depend on the angle θ, and θ with respect to incline plane of applied load.

Stresses on inclined plane in axially loaded member 

Graph below show the magnitudes of σ n and τ n as a function of θ .

 σn  

max when θ is 0O or 180O τ n max when θ is 45O or 135O Therefore, the maximum normal and shearing stress for axial tensile or compressive loading are

σ max

P = , A

τ max

P = 2A

Bolt failure at 45°

Example Problem 7 

A plastic bar with circular cross section will be used to support an axial load of 1000 lb. The diameter of the bar is 1.25 inch. Determine the normal and shearing stresses on section a—a .

Note: Angle θ is measured counterclockwise from positive x-axis to positive n-axis.

Allowable Stress 

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Reasons for Failure of structures: ◦ Actual Load is different from the designed load ◦ Difference in Measurements due to errors in fabrication or in assembly ◦ Unknown Vibrations, impact or accidental loading ◦ Deterioration due to corrosion, decay and weather. ◦ High variability in mechanical properties. An allowable stress is a maximum stress used for sizing structural members to guard against failure of the member when in service. The Factor of Safety takes the allowable stress into account. F.S must be greater than 1 to avoid failure.

σ fail F .S . = σ allow

τ fail F .S . = τ allow

Stress Transformation - Chapter 9 Learning Outcomes Understand the concept on how to transform the stress components that are associated with a particular coordinate system into components associated with a coordinate system having a different orientation.  To identify In-Plane Principal Stresses, σmax,min & Planes Orientation, θp with (τxy = 0)  To calculate Maximum In-Plane Shear Stress & Orientation Planes, θs with ( σaverage ) 

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For analysis purpose, it is convenient to use coordinate system. Consider a small element to show stresses on +ve and –ve surfaces. Stresses σ on planes have outward normals in the x, y, and z directions. Single subscript indicate plane which the stress acts. Shearing stress τ has 2 subscripts (normal plane and parallel plane)

2-Dimensional or Plane Stress 

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Take 2 parallel faces of the small element, perpendicular to the z-axis. The only component of stress present are σx, σy, and τxy = τyx . 2D sketch is shown. Stresses σ and τ on an arbitrary plane such as a—a can be obtained using FBD and equation of equilibrium.

Stress Transformation Equations 

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The equation relates normal and shearing stress on one plane (whose normal is oriented at an angle θ wrt a reference x-axis), through a point and known stress σx , σy , and Ƭxy =Ƭyx on the reference plane that can be developed using FBD. All stress are positive Dotted line a—a represent any plane through the point Stress are multiplied by the areas over which they act.

Counter clockwise angle θ is positive, where θ is measured from the positive x-axis to the positive n-axis. The xy and nt axes shown are positive.

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For the wedge-shaped element, the areas are: ◦ dA for the inclined face (plane a—a) ◦ dA cos θ for the vertical face, ◦ dA sin θ for the horizontal face.

The n-axis is perpendicular to the incline face.  The t-axis parallel to the inclined face. 

The stresses are multiplied by the area.  Summation of force in ndirection and t-direction: 

Sign conventions Tensile normal stress is positive.  Compressive normal stress is negative,  Shearing stress is positive if point in positive direction of the coordinate axis of the second subscript,  Angle is measured counterclockwise from reference +ve x axis. 

FACULTY OF CHEMICAL ENGINEERING, FKK, UiTM SHAH ALAM, 2019

Example Problem 8 At a point on structural member subjected to plane stress there are normal and shearing stresses on horizontal and vertical planes through a point. Use stress transformation equation to determine: a) Normal and shearing stress on plane a—b b) Normal and shearing stress on plane c—d (perpendicular to a—b)

Principal Stresses and Maximum Shearing Stress     

The stress transformation equations provide a means to determine σn & τnt on different planes. As angle varies, 0<θ<360 , σn & τnt varies. At θ = 0º, σn = σx ,τnt = τyx At θ = 90º, σn = σy ,τnt = -τyx = -τ xy . For design purpose, critical stresses at the point are usually the maximum tensile stress and the maximum shearing stress.

General Equation of Plane-Stress Transformation 

Knowing that :σ n @ σ x′ =

σ x +σ y σ x −σ y + cos 2θ + τ xy sin 2θ ....eq(1) 2 2

τ nt @τ x′y′ = −

σ x −σ y sin 2θ + τ xy cos 2θ ........(eq3) 2

To find in y’ direction, substitute θ= θ+90o in eq 1 σ y' = 

σ x +σ y σ x −σ y − cos 2θ − τ xy sin 2θ ........eq(2) 2 2

Maximum and minimum values of σn occur at value of θ for which d σn /d θ=0, the differentiation yields :- d σ n = −(σ x − σ y ) sin 2θ + 2τ xy cos 2θ dθ

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Equal to zero and solving gives, Orientation of the element tan 2θ p =

2τ xy

σ x −σ y

Occurs at Principal planes (planes free of shear stress)

Principal stresses & Orientation of element tan 2θ p = σ 1, 2 =

σ x +σ y 2

2τ xy

Orientation of the element (at σmax, min)

σ x −σ y 2

 σ x −σ y   + τ xy2 ±   2 

Principal stresses on principal planes.

The solution has 2 roots, θp1 and θp2. 2θp1 and 2θp2 are 180° apart. So θp1 and θp2 are 90° apart.

Maximum In-Plane Shear Stress and average stress 

From equation

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Maximum in-plane shear stress occurred when dτ xy / dθ = 0

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Thus, gives the orientation maximum in plane shear stress = tan 2θ s = −

τ nt @τ x′y′ = −

(σ x − σ y ) 2τ xy

σ x −σ y sin 2θ + τ xy cos 2θ 2

Maximum In-Plane Shear Stress and average stress • Maximum in plane shear stress 2

τ max in plane

 σ x −σ y   + τ xy2 =   2 

• Average plane normal stress

σ avg =

σ x +σ y 2

IMPORTANT POINTS The principal stresses represent max and min normal stress. No shear stress acts on it.  For max in-plane shear stress, oriented 45°, an average normal stress will also act on it. 

Summary…. Principal Stress

σ max,min =

σ x +σ y 2

2

2

 σ x −σ y   + τ xy2 ±   2 

τ =0

tan 2θ p =

Maximum In-plane Shear Stress τ max in plane

σ avg = 2τ xy

σ x −σ y

 σ x −σ y   + τ xy2 =   2 

σ x +σ y 2

tan 2θ s = −

(σ x − σ y ) 2τ xy

FACULTY OF CHEMICAL ENGINEERING, FKK, UiTM SHAH ALAM, 2015

Example Problem 9 

Calculate the orientation of element, maximum in plane shear stress and average normal stress.

Mohr’s Circle: Lesson Outcome 

To understand the concept of MOHR’S CIRCLE in solving principle stresses and calculating maximum shearing stress of plane transformation.

Mohr’s Circle 

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Equation for plane stress transformation presented in a graphical solution so that is often convenient to use and easy to remember. No formulas needed. Visualize how the normal and shear stress components vary as the plane which they act is oriented in different directions. Mohr’s Circle was developed by a German civil engineer, Otto Mohr. Rewriting stress transformation equations results in a circle.

Mohr’s Circle: Procedures 

Establish coordinate axes: ◦ X axis: σ positive to the right ◦ Y axis: τ positive downwards.

Plot center of circle, C (σave, 0) where σave is (σx + σy)/2 from the origin.  Establish a point on the outer radius, A (σx , τxy) to the right. Length AC is radius of circle.  Draw circle. 

Principal stresses are at τ = 0, represented by B and D. Maximum in-plane shearing stress Ƭp is point F and E, its accompanying normal stress σaverage that act on plane.

B = σmax D = σmin

Example Problem10 Determine and show on sketch : a) The principal and maximum shearing stress at the point b) The average normal stress on plane