Chapter 1 Combined stresses PDF

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CHAPTER 1 COMBINED DIRECT AND BENDING STRESSES

1.1 INTRODUCTION Direct stress alone is produced in a body when it is subjected to an axial tensile or compressive load. Bending stress is produced in a body when it is a bending moment. But if a body is subjected to axial loads and also bending moments, then both stresses (i.e. direct and bending stresses) will be produced in the body. In this chapter, we shall study the important cases of members subjected to direct and bending stresses. Both these stresses act normal to a crosssection, hence the two stresses may be algebraically added into a single stress.

1.2 COMBINED BENDING AND DIRECT STRESSES Consider a column subjected to compressive load P acting along the axis of the column as shown in Figure 6.1 below. This load will cause a direct compressive stress uniform across the cross section of the column.

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Figure 1.1 Direct compressive stress  d will be given as   d  

P (N/m2) A

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Where A = cross section area

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1.3  ECCENTRIC LOAD WITH RESPECT TO ONE AXIS Many cases occur in which the applied load causes not only a stress due to bending, but also exerts a pull or a thrust on the section, creating direct tensile or compressive stress. Consider a short column or strut on which an eccentric load P acts at a distance e from the centre line through the centroid G of the section as shown in Figure 6.2

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Figure 1.2

To calculate the stresses in the strut, the eccentric load P must be replaced by a direct load P which acts through the centroid of the cross-section and a couple or moment, M = P.e., where e is the eccentricity from G . The resultant stress at any point in any normal section XX, say, consists of a direct stress due to the direct load, P and a bending stress due to the moment P.e., and is the algebraic sum of these stresses. To calculate the resultant stress at the corners of the section:

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1. Calculate the direct stress due to axial load P : P - d   (Compressive) A 2. Calculate the bending stress due to moment P.e. :

M. y I yy

b  but

M = P.e.,

Hence  b  

b 

P.e.b / 2 I yy

P .e .b / 2 I yy

(Tensile) at A and D

(Compressive) at B and C

3. If required, the stress due to the column’s self-weight can also be calculated, if the density in kg/m3 is given. This produces a compressive stress. 4. The resultant stress at faces of the section is the algebraic sum of the stresses i.e. Resultant stress at the faces of the section:

f Q1  

P P.e.b / 2  A I yy

(minimum)

fQ  

P P.e.b / 2  I yy A

(maximum)

Convention: Tensile stress is positive and compressive stress is negative.

EXAMPLE 1.1 (to be done in class) A rectangular column of width 200 mm and thickness 150 mm carries a point load of 240 kN at an eccentricity of 10 mm, as shown in Figure (1.3) below. Determine the maximum and minimum stresses on the section.

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Figure 1.3

1.4 MIDDLE THIRD RULE FOR RECTANGULAR SECTIONS (Condition of no tension) Concrete columns are weak in tension, therefore, the load must be applied on these columns in such a way that there is no tensile stress anywhere in the section. But when an eccentric load is acting on a column, it produces direct stress as well as bending stress. The resultant stress at any point in the section is the algebraic sum of the direct stress and bending stress. Consider a rectangular section of width b and depth d as shown in Figure (1.4). Let this section be subjected to a load which is eccentric to the axis Y – Y.

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Figure 1.4

Direct stress:

d 

 P A

Bending stress on faces AD and BC:

b  

Myx I yy

: x 

b 2

M y  Pe I yy 

db 3 12

b    

Pe . b / 2 db 3 / 12 6 P.e db 2

Total stress: =

T 

6 Pe P  bd db2

P bd

6 e  1  b   

At face AD : we have the minimum stress

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5 

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

P bd

 6 e 1  b   

For no tensile stress

σT  0. 

P bd 1

 6e  1  bd   0. 6e  0 b

b b  e : e  6 6

The above result shows that eccentricity e must be less than or equal to b/6. Hence, the range within which the load can be applied so as not to produce any tensile stress is within

𝑏 𝑏 𝑏   6 6 3 called the middle third of the base. Similarly, if the load had been eccentric with respect to the x – x, the condition that tensile stress will not occur is when the eccentricity of the load with respect to the axis x – x does not exceed d/6, i.e. e  d/6. Also, the range is d/6  d/6  d/3.

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SELF-EXERCISE 1 (A) 1. A rectangular tie-bar 10 m long, 300 mm deep by 100 mm wide is horizontal and subjected to an axial tensile load of 3 750 kN. If a concentrated load of 15 kN acts vertically downwards at mid-span, calculate the maximum and minimum stresses in the bar neglecting its self-weight and assuming the beam to be simply supported at its ends.

2. A short circular column is 4 m long and 0,6 m in diameter. It carries a load of 2 500 kN at a distance 200 mm from the centre. Calculate the maximum and minimum stresses in the column. Sketch the stress distribution diagram and find the position of the neutral axis. 3. A short time-bar 50 mm by 50 mm is subjected to a tensile load of 400 kN, having its point of application 5 mm from a face towards the centre and on a symmetry axis. Calculate the maximum and minimum stresses set up in the bar and state their nature. (+ 544; - 224 MPa)

4. A 305 x 305 x 97 kg/m H-section is used as a short column. It supports a load of 300 kN somewhere along the ‘YY’-axis. Calculate its point of application from the ‘XX’-axis, such that there is no stress in one of the flanges.

5. A universal column section acting as a stanchion carries an axial load and two further loads from incoming beams supported by brackets as shown in figure 6.5 below. Determine the extreme fibre stresses acting at the corners A. B, C and D given the properties of the universal column are: A  11400mm 2

I xx  143 106 mm4

I yy  48 10 6 mm 4

(7.6, 31.6, 57.8, 81.8 MPa)

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Figure 6.5

6. A longitudinal tie is 5 m long and 75 mm in diameter. An axial load of 90 kN acts on its ends. Determine the maximum tensile and minimum compressive stresses in the stay. Assume the stay to be horizontal and simply supported at its ends and that the density of the stay material is 7 850 kg/m3. Take g as 9,81 m/s2 (+ 46,04; - 5,3 MPa)

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1.5

ECCENTRIC LOAD WITH RESPECT TO BOTH AXIS

Consider a rectangular section b x d of a column subjected to a vertical load P at the point that is at a distance ex from the y – y axis and ey from x – x.

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Figure 1.6 Direct stress: σd = -

P A

Bending stress: a. P creates a moment about the YY-axis: My = P.ex

b  

My. Ixx

b 2 (  at A and D;  at B and C )

b. P creates a moment around the XX-axis; Mx = P.ey

b  

M x. Ixx

d 2 (  at A and B;  at C and D)

Resultant stress:

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A

d b M x. M y. P 2 2     I xx I yy A

B

d b M x. M y. P 2 2     I xx I yy A

9 

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C

d b M x. M y. P 2 2     I xx I yy A

D

P    A

d b M x. 2  2 I xx I yy

My.

Note: Method is only valid when x and y axes are axes of symmetry.

EXAMPLE 1.2

Figure 1.7 shows a cross-section of a column with width 300 mm and depth 400 mm. A compressive load P = 360 kN is applied at point 75 mm from y – y axis and 100 mm from x – x axis. Find the stresses at the corners of the column.

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Figure 1.7

SELF-EXERCISE 1 (B)

1. A sort hollow pier 1.2 m square on the outside and side thicknesses of 225 mm, supports a vertical load of 120 kN located on a diagonal at 0,690 m from the vertical axis of the pier. Neglecting the self-weight of the pier, calculate the stress set up at the four outside corners on a horizontal section of the pier (-616,7; -136,8; +343,2 kPa)

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2. A rectangular pier is subjected to a compressive load of 450 kN. Its width is 1500 m and depth is 1000 mm.

If ex = ey = 250 mm Find the stresses on the faces of the rectangular pier. 

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