4.1 Rolling loads and influence line diagrams for determinate beams and trusses

1. Influence line:

An influence line for a   given characteristic, which includes a reaction, axial pressure, shear pressure, or bending moment, is a graph that indicates the version of that characteristic at any given factor on a shape because of the software of a unit load at any factor on the shape. Influence line is additive and scalar. The scaled most and minimal are the essential magnitudes that need to be designed for within side the beam or truss.

Examples:

1] A simply supported beam is subjected to loads as shown in fig.

Using influence lines find

Support reactions

Shear force & bending moment at ‘C’

Solution:

1) VA&VB

VA=20+251=16+10=26KN

VB=20+25+=4+40=44KN

=0.6

SF @c=20(-0.2) +25=5KN

BM @c=20(0.6) +(20.8) (25) =32KN.M

2. Rolling loads:

Loads which roll over the given structural detail from one quit to the every other quit. In a rolling load most shears force, bending moments.

Example: Train on the railway track, automobiles at the bridges or roads are rolling loads etc.

Rolling loads there are 3 types:

1. Single factor rolling load

2. Uniformly distributed load

3. Two factor rolling load

Uniformlydistributedloadtherearemaintypes:1)Longerthanspan2)Smallerthanspan

ILD for beams and rolling load:

It is a diagram which shows version of AF, SF and BM on the sort of characteristic at a unique section for numerous function of moving load.

It is a diagram which indicates variation of AF, SF and BM on any such function at a particular section for various position of moving load.

AFD, SFD, BMD    ILD

Load position fixed    (for particular%)

How to find value of function from ILD

i) Structure subjected to point loads

Meg. Of function=W1Y1+W2Y2+W3Y3+W4Y4

Ii) Structure subjected to udl

Magnitude of function= {Area of ILD under UDL}𝓌

i)VA=?(ii)MA=?(iii)SF=(iv)BM=

1)     ILD’s for VA & MA

Consider unit load at a distance

MA=MA-1(x)=0

MA=x

X  VA  MA

0  1  0

L  1  0

ILD for bending moment at A is as below

ILD for shear force & BM at ‘C’

SF @c=0, BMC=0

SFC=1; BMC=-1(x-a)

For x=a; BM@ c=0X=L;

BMC=-1(1-a) cb

2)

1)ILD’s for VA&VB

x  VA  VB

0  1  0

L  0  1

2) SF c & BM c

1)

2) 

Example:

1]A simply supported beam subjected to moving udl

Solution:

1)

VA max= (20) =75KN

VA max=20=75KN

Note: Absolute maximum shear force is nothing but greater 07VA, max & VB, max

Abs. Max SF=75KN

2)

Max.(–ve) SFc= (-) (20)

= ()16KN

Max. +ve SF =35KN

Max. SF c=35KN

Note:

When length of udl is smaller than bas of triangle, placed udl such that load per on LHS & RHS of height of triangle is same.

x=2m

=(5)(20)=180KNm

Abs. Max BM=(5)(20)=187.5KNm

Diagonal members design

Top camp

Key takeaways

1. Find VAVB

2. Find SF

2. Find BM

3. Draw influence line

4.2 Absolute maximum bending moment and shear force

A. Maximum Bending moment:

Consider a section x-x at the distance x- from left support

∑ (BM)XX = WL/2. X- (W.X) X/2 = WL/2 = WX2/2

Bending moment equation shows that BM curve is parabolic

At x=0    BMA = 0

At X= L      BMB = WL/2 X L – WL2/2 = 0

Position and magnitude of maximum BM

Maximum bending moment occurs where shear force is zero.

Now location of point where SF = 0

∑ (SF) XX = 0

WL/2 – WX = 0

Therefore x= L/2

Bending moment at x = L/2

BM = WL/2 (L/2) – W/2 (L/2)2 = WL/4-WL2/8

B.  Maximum Shear force calculation:

Consider a section x-x at the distance X from A

∑ (SF)XX = WL/2 –WX

At x = 0, SFA = WL/2

At x= L,    SFB = WL/2- WL = -WL/2

Example:

A simply supported beam of span L is carrying uniformly distributed load w over its entire span calculate S.F. And B.M. Also draw SFD and BMD.

Solution:

1. Support reaction:

Since the beam is subjected to the symmetrical loading support reactions is half of the total load

2. Shear force calculation:

Consider a section x-x at the distance X from A

∑ (SF)XX = WL/2 –WX

At x = 0, SFA = WL/2

At x= L,    SFB = WL/2- WL = -WL/2

3. Bending moment:

Consider a section x-x at the distance x- from left support

∑ (BM)XX = WL/2. X- (W.X) X/2 = WL/2 = WX2/2

Bending moment equation shows that BM curve is parabolic

At x=0    BMA = 0

At X= L      BMB = WL/2 X L – WL2/2 = 0

Position and magnitude of maximum BM

Maximum bending moment occurs where shear force is zero.

Now location of point where SF = 0

∑ (SF) XX = 0

WL/2 – WX = 0

Therefore x= L/2

Bending moment at x = L/2

BM = WL/2 (L/2) – W/2 (L/2)2 = WL/4-WL2/8

Key takeaway:

1. Find support reaction

2. Find shear force

3. Find bending moment

4. Draw SFD

5. Draw BMD.

4.3 Muller-Breslau’s principal & its applications for determinate structures.

Muller- Breslau principle:

Influence liens for simply supported beam

Find reactions at support, S.F.& B.M. At a section _______ at A  RA=?

10  y1 +   2  30 + 60  y4

= y1 =0.8       =  y3=0.4

  =       y2 =0.6       =  y4=0.2

RA = 10  0.8 +  2  30 + 60   0.2

 RA =KN

RB= 10  y1 +  30  2 + 60  y4

  =       y1 =0.2       =  y3=0.6

  =       y2 =0.4       =  y4=0.8

RB = 10  0.2 +  2  30 + 60   0.2

 RA = 62    KN

Near for at E

= = 0.5           =  y3

= = 0.5           =  y4

  =  y2          =  y1

At E = 10  y1 +30  1+ 30  1 + 60  y4

At E =     8   KN

-ve S.F.at E = 10  -y1 + 30  1 

   =      =      =2.5

  =  =  =

  =  y2  =

B.M. At E = 10  y1 + 30  1         +30  1  + 60  y4

Find the position of loads to get max-ve S.F. Of B.M.

Two wheel loads of 80 KN & 100 KN at 3 m find max.m the S.F. & maxm -ve S.F. And B.M. At 5 m from left

Application of determinate structure:

A simply supported beam is subjected to loads as shown in fig.  using influence lines find

1) VA& VB

VA = 20   + 25  1   = 16 + 10 = 26 KN

VB = 20   + 25    +     = 4 + 40 = 44 KN

    =   =0.6

SFc = 20 (-0.2) +  25= 5KN

BMc = 20 (0.6) + (2  0.8) (25)

= 32 KN/M.

References: