Yesterday,there are 2 question from one U student regarding this matter.Q1,given one horizontal beam with 4 force from the to and have distance one and other that 3 forces.Q2 also given same withe the Q1 but more forces from the top.Determine the reactions value and produced SFD and BMD.
What SFD(shear force diagram) and BMD (Bending Moment Diagram)?According to the sources SFD is defined the grapical representation of the variations of the shear force along the length of the beam and is abbreviated as SDF
BMD is the grapical representation of the variations of the bending moment along the length of the beam and is abbreviated a BMD.Before that shear force is defined as the algebratic sums of all the vertical forces either to be left or to be right hand side of a section.For bending moment is defined as the algebric sum of the moments of all the vertical forces either to be left or to be right of the section.
Moment= Force x distance ,where the ∑Mx or y is zero.Let example of one a simply supported beam A and force 20 N at B and 2 m distance from A,30 N at C and another 3 m distance from B and last 20 N at D and 4 m apart from C and also at support at D. So how to detemine the reactions and SFD and BMD.
First take a moment from the left or right of the beam to get the Ra or Rc value.Let we take at A is a moment zero.
∑Ma=20 x 2 + 30 x 5 + 20 x 9-Rd x 9
Solve the equation Rd=40 +150 +180/(9)
Rd=41.11 N
∑Fy=0,Ra +Rd= 20 +30 +20
Ra=70 -41.11 =28.89 N,finish or take moment at D
∑Md= 0,30 x 4 + 20 x 7-Ra x 9
Ra=120 + 140/(9)=28.89 N same as result on the above
and Rd= 41.11 N.These equation is the basic formulae for calculation to get the value of the reactions.All the type of the beam are applying the same concepts and formulas to get the reactions value.
From the two reactions are known now we can draw the SFD and BMD for the above proble.Take a garph papers x axis denotes for distance and y axis denoted as shear force value or bending moment value.The right scale have to apply to the two axis.
The x we mark A,B,C and D follow the beam dimensions and forces distance.Can start from two direction,from A or D,let we start from A,+ 28.89 N we mark at y axis and draw at straight line from the 28.89 n value until meet the vertical line at B,at B - 20 must deduct from 28.89 N,so the value of shear at B is 28.89 - 20=8.89.Maerking straight line from B until meet the vertical line at C,-30 deduct from 8.89 ,so -30 + 8.89=--21.11N,at D,the clear shear force is 41.11-20= 21.11 N.to check wheather our answer is correct,the value of the end D must be same value.
So the SFD can be drawn.For the BMD ,just start from A or D,if start from A ,draw line from A is 0 and calculate BMD for A and B, 28.89 x 2= 57.78 Nm,at + side and draw from line D is 0 and to the calculate BMD for C and D, -21.11 x 4=84.56 N so at C the value of BM are -84.56 N and value of BM at B is 57.78.From point at B draw a line to C.So we get the BMD and the value of BM along the beam.
So many type of beam can solve by this method but for a long beam ,we use a soft ware or excel to get the value but now,the beam calculator application at handphone can calculate the reactions on the spot and from there can draw a SFD and BMD.Thank you