SFD and BMD for centrally point loaded beam

Draw SFD and BMD for centrally point loaded beam.

Solution:

Step 1: Calculate support reactions

$$\sum Fy = 0$$

Upward forces = Downward forces

R_A + R_B = W____(Eq, 1)

$$\sum Ma = 0$$

Now, we are considering loads on right side of A

By sign conventions of BM

Moment due to load "W" is -ve because this moments rotate in clock-wise direction and clock-wise direction moments on right sides are considered as -ve.

Moment due to load "R_B" is +ve because this moment rotate in anti clock-wise direction and anti clock-wise direction moment on right side are considered as +ve.

$\sum Ma$	$= 0$
$-\left (w\times\frac{L}{2}\right )+\left ( R_{B} \times L\right )$	$= 0$
$\left ( R_{B} \times L\right )$	$=\left (w\times\frac{L}{2}\right )$
$R_{B}$	$= \frac{\left ( w \times \frac{L}{2} \right )}{L}$
$R_{B}$	$= \left ( w \times \frac{L}{2} \right )\times \frac{1}{L}$
$R_{B}$	$= \frac{W}{2}$

Put this value in Eq. (1)

$R_{A} + \frac{W}{2}$	$= W$
$R_{A}$	$= W - \frac{W}{2}$
$R_{A}$	$= \frac{W}{2}$

Step 2: Calculate shear forces

While calculating shear force; it may be calculated for left or right side of point. For good practice here we calculated it for left side of point. At each point SF is calculated just left and just right side of each point.

E.g.

SF at just left point of A, SF at just right point of A,

SF at just left point of B, SF at just right point of B,

SF at just left point of C, SF at just right point of C,

Here we mention SF at just left point of A as SF_LA. And SF at just right point of A as SF_RA.

Now, start calculating SF form left to right the sign conventions are as mentioned above.
Here we calculating SF form left to right so forgot about right side signs.

1) Shear force at just right of A
SF_LA = 0 (There is no any force in left of beam there for SF at left of A is zero)

2) Shear force at just right of A

Only reaction due to A support is present in left side of section X-X and just right of point "A". And this reaction is upward. Upward forces at left of section are consider +ve. Therefor

$SF_{RA}=+R_{A}=+\frac{W}{2}$
3) Shear force at just left of C

Only reaction due to A support is present in left side of section X-X and just left of point "C". And this reaction is upward. Upward forces at left of section are consider +ve. Therefor

$SF_{LC}=+R_{A}=+\frac{W}{2}$
4) Shear force at just right of C

Here reaction due to A support and downward force "W"(Point load at C) is present in left side of section X-X and just right of point "C". The reaction "R_A" is upward hence it is taken as +ve and point load acting on point "C" is downward hence it is taken as -ve. Therefor

$SF_{RC}=+R_{A}-W=+\frac{W}{2}-W=-\frac{W}{2}$
5) Shear force at just left of B

Here reaction due to A support and downward force "W"(Point load at C) is present in left side of section X-X and just left of point "B". The reaction "R_A" is upward hence it is taken as +ve and point load acting on point "C" is downward hence it is taken as -ve. Therefor

$SF_{LB}=+R_{A}-W=+\frac{W}{2}-W=-\frac{W}{2}$

6) Shear force at just right of B

Here reaction due to A support, downward force "W"(Point load at C) and reaction due to support B is present in left side of section X-X and just right of point "B". The reaction "R_A" is upward hence it is taken as +ve, point load acting on point "C" is downward hence it is taken as -ve and reaction due to support B is upward hence it is taken as +ve. Therefor

$SF_{RB}=+R_{A}-W+R_{B}=+\frac{W}{2}-W+\frac{W}{2}=0$
All these SF calculated summarized as below,

$SF_{LA}=0$
$SF_{RA}=+\frac{W}{2}$
$SF_{LC}=+\frac{W}{2}$
$SF_{RC}=-\frac{W}{2}$
$SF_{LB}=-\frac{W}{2}$
$SF_{RB}=0$

Step 3: Draw SFD

Now, SFD is drawn as below
(Note: SFD must be drawn below the beam)

Step 4: Calculate bending moments

While calculating bending moments; it may be calculated for left or right side of point. For good practice here we calculated it for left side of each point. At each point BM is calculated.

E.g.

BM at point of A,

BM at point of B,

BM at point of C,

Here we notation for BM at point of A as M_A.
At any point
Bending Moment = Force * Perpendicular Distance

Now, start calculating BM form left to right the sign conventions are as mentioned below,

Here we calculating BM form left to right so forgot about right side signs.
1) Bending moment at point A

There is no any force at left of point A. Therefore

$BM_{A}=0$

2) Bending moment at point C

At this point only reaction due to support A is present at left side of point C. This moment due to RA is Clock wise. By sign convention clock wise moment at left side of point is +ve; hence moment due to RA is +ve. There fore,

$M_{C}=R_{A}\times Perpendicular\: Dist.\: between\: A\: and\: C$

$M_{C}=\frac{W}{2}\times \frac{L}{2}$

$M_{C}=\frac{WL}{4}$

3) Bending moment at point B

At this point there are two forces acting at left side of point B.

1. Reaction due to support A.

2. Point load at C.

1. Reaction due to support A : Reaction due to support A is clock wise and it is taken as +ve because of by sign convention clock wise moment at left side of point is taken as +ve. It acting at L distance form point B.

2. Point load at C : Point load at C is anti-clock wise and it is taken as -ve because of by sign convention anti-clock wise moment at left side of point is taken as -ve. It acting at L/2 distance form point B.
Therefor,
$M_{B}=+\left (R_{A}\times Distance\: bet.\: A\: and\: B \right )-\left ( W\times Distance\: bet.\: C\: and\: B \right )$
$M_{B}=+\left (\frac{W}{2}\times L \right )-\left ( W\times \frac{L}{2} \right )$
$M_{B}=+\left (\frac{WL}{2} \right )-\left ( \frac{WL}{2} \right )$

$M_{B}=0$