Draw a Shear and Moment Diagram for the Beam
The vertical force acting o the axle or any construction is known as shear force and the turning outcome on the structure is known as the bending moment.
The diagram which represents the different effective forcefulness on the axle is known as Shear and Moment diagrams. I have given in the example beneath.
Let us see some numerical example of beam for better understanding.
Also Read, Shear force diagram and Bending Moment diagram for Different Load
Draw the Shear and Moment diagrams for the beam
1) Calculate the shear strength and bending moment for the beam subjected to concentrated load every bit shown in the effigy. Likewise, draw the shear forcefulness diagram (SFD) and the angle moment diagram (BMD).
Solution;
Gratis body diagram of the given figure,
Taking moment almost betoken B,
RAy x 4 – xx ten 2 = 0
4 RAy = 40
RAy = xl/iv = x KN
Sum of vertical force is equal to nada i.eastward. Ʃ V = 0
RAy + RBy – xx = 0
10 + RBy -twenty = 0
RBy = 10 KN
Sum of horizontal strength is equal to zero i.eastward. ƩH = 0
RAy + RBy – 20 = 0
10 + RBy -20 = 0
RBy = 10 KN
Sum of horizontal force is equal to cipher i.e. ƩH = 0
RAx = 0
Shear strength calculation
Shear strength at A left (FAL) = 0
Shear force at A right ( FAR) = 10 KN
Shear forcefulness at C left (FCL) = 10 KN
Shear force at C right (FCR) = 10 – twenty = -10 KN
Shear strength at B left (FBL) = -10 KN
Shear force at correct (FBR) = 10 – x = 0
Bending moment calculation
Moment at A (MA) = 0
Moment at C (Mc) = 10 x 2 = 40 KNm
Moment at B (MB) = 10 x iv – 20 x 2 = 0
two) Calculate the shear force and bending moment diagram of the beam as shown in the figure. Also draw shear force diagram (SFD) and bending moment diagram (BMD).
Solution;
Free body diagram of the given effigy is given below;
Taking moment about betoken B , we get
ƩMB = 0
RAy x 6 – 10 ten 4 – x x 2 = 0
half-dozen RAy = 40 + twenty
RAy = 60/6 = ten KN
Sum of vertical force is equal to nothing i.e. Ʃ V = 0
RAy + RPast – x – 10 =0
10 + RPast -20 = 0
Rpast = 10 KN
Sum of horizontal force is equal to zero i.e. ƩH = 0
RAx = 0
Shear force calculation
Shear force at A left ( FAL ) = 0
Shear force at A right (FAR) = 10 KN
Shear forcefulness at C left (FCL) = x KN
Shear force at C right (FCR) = x – 10 = 0
Shear force at D left ( FDL) = 0
Shear force at D right (FDR) = 0 – 10 = -10 KN
Shear force at B left ( FBL) = -x KN
Shear force at B correct (FBR) = – 10 + 10 = 0
Bending moment calculation
Moment at A (MA) = 0
Moment at C(GrandC ) = x x 2 = twenty KNm
Moment at D (MD) = 10 x 4 – ten x two = xx KNm
Moment at B (MB) = 10 ten half dozen – 10 x iv – 10 ten two = 0
iii) Observe the shear force and angle moment of the given effigy. Besides, summate shear strength diagram and bending moment diagram.
Solution;
Gratuitous body diagram of the given figure is given below;
Taking moment about bespeak B, we go
RAyx 5 – ten x iv – 5x ii x 2/2 = 0
5 RAy -50 = 0
RAy = 10 KN
Sum of vertical force is equal to aught i.e. Ʃ V = 0
RAy + RBy– 10 – 5 ten 2 = 0
10 + RPast – 20 = 0
RPast = 10 KN
Sum of horizontal force is equal to goose egg i.e. ƩH = 0
RAx = 0
Shear strength adding
Shear force at A left (FAL) = 0
Shear force at A right (FAR) = x KN
Shear strength at C left (FCL) = 10 KN
Shear force at C right (FCR) = ten -10 = 0
Shear force at D left (FDL) = 0
Shear force at D right (FDR) = 0
Let, 10 be the distance from point D.
Shear force at (x = ane) = 0 – five 10 1 = -5 KN
Shear force at (x = 2 ) left = -5 x 2 = -10 KN
Shear force at B correct (FBR) =-ten + 10 = 0
Bending moment adding
Moment at A = 0
Moment at C = 10 x 1 = x KNm
Moment at D = 10 x 3 – 10 x 2 = 10 KNm
Moment at B = 10 10 5 – x x 4 – 5 x 2 x2/2 = 0
4) Calculate the bending moment and shear forcefulness of the given beam. Also draw shear force diagram and bending moment diagram.
Solution,
Free body diagram of the given figure is given below;
Taking moment near point B, we go
ƩMB = 0
RAy ten 4 – 10 x four x 4/2 = 0
4 RAy – 80 = 0
RAy = 80/4 = 20 KN
Sum of vertical forcefulness is equal to nothing i.e. Ʃ V = 0
RAy + RBy -10 x four = 0
20 + RBy – 40 = 0
RPast = 20 KN
Sum of horizontal force is equal to aught i.eastward. ƩH = 0
RAx = 0
Shear force calculation
Let, 10 exist the distance from point A.
Shear forcefulness at A left ( FAL) = 0
Shear force at A right ( FAR) = 20 KN
Shear forcefulness at (ten =i ) = 20 – 10 x ane = x KN
Shear force at (ten = 2 chiliad) = 20 – x ten 2 = 0
Shear force at ( x = iii ) = 20 – 10 x 3 = – 10 KN
Shear force at B left ( FBL) = 20 – x 10 four = -xx KN
Shear force at B right (FBR) = -twenty +twenty = 0
Bending moment calculation
Moment at A ( MA) = 0
Moment at (10 = one) = 20 x 1 – 10 x one x ½ = fifteen KNm
Moment at (x = 2 ) = 20 x 2 – ten ten 2 10 2/ii = 20 KNm
Moment at (ten = 3) = twenty x three – 10 x 3 10 3/ii = 15 KNm
Moment at B (MB) = 20 x four – 10 x iv ten four/two = 0
In this way nosotros can Draw the Shear and Moment diagrams for the axle.
Read Also,
Point of contraflexure – Zero bending moment at a section of Axle
Moment of inertia formula | Definition for moment of inertia
Shear Force Diagram and Bending Moment Diagram
Kinematic indeterminacy and Static indeterminacy – Axle, Frame etc
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