numerical-problems-on-sfd-and-bm

Problem: Draw the shear force and bending moment diagrams for the loaded beam.

From the load diagram: using Σ MD=0, we get 30 x 6 + 50 x 2 = RB x 5 RB=56kN RD=30 + 50 – 56 =24 kN Consider a portion of Segment AB at distance x from A

Vx= –30 kN Mx = –30x kNm Consider a section in Segment BC at distance x from A

Vx= –30+56, Vx=26 kN Mx= –30x+56(x–1) Mx =26x–56 kNm Consider a section in Segment CD at a distance x from A

VCD=Vx=–30+56–50

VCD= –24 kN

MCD= Mx = –30x+56(x–1)–50(x–4) = –30x+56x–56–50x+200 MCD=–24x+144

kNm

To draw the Shear Diagram: 1. In segment AB, the shear is uniformly distributed over the segment at a magnitude of –30 kN. 2. In segment BC, the shear is uniformly distributed at a magnitude of 26 kN. 3. In segment CD, the shear is uniformly distributed at a magnitude of –24 kN. To draw the Moment Diagram: 1. The equation MAB = –30x is linear, at x = 0, MAB = 0 and at x = 1 m, MAB = –30 kN·m. 2. MBC = 26x – 56 is also linear. At x = 1 m, MBC = –30 kN·m; at x = 4 m, MBC = 48 kN·m. When MBC = 0, x = 2.154 m, thus the moment is zero at 1.154 m from B. This point where bending moment is changing sign is called POINT OF CONTRAFLEXURE. 3. MCD = –24x + 144 is again linear. At x = 4 m, MCD = 48 kN·m; at x = 6 m, MCD = 0.

These must be on the same page and one above the other in the order shown. POINTS OF IMPORTANT OBSERVATION: (i) Where shear force changes sign are the points of maximum bending moment clear from SFD and BMD. (ii) Numerically highest will be maximum shear force from SFD (iii)Numerically highest will be maximum bending moment from BMD

Problem: Draw the SFD and BMD for the loaded beam shown.

Solution: Σ MA=0, 10RC=2(80)+5[10(10)], RC=66kN Σ MC=0, 10RA=8(80)+5[10(10)], RA=114kN A section in Segment AB:

VAB=114–10x kN MAB=114x–10x x/2 MAB=114x–5x2 kNm A section in Segment BC:

VBC=114–80–10x VBC=34–10x kN MBC=114x–80(x–2)–10x(x/2) MBC=160+34x–5x2

kNm

Problem: Draw the SFD and BMD for the loaded beam. 30 kN/m

Find the reactions by using the equations of equilibrium. Σ MA=0, 6RD=4[2(30)], RD=40 kN MD=0, 6RA=2[2(30)],

RA=20 kN

Consider segment AB

VAB=20 kN MAB=20x kNm Consider segment AC

VBC=20–30(x–3) VBC=110–30x kN MBC=20x–30(x–3)(x–3)2 MBC=20x–15(x–3)2 kNm Consider segment AD

VCD=20–30(2) VCD=–40 kN MCD=20x–30(2)(x–4) MCD=20x–60(x–4) kNm

PROBLEM: Draw the SFD and BMD for the loaded beam.

The maximum bending moment occurs just below where SF changes sign in SFD.

PROBLEM: DRAW THE SFD AND BMD FOR THE LOADED CANTILEVER BEAM.

Consider segment AB:

VAB=–20 kN MAB=–20x kNm Consider segment AC

VAC=–20

kN

MAC=–20x+80 kNm

Problem: Draw the SFD AND BMD for the loaded beams shown.

Problem: Draw the BMD for the loaded beam shown. Also show the actual shape of the bent beam. Also mark the point of contra-flexure.

Fig : Loaded Beam

Fig. Free Body Diagram

Fig. Bending moment diagram

Fig. Shape of the bent beam

PROBLEM: Draw the SFD and BMD for the loaded beam.

PROBLEM: Draw the shear force and bending moment diagrams.

Fig. BMD •

Identify the maximum shear force and bending-moment from SFD and BMD. Vmax=26 kN and Mmax = 50 kNm

PROBLEM: A simply supported steel beam is to carry the distributed and concentrated loads shown. Draw the SFD. From SFD find the location of the maximum bending moment.

Maximum bending moment occurs where the shear force changes sign i.e. at point E. Mmax=M2.6 =52 x 2.6 –20 x 2.6= 135.2—52 = 83.2 kNm.

PROBLEM: Draw the SFD and BMD for the loaded beam with one end hinged and other end is simply supported. Four steps: Draw free body diagram. Find reactions by using equations of equilibrium. Draw SFD. Draw BMD.

Problem: Draw The SFD and BMD for the loaded beam shown.

Problem: Draw the SFD and BMD for the loaded beam shown.

PROBLEM: Draw the SFD and BMD for the loaded beam.

Problem: Draw the SFD and BMD for the loaded beam shown below. Indicate point of contra-flexure if any.

PROBLEM: Draw the SFD and BMD for the loaded cantilever beam.

HEAR FORCE DIAGRAM AND BENDING MOMENT DIAGRAMS IN THE SHORTEST WAY Draw loaded beam Below this draw free body diagram. Below this draw shear force diagram. Below this draw bending moment diagram.

LOADED BEAM, FREE BODY DIAGRAM, SFD and BMD NOTE: When a concentrated force acts downward then the shear force diagram will jump downward at that particular point. When a concentrated force acts upwards then the shear force diagram will jump upward at that particular point. The UDL will be with an inclined line. •When a moment M is applied clockwise on left, the moment diagram will jump upward. When M acts counterclockwise on left, the moment diagram will jump downward. The points of concentrated loads are joined by straight lines and UDL by smooth curves.