Test: SFD & BMD Level - 3 - Mechanical Engineering MCQ

When SF = Constant BM should have a linear variation.

When SF linearly varying BM should vary parabolically.

Both the conditions are satisfied for option (A).

The maximum shear force will occur at A.

We know RA + RB = wL / 2

Taking moments about A

For maximum B.M, S. F must be change sign shear force at

x = R_A − wx = 0

∑Fy = 0

= wl / 2

∴ RA = wL / 4

For maximum B.M, S.F must be changes sign Maximum B.M @ c

= 240Nm

Torque = 400 × 0.25 = 100Nm

15 + w x 3 = 14 + 19 = 33 ⇒ w = 6 kN/m

= 16.33

Load of 8 kN at E is equivalent to a load 8 kN at C and a CCW moment of 8 kN at C as shown in equivalent loading diagram as shown in figure.

Reaction Moments about A

8R_B + 8 = 8 × 4 + 5 × 4 × (4 + 2)

= 32 + 20 × 6 = 152

R_B = 15 kN

R_A = 20 + 8 − 18 = 10 kN

BM diagram

MA = 0

MC = 10 × 4 = +40 kN

MC ′ = 40 − 8 = 32 kN M6 = 10 × 6 − 8 − 8 × 2 − 5 × 2 × 1

= 60 − 8 − 16 − 10 = 26 kN/m

M8 = 0

In portion CB

Fx = 10 − 8 − (x − 4) = 0

2 − 5(x − 4) = 0

2 = 5(x − 4)

x = 4.4 m

M = bending moment where SF = 0

= 10 × 4.4 − 8(4.4 − 4) − 8 − 5(4.4 − 4)(0.2)

= 44 − 3.2 − 8 − 5(0.4)(0.2)

= 32.8 − 0.4 = 32.4 kN/m

Reaction at C, RC = −8 + 12 = +4 kN (upwards) ↑

SF diagram: Portion AB(x = 0 to 2 m)

F_AB = +8 kN (constant)

(Taking upward forces on the left side of the section to be positive)

Portion BC (x = 2 to 5 m)

Fx = 8 − 4 (x − 2)

= 0 at x = 4 m

= −4 kN at x = 5

Figure (b) shows SF diagram.

BM diagram: Portion AB (x = 0 to 2 m)

M_x = +8x

= 0 at x = 0 m

= 16 kNm at x = 2 m

Portion BC (x = 2 to 5 m)

There is no point of contra-flexure in the cantilever.

The trapezoidal loading on the beam consists of a uniformly distributed load and a triangular loading as shown in figure.

Reactions: Taking moments about the end A, we have

Consider any section XX at a distance x from A Load intensity at the section XX

= 8 +x / 5 × 8 = 8 + 1.6x

S.F. Analysis S.F. at the section XX

The S.F. diagram follows a parabolic law At x = 0, i. e. , at A

Section of Zero Shear Equating the general expression for shear force to zero,

80 / 3 − 8x − 0.8x2 = 0

∴ 3x2 + 30x − 100 = 0

Solving, we get = 2.637 m.

B.M. Analysis

B.M. at any section XX

The B.M. diagram follows a cubic law.

Maximum bending occurs at the section of zero shear, i.e., at a distance of 2.637 meters from A.

∴ Maximum bending moment

= 37.615 kNm

Let the supports be at A and B so that the overhang BC = a metre ∴ AB = (10 − a)metre

Taking moments about A,

Maximum positive B.M. will occur at a section in AB. Let this section be x metres from A.

∴ S. F. at this section = 0

∴ V_a − 10x = 0

For the condition that the maximum bending moment shall be as small as possible, the maximum positive bending moment and the hogging moment over the support should be numerically equal.

Solving by trial and error, we get a = 2.33 m.

Maximum B.M. with the above value of a

= 5a(a + 2)

= 5 x 2.33(2.33 + 2)

= 50.44 kNm

Figure shows the beam showing the actual position of the supports.

SF and BM diagrams

78.23 kN

Reaction at A = V_a = 110 − 78.23

= 31.77 kN.

Now the SF and BM diagram can be drawn easily.

Taking moments about the left end A,

V_b × 10 + 100 + 150 = 10 x 10 x 5

∴ Vb = 25 kN ↑

∴ Va = (10 × 10) − 25 = 75 kN ↑

S.F. Diagram. At any section distant x from A,

the shear force is given by,

S = 75 − 10x At x = 0, S = +75 kN,

At x = 10, S = −25 kN

Section of zero shear. Equating the general expression for shear force to zero,

75 − 10x = 0

∴ x = 7.5 m

B.M. Diagram. At any section distant x from the end A the Bending Moment is given by,

M = 75x − 5x2 − 150

At x = 0, M = −150 kNm At x = 10,

M = 750 − 500 − 150 = +100 kNm

At x = 7.5 m,

M_max = 75 × 7.5 − 5 × 7.52 − 150

= +131.25 kNm

Point of contra-flexure. Equating the general expression for B.M. to zero,

75x − 5x2 − 150 = 0x2 − 15x + 30 = 0

Solving we get x = 2.375 m.

From right hand side support = 10 − 2.375

= 7.625m

Say reaction at B is RB, for beam AB this reaction will act load for beam BCDE.

Beam BCDE: Taking moments about point E.

6 x 2 + 3 x 5 = 3R_C

Reaction, R_C = 9 kN

Reaction, R_E = 3 + 6 = 9 kN = 0 kN

Maximum shear force in the beam is 6 kN.

V_b x 6 = 30 × 5 + 1 / 2 x 3 x 20 x 3

V_b = 40 kN ↑

Total load applied = 30 + 1 / 2 x 3 x 20 = 60 kN

V_a = 60 − 40 = 20 kN

Consider any section × in CD distance x from C.

Load intensity at the section x = 20x / 3

S. F at the section X = Sx = 20 − 1 / 2x − 20x / 3

S.F at any section in DE = 30 − 40 = −10 kN

S.F at any section in EB = −40 kN

V_b × 3 + 20 × 1.2 × 0.6 = 20 × 1.8(3 + 0.9)

3. V_b + 14.4 = 140.4

V_b = 42 kN

V_a = (20 × 1.2) + (20 × 1.8) − 42 = 18 kN

BM at C = 0

B. M at A = −20 × 1.2 × 0.6

= −14.4 kNm

BM at B = −20 × 1.8 × 0.9

= −32.4 kNm

BM at D = 0

The 10 kN load on the bracket is transmitted to B along with a clockwise couple of

10 × 1 = 10 kNm.

Taking moments about A,

V_d × 6 = (10 × 3) + (3 × 4 × 6)

∴ V_d = 17 kN ↑ and

V_a = 10 + (3 × 4) − 17 = 5 kN ↑

S.F. Analysis

S. F. at any section in AB = +5 kN

S. F. at any section in BC = +5 − 10 = −5 kN

S. F. at any section in CD,

distant x from C = S = 5 − 10 − 3x

∴ S = −5 − 3x

At x = 0, S = −5 kN, At x = 2,

S = −5 − 3 × 2 = −11 kN

S. F. at any section in DE,

distant x from E = S = 3x

At x = 0, S = 0 and at x = 2, S = +3 × 2 = +6 kN

The S. F. at D changes suddenly from

−11 kN to + 6 kN.

B. M Analysis B. M. at A= 0

B. M. on LHS of B = +5 × 2 = +10 kNm

B. M. on RHS of B = +10 + 10 = +20 kNm

B. M. at C = +5 × 4 − 10 × 2 + 10 = +10 Nm

B. M. at D = −3 × 2 × 1 = 6 kNm

B. M. at E = 0

From SF diagram Reaction at A, RA = 6.5 kN ↑

Reaction at C, RC = 5.5 + 4 = 9.5 kN ↑

Over AB = 3 m

Overhang, CD = 1 m

Load at point, D = 4.0 kN ↓

Fig. (b) shows the loading diagram of the beam.

BM diagram

Position AB(x = 0 − 3 m)

M max.

SF in portion AB,

6.5 − 4(x) = 0 = SF

x = 1.625 m

M_max = 6.5 × 1.625 − 2 × 1.625²

= 10.5625 − 5.28125

= 5.28 kNm

Point of contra-flexure Less in portion BC,

Mx = 6.5x − 4 × 3(x − 1.5)

Note that CG of UDL lies at 1.5 m from A

6.5x − 12(x − 1.5) = 0

6.5x − 12x + 18 = 0

18 = 5.5x

x = 3.27 m

Fig (c) shown BMD of the beam.

RA + RB = 12 × 6 + 36 = 108kN

Taking moment about A

R_B = 84kN, RA = 24kN

B.M @ section XX is zero

⇒ x = 4m