ESO 202A/204: Mechanics of Solids (2016-17 II Semester) Solution of Assignment No. 6

6.1 6.2

6.3

6.4

Let us use superposition of loads.

Note: Net B.M.D will be superposition of two diagrams.

6.5

q(x) = -q0sin(πx/l) dV

+ q(x) = 0 dx dM

+ V (x) = 0dx

} M=0 at x=0, l

V=-q0(l/π)cos(πx/l)+C1

M=q0(l2/π2)sin(πx/l)-C1x+C2

Using B.C we find that C1=C2=0

(i) For no +ve B.M in the beam: qol2/8 ≤ q0a2/2 i.e. a ≥ l/2

(ii) For Max. + ve B.M = Max. - ve B.M: Net + ve B.M = (qol2/8 - q0a2/2)

Therefore, (qol2/8 - q0a2/2) = q0a2/2 Or, l2/8 = a2 or, a = l/(2√2) For a < l/2, the SFD & BMD are:

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6.6

Total drag = D

Drag per unit length = D/L

tanθ = (540/1800) gives, θ0 = 16.70

L = 540mm

∑Fx = 0 gives, Rx+Tx = D which gives, Rx = (D-Tx) ∑Fy =0 gives, Ry = Ty

B

∑MB = 0 gives, Tx.L-D.(L/2)-M0 = 0, which gives, M0 = (TxL-DL/2) = moment at the base of antenna (which is negative B.M. and is maximum)

Now, the bending moment at a section x is:

Mb = -Tx.x +Dx2/(2L) (which is positive B.M.)

For this moment to be maximum,

0=dM b gives, -Tx + Dx /(L) =0, gives, x/L =Tx/D dx

Mb(max)= -Tx2.L/D + Tx

2.L/2D = - Tx2.L/2D

For maximum moment to be minimized, we must have, Tx.L – DL/2 = - Tx

2.L/2D or, Tx

2 + 2D.Tx – D2 = 0 or, Tx = 0.414D

Tx = T cosθ = 0.9578T hence, T = 0.432D

6.7 Horizontal reaction at A will be zero.

(1) ∑Fy = 0 gives RA+RB +RC = 8x100

∑MA = 0 gives 8 RB +16 RC =100x8x4 (2)

Consider a section at B, given that MB = - 400 KNm.

Therefore, 8 RC + 400 =0 RC = -50 KN. From 1 and 2, RB = 500 KN RA = 350 KN

One can now draw the SFD and BMD

400(Given)

RC

8 m

BM where SF is zero, i.e at 3.5 m from support A = RA x 3.5 -2

5.35.3100 xx = 612.5KNm

To find X (Where BM is zero):

RA.X - 2

100 .X2 = 0 or 350X - 50 X2 = 0,

X = 7m