Why are two arrangements (case I and case II) checked for the 20-25-30 kN train?

The train may move in either direction, so either 30 kN or 20 kN may lead

The beam has two supports so two cases are needed

Shear and moment need different loads

The overhangs are of different length

Where does the negative ILD for M_D lie, and what is the maximum negative moment?

On the overhangs; 30 kN at C gives -37.50 kN m

On the overhangs; 20 kN at E gives -31.25 kN m as the governing value

What is special about the influence line for shear at D?

It has a jump of 1 at D, from -0.375 to +0.625

It is a single triangle with peak at D

It is zero for loads on the overhangs

It has equal ordinates on both sides of D

Which results are obtained for maximum positive and negative shear at D?

V+ = 25.00 kN (case II), V- = -11.25 kN (case I)

V+ = 11.25 kN (case I), V- = -25.00 kN (case II)

V+ = 25.00 kN and V- = -10.62 kN both from case I

2069 Poush · 8 marks

Overhang beam: max BM and SF at D, either direction

Short answer

Max positive BM at D = 81.875 kN m (25 kN over D), max negative = −37.50 kN m (30 kN at C). Max positive shear = 25.00 kN (20 kN leading, 30 kN just right of D), max negative shear = −11.25 kN (30 kN just left of D).

For the overhang beam determine the maximum positive and negative bending moment and shear force at D due to the three concentrated loads as shown in figure which moves in either directions. [Figure: overhanging beam C-A-D-B-E; CA = 2 m, AD = 3 m, DB = 5 m, BE = 2 m; supports at A and B; loads 20 kN, 25 kN, 30 kN with spacings 2 m and 3 m.]

Data: same overhanging beam $C$ – $A$ – $D$ – $B$ – $E$ ( $CA = 2$ , $AD = 3$ , $DB = 5$ , $BE = 2$ m; supports $A$ and $B$ ). Loads 20, 25 and 30 kN with spacings 2 m (20–25) and 3 m (25–30). Because the train may move in either direction, both arrangements are checked: 30 kN leading (case I) and 20 kN leading (case II). $x$ is measured from $C$ .

ILDs for D

$M_D$ ordinates (m): $C$ : −1.25; $A$ : 0; $D$ : +1.875; $B$ : 0; $E$ : −0.75.
$V_D$ (shear just at $D$ ): for a load between $C$ and $D$ , $V_D = -R_B = -\dfrac{x-2}{8}$ ; for a load between $D$ and $E$ , $V_D = R_A = \dfrac{10-x}{8}$ .

Position	$C$ (0)	$A$ (2)	$D^-$ (5)	$D^+$ (5)	$B$ (10)	$E$ (12)
$V_D$ ordinate	+0.25	0	−0.375	+0.625	0	−0.25

The ILD of shear has a jump of 1 at $D$ .

Bending moment at D

Case I (30 kN leads): maximum $+$ : 30 kN at $x=8$ , 25 kN over $D$ ( $x=5$ ), 20 kN at $x=3$ :

M^{+} = 30(0.75) + 25(1.875) + 20(0.625) = 81.875\ \text{kN m}

Maximum $-$ : 30 kN at $C$ : $M^{-} = 30(-1.25) = -37.50$ kN m.

Case II (20 kN leads): the best positive value is only 75.00 kN m (20 kN at $B$ , 25 kN at $x = 8$ , 30 kN over $D$ ) and the negative value -31.25 kN m, both smaller in magnitude.

Shear force at D

Positive shear (case II, 20 kN leading): 30 kN just to the right of $D$ ( $0.625$ ), 25 kN at $x = 8$ ( $0.25$ ), 20 kN at $B$ (0):

V^{+} = 30(0.625) + 25(0.25) + 20(0) = 25.00\ \text{kN}

Negative shear (case I): 30 kN just to the left of $D$ ( $-0.375$ ), 25 kN at $A$ ( $0$ ), the 20 kN not yet on the beam:

V^{-} = 30(-0.375) = -11.25\ \text{kN}

(Case II gives only -10.62 kN for the negative shear.)

Answer (either direction): $M_D^{+} = 81.875$ kN m; $M_D^{-} = -37.50$ kN m; $V_D^{+} = 25.00$ kN; $V_D^{-} = -11.25$ kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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