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# Find the maximum positive and negative shear force Assignment Help

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Influence Line Diagram For Shearing Force - Find the maximum positive and negative shear force
**Find the maximum positive and negative shear force:**

Find the maximum positive and negative shear force at point P in beam of Figure which is crossed by two connected wheel loads 3 m aparts moving from right to left. The front wheel carries a load of 20 kN and the rear wheel 10 kN.

**Solution**

**Maximum Negative SF**

For maximum negative shear force at P, the heavier wheel (20 kN) should be placed just to the left of P, the other wheel (10 kN) will then lie at Q which is 3 m to left of P (Figure 8(a)). The ordinate of IL diagram at P is - 0.4 and in which at Q is - 0.1 (by same triangles).

Hence, maximum negative shear force at P,

V_{P} = 20 × (- 0.4) + 10 (- 0.1) = - 9 kN.

**Maximum Positive SF**

Here, two cases need to be examined

(a) When the heavier wheel 20 kN is just crossed to right of P, and the lighter wheel (10 kN) is at Q 3 m behind it (Figure 8(b)). Hence the shear force

V_{P} = 20 × (+ 0.6) + 10 × (- 0.1) = 11 kN.

(b) When the lighter wheel (10 kN) has just crossed to right of P and the front wheel (20 kN) is 3 m to right of it at R as in Figure 8 (c).

The ordinate of IL diagram at P is + 0.6 and at R it is + 0.3.

Hence, shear force V_{P} = 20 × (+ 0.3) + 10 × (+ 0.6) = 12 kN.

The second position gives the higher value; hence, the maximum positive SF will be 12 kN.

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**Find the maximum positive and negative shear force:**

**Solution**

**Maximum Negative SF**

_{P}= 20 × (- 0.4) + 10 (- 0.1) = - 9 kN.

**Maximum Positive SF**

_{P}= 20 × (+ 0.6) + 10 × (- 0.1) = 11 kN.

_{P}= 20 × (+ 0.3) + 10 × (+ 0.6) = 12 kN.