A square footing 4m x 4m carries an udl of 628 KN/m2 Find σz at 8 m depth below a point 1 m inside each of the two adjacent sides of the footing. Use equivalent point load method.A square footing 4m x 4m carries an udl of 628 KN/m2 Find σz at 8 m depth below a point 1 m inside each of the two adjacent sides of the footing. Use equivalent point load method.
Answers
Answer:
1.A simply supported beam of span 10m carries a udl of 20 kN/m over its central 4m length. With the help of influence line diagram, find the shear force at 3m from the left support.
2.A single rolling load of 100 kN moves on a girder of span 20m.
(a) Construct the influence lines for (i) shear force and (ii) bending moment for a section 5m from the left support. (b) Construct the influence lines for points at which the maximum shears and maximum bending moment develop. Determine these values.
Solution:
To find maximum shear force and bending moment at 5m from the left support:
(i) Maximum positive shear force
By inspection of the ILD for shear force, it is evident that maximum positive shear force occurs when the load is placed just to the right of D.
Maximum positive shear force = load * ordinate = 100 * 7.5
At D, SFmax + = 75 kN. (ii) Maximum negative shear force
Maximum negative shear force occurs when the load is placed just to
the left D.
Maximum negative shear force = load * ordinate = 100 * 0.25 At D, SFmax = -25 kN.
(iii) Maximum bending moment
Maximum bending moment occurs when the load is placed on the section D itself.
Maximum bending moment = load * ordinate = 100 * 3.75 = 375 kNm
(b) Maximum positive shear force will occur at A. Maximum negative shear force will occur at B. Maximum bending moment will occur at mid span. The ILs are sketched in fig.
(i) Positive shear force
Maximum positive shear force occurs when the load is placed at A. Maximum positive shear force = load * ordinate = 100*1
SFmaxmax + = 100 kN (ii) Negative shear force
Maximum negative shear force occurs when the load is placed at B. Maximum negative shear force = load * ordinate = 100 * (-1)
SFmaxmax = - 100 kN (iii) Maximum bending moment
Maximum bending moment occurs when the load is at mid span Maximum bending moment = load * ordinate = 100 * 5 = 500 kNm
3.Draw the ILD for shear force and bending moment for a section at 5m from the left hand support of a simply supported beam, 20m long. Hence, calculate the maximum bending moment and shear force at the section, due to a uniformly distributed rolling load of length 8m and intensity 10 kN/m run.(Apr/May 05)
Solution:
(a) Maximum bending moment:
Maximum bending moment at a D due to a udl shorter than the span occurs when the section divides the load in the same ratio as it divides the span