Video 1

In this lecture we will discuss about the eccentrically loaded base plate. So when the base plates are loaded eccentrically or subjected to axial load as well as bending moment the pressure distribution from the concrete does not remain uniform. So if concentric load is there then only the uniform pressure from the concrete will act on the base plate, as P/A. But here if the load is eccentrically acting or certain moment is acting then the stress in one side it will be tension, in other side it will be compression. So the stress development will be linearly varying about its neutral axis and the stress will be M/Z where Z is the section modulus. Therefore the stress at two different extreme ends will be different, somewhere it will be P/ A+M /Z , in other direction it will be P/ A−M / Z. Therefore the magnitude of the stress in two direction in two points will be different, therefore we have to find out the thickness of the base plate on the basis of maximum stress developed on the on the base plate due to this not uniform pressure from the concrete block.

So if we see here that if the column is eccentrically loaded say if the load is P is with a distance of e then the load distribution below the column will not be uniform. Due to axial load P, the pressure distribution is uniform and stress diagram is rectangular.The direct stress ¿ PA1=PLBWhere, A1=¿ area of base plate provided.L=¿ length of the base plateB=¿ width of the base plateDue to the bending moment the bending stress developed is¿± MIy¿± M112B L3×L2=±6 M B L2=± 6PeB L2The combined stress due to axial load and bending momentf = PLB ±6 PeB L2 =PLB(1± 6Le)

Depending upon the eccentricity and length of the base plate three cases are possible.Case I when there is no tension, 1−6 e L=0⇒e= L6Case II when the tension developed is small e= L6 to L3

Case III When the tension developed is substantial e> L3

Video 2

Case I (e< L 6)The entire base plate is under compression. The maximum pressure from concrete must not exceed the bearing strength of concrete ( 0.45 f ck ).1. Combined stresses, f = P LB(1 ± 6Le) When e= L6 , f =2LBP , 0 respectively.

2. The combined stress should be less than or equal to 0.45f ck .2 PLB ≤ 0.45f ck ⇒ B= 2P 0.45 Lf ck3. The thickness of base plate is computed by equating the moment capacity of the base plate to the moment at the critical section, which is assumed to be at the outer edge of the column flange.

Case II e=L6(¿ L3)Most part of the base plate is in under compression, with little or negligible tension on the remaining part.1. Calculation of the length of the plate under compression (x):x3 +e= L2⇒ x=3(L 2 −e)2. Calculation of the width of the base plate:Compressive force(C) in concrete = area of stress triangle × widthC= 0.45f ck x 2× B=P ⇒ B= 2 P0.45f ck x ⇒ B= 2P0.45 f ck× 3(L 2 −e)3. The thickness of base plate is computed by equating the moment capacity of the base plate to the moment at the critical section, which is assumed to be at the outer edge of the column flange.

Case III (e> L 3 )Part of the base will be in compression with substantial tension in the remaining part and will be taken up by anchor bolts.1. The size of the base plate (L×B) is determined as before or assumed.2. From the equilibrium of forces,P=0.45 f ck x B−Fb And M=0.45 f ck x B(L2−2x)−Fb aWhere, P=¿ axial compressive force Fb=¿ Tensile force in the bolta=¿ Distance of line of anchor bolts from c.g. of the column.

The value of x may be determined from the above force and moment equations, x= L2 +a−[(L 2 +a)2−{2× 0.45 (M +fPa ck B) }]0.5

The maximum moment is determined at the critical section.M=0.45 f ck x B(c2−2x) Where, c2=¿ outstand of base plate from the column flange.3. The thickness of base plate is computed by equating the moment capacity of the base plate to the moment at the critical section, which is assumed to be at the outer edge of the columnflange.4. Calculation of design tensile force in the bolt: P=0.45 f ck x B−Fb⇒ Fb=0.45f ck x B+P

5. Welded connection is designed to join the column section with the base for the maximum tension in the column flange due to the applied moment. So this is whole about the eccentrically loaded base plate and the slab base with these three lecture we have covered. In next lecture we will go to the gusset base where we will see whenthe moment is also coming into picture then how to design the gusset base, one is design of the base plate and another is the design of the angle which is connected to the base plate or gusset base plate with the column.