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I have been trying to solve the problem below involving a concrete anchor block subject to two equal horizontal forces. Normally I would just apply the force as a moment acting about the centroid of the concrete block which would produce a linear bearing pressure distribution at the base. Adding the self weight of the block as a compressive bearing pressure and I would then have an idea of the complete bearing pressure distribution. Am I missing something? taking moments about the centroid with the forces turned into vectors via the position vector and using the cross product only gets me so far.

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Usually, these kinds of problems can be handled as a footing with P load and M overturning moment. I don't quite get those assumptions though.

The overturning moment will cause the anchor block to triangular positive and negative stresses on the soil, which can be calculated as, call this stress Q at the maximum point the end of the triangle:

$$ \Sigma M_y=0,\quad (1500+1500)5kNm=2(Q*7.5/2)_{total\ stress}*(15*1/3)_{lever\ arm}$$ $$Q=15kNm/37.5=400N$$ but this is for the 5 meters base width, therefore $$400/5=80N/m^2$$

The slope of this stress surface is $80/7.5=10.66\text{ N}$ per meter. With 0 stress at the middle of the 15m length.

So the total stress distribution is $$P_{total}=(P-80+10.7*x)Nm2$$

please check my arithmetic.

diagram

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