Problem 4.10. Shear resistance of a masonry wall

Determine the shear resistance of the wall of the previous problem if the normal force is N = 500 kN, with an eccentricity of e = 0.76 m. Length of the wall is 4 m, thickness of the wall is 30 cm. The compressive strength of masonry is f_c = 1.74 N/mm², shear strength is f_v= f_v0 + 0.4σ_c, where the initial shear strength is f_v0= 0.14 N/mm² and σ_c is the average compressive normal stress.

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Shear resistance, V_R [kN]=

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Steps

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Follow steps of Example 4.2.

Step 1. Determine normal stress distribution from the given normal force, N. Calculate the average normal stress.

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$N = \frac{σ^{\max} h_{c} b}{2} \to σ^{\max} = \frac{2 N}{h_{c} b} = \frac{2 \times 500 \times 10^{3}}{3720 \times 300} = 0.896 \frac{N}{{mm}^{2}} where h_{c} = 3 (\frac{h}{2} - e) = 3 (\frac{4.0}{2} - 0.76) = 3.72 m$

Normal stress is below the compressive strength (f_c = 1.74 N/mm²), thus the assumption holds true, the material is elastic (see stress diagram in the Figure). The average normal stress is

$σ_{c} = \frac{σ^{\max}}{2} = \frac{0.896}{2} = 0.448 \frac{N}{{mm}^{2}}$

Step 2. Determine the shear resistance of the cross section.

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$V_{R} = h_{c} b (f_{v 0} + 0.4 σ_{c}) = h_{c} b (f_{v 0} + 0.4 \frac{2 N}{2 h_{c} b}) = h_{c} b f_{v 0} + 0.4 N = = 3720 \times 300 \times 0.14 + 0.4 \times 500 \times 10^{3} = 256.2 \times 10^{3} N = 356.2 kN$

Results

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Follow steps of Example 4.2.

Assuming elastic material and linear normal stress distribution, the maximum normal stress arises from the given normal force, N is

$N = \frac{σ^{\max} h_{c} b}{2} \to σ^{\max} = \frac{2 N}{h_{c} b} = \frac{2 \times 250 \times 10^{3}}{3720 \times 300} = 0.896 \frac{N}{{mm}^{2}} where h_{c} = 3 (\frac{h}{2} - e) = 3 (\frac{4.0}{2} - 0.76) = 3.72 m$

Normal stress is below the compressive strength (f_c = 1.74 N/mm²), thus the assumption holds true, the material is elastic (see stress diagram in the Figure).

The average normal stress is

$σ_{c} = \frac{σ^{\max}}{2} = \frac{0.896}{2} = 0.448 \frac{N}{{mm}^{2}}$

Shear resistance of the cross section results in

$V_{R} = h_{c} b (f_{v 0} + 0.4 σ_{c}) = h_{c} b (f_{v 0} + 0.4 \frac{2 N}{2 h_{c} b}) = h_{c} b f_{v 0} + 0.4 N = = 3720 \times 300 \times 0.14 + 0.4 \times 500 \times 10^{3} = 356.2 \times 10^{3} N = 356.2 kN$