A rigid block of mass M is supported by three symmetrically spaced rods as shown in figure. Each copper rod has an area of 900 mm^2; E = 120 GPa; and the allowable stress is 70 MPa. The steel rod has an area of 1200 mm^2; E = 200 GPa; and the allowable stress is 140 MPa. (a)Compute the stress of the copper rod. (b)Compute the stress of steel rod. (c)Compute the largest weight which can be supported.

Q: Calculate the normal stress in the brass rod in MPa. Calculate the normal stress in the bronze rod…

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Q: 3. Given the following stress functions, = 3x² + 3y² − σ₁ = 3y² o₂ = 3x+y-z+ 6x = -Z Tay=z-6xy-² T₁₂…

A: Given data,

Q: 3. Determine the principal stress, the maximum in-plane shear stress, and average normal stress.…

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Q: The rod has a diameter of 40 mm. Determine the state of stress at point A and draw the resulting 3-D…

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Q: A rigid bar AB is pinned at end A and issupported by a wire CD and loaded by a force P atend B (see…

A: Given:- E = 210 GPa σY =820 MPa L =1.0 m diameter is d =3 mm To find:- (a) Assuming n = 0.2,…

Q: For the stress block shown, what is OP2 if OP1 = 140 MPa (T)? A.60 MPa (T) 40 MPa B.60 MPa (C) C.20…

A: Calculations:σx=+120mpaσy=-40Mpaσ1=140mpagivenσ1=σx+σy2+σx-σy2+ζxy2140=120-402+120+4022+ζxy2140=40+8…

Q: 3. The stress element is shown in the figure. Determine the equivalent stress according to the third…

A: Solution: Given: σx=40 MPaσy= 40 MPaτxy=60 MPa Equivalent stress according to third and fourth…

Q: Obtain the maximum bending moment Mx that the cross section given below can safely carry. The…

A: Sigma=100MPa

Q: Q.2) A rigid bar ACD is supported by two circular steel (E= 30x10° psi) rods DE and a pin support at…

A: Draw free body diagram From deflected shape δAB8=δP4=δDE6=tanθ Consider the moment equilibrium…

Q: A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is…

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Methods to Determine Vertical Stress

The load applied to the soil is transferred to the underground. The presence of water at soil pores and solid grains above the soil are the primary components that are responsible for the effective load transfer. Due to the load acting, the internal component of the soil or the underground formations undergoes deformations or a tectonic shift. The amount of deformation depends upon the amount of load acting on the soil surface, which indeed, depends on the weight of everything present above the soil. Due to this, the underground soil element develops internal stresses which try to resist the tectonic changes.