ENGR 225 Section 1.3 – 1.6

Internal Loadings

Resultant Force and Moment

Coplanar Loadings

Stress

Stress Derivation A very small finite force ΔF is acting on an associated area ΔA. Replace this force with its directional components, ΔFx, ΔFy, and ΔFz. Take the limit as ΔA approaches zero of the quotient of the component forces and the areas This quotient is called, stress. It represents the intensity of the force on a specified plane.

Normal Stress, σz The intensity of force per unit area acting normal to ΔA.

Shear Stress, τzx τzy The intensity of force per unit area acting tangent to ΔA.

Stress

Normal Stress in an Axially Loaded Bar

Normal Stress Distribution dF =  dA dF =   dA Resultant Internal force P =   dA

Average Normal Stress Distribution If we assume  is constant throughout the area or Averaged over the cross sectional area, then P =   dA

Assumptions in using relation for average normal stress 1. Uniform cross section throughout the length 2. Uniform deformation. Applied load is along Centroidal axis Homogenous and Isotropic material Only axial load applied. Weight of the bar is neglected. Uniaxial Stress – Tensile or Compressive

Example 1. 7: The 80 kg lamp is supported by two rods AB and BC Example 1.7: The 80 kg lamp is supported by two rods AB and BC. If AB has a diameter of 10 mm and BC has a diameter of 8 mm determine the average normal stress in each rod.

Lecture Example: Determine the stress on the floor of a 135 lb Lecture Example: Determine the stress on the floor of a 135 lb. woman standing still wearing the following shoes that she purchased on sale. Assume that her weight is equally distributed between the heels and toes of her feet.

Average Shear Stress

Single Shear

Double Shear

Lecture Example: Determine the average normal and average shear stress developed in the wood fibers that are oriented along the section a-a.

Lecture Example: Determine the average compressive stress along the smooth areas of contact defined by AB and BC.

Lecture Example: Determine the average shear stress along the horizontal plane defined by EDB.

Complementary Property of Shear Stress

Complementary Property of Shear Stress Two- dimensional case zy Force balance gives zy = ’zy Moment balance about x axis gives zy = yz Hence zy = ’zy = yz = ’yz yz ’yz ’zy z y

Allowable Stress F.S. = Ffailure / Fallowable Failure Load : Material testing To ensure safe working : Allowable Load < Failure Load F.S. = Ffailure / Fallowable Factor of Safety : accounts for Unknown natural factors Errors in manufacturing and assembly Errors in load estimation Material weathering

Design of Simple Connections If  = P/A holds F.S. = fail / allow Hence A = P / allow

Rods A and B are made of steel having failure stress 510 MPa Rods A and B are made of steel having failure stress 510 MPa. Use Factor of Safety 1.75, determine the smallest diameter to support the given load. Beam is assumed to be pin connected at A and C.

Determine the intensity w of the maximum distributed load that can be supported by the hanger assembly so that an allowable shear stress of 13.5 ksi is not exceeded in the 0.4 in diameter bolts at A and B and an allowable tensile stress of 22 ksi is not exceeded in the 0.5 in diameter rod AB.