A simple span member is 6.65 m in length is made up of Apitong 185 mm x 270 mm wooden section, with an allowable stress based on 80% stress grade of 16.5 MPa in bending and tension parallel to grain. The beam carries a concentrated load of 18 kN at the center and neglecting its own weight. Weight of wood = 7.5kN/m3. The beam carries an axial tensile load of 192 kN.

Bending and tension parallel to the grain = 16.5 MPa Modulus of Elasticity in Bending = 7310 MPa Compression parallel to the grain = 9.56 MPa Compute for the following:

a) actual tensile stress if only tensile force is acting.

b) interaction value for both bending and tensile stress

c) ratio of the difference between its actual bending and tensile stress to the adjusted bending

stress for slenderness

Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

Given :length = 6.65 m185 mm x 270 mm wooden sectionallowable stress = 16.5 Mpaconcentrated load = 18 kNWeight of wood = 7.5kN/m3axial tensile load = 192 kN.Bending and tension parallel to the grain = 16.5 MPaModulus of Elasticity in Bending = 7310 MPaCompression parallel to the grain = 9.56 MPaSolution:a) actual tensile stress if only tensile force is acting.actual tensile stress = applied load/ cross sectional area=18 * 10^3 / 185 mm x 270 mm=18000/49950= 36.03 *10^-1 Nmm^2b) interaction value for both bending and tensile stressInteraction equation ,  Rb +Rτ = 1 ...