Please solve the problem correctly P9.012 A 4.0-m-long simply supported timber b
ID: 1710836 • Letter: P
Question
Please solve the problem correctly
P9.012 A 4.0-m-long simply supported timber beam carries two concentrated loads as shown. The cross-sectional dimensions of the beam are also shown. Assume L1=0.8 m, L2-1.6m, p-5 kN, Q-35 kN, a=170 mm, b-165 mm, c=70 mm, and d=410 mm. (a) At section a-a, determine the magnitude of the shear stress in the beam at point H. (b) At section a-a, determine the magnitude of the shear stress in the beam at point K. ength (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 4.0-m span (d) Determine the maximum compression bending stress that occurs in the beam at any location within the 4.0-m span length Answers: (a) TH= (b) TK= (c) Tmax ikPa 2kPa kPa 4kPa
Explanation / Answer
Let us analysis the beam to determine the shear force at section a-a
From moment equilibrium equation,total moment about support A should be equal to zero.Let vertical reaction at suppprt D be equal to Dy
(5*1.6)+(35*3.2)-4Dy=0
Dy=30 kN
Shear force at section a-a = Dy=30kN
a) shear stress at a point is given by VQ/IB
V=shear force
Q=static moment of area at the point where shear stress is needed
I=moment of inertia of the cross section
B=width of section
Here,V=30kN=30000N
Q=a*b*(d-a)/2=0.17*0.165*(0.41-0.17)/2=0.003366 m^3
I=bd^3/12=0.165*0.41^3/12=0.000948m^4
B=0.165
Therefore, shear stress =645.57kPa
b)at point K, Q=b*c*(d-c)/2=0.0019635m^3
Shear stress=376.58kPa
c)Maximum shear force along entire beam span=30kN=30000N
Max shear stress=shear stress at mod depth of beam=1.5*V/(b*d)
=1.5*30000/(0.165*0.41)=665.19kPa
d)Maximum bending moment in beam=bending moment at C=30*0.8=24kNm=24000Nm
Section modulus of beam=bd^2/6=0.165*0.41^2/6=0.00462275m^3
Maximum compressive stress due to bending=24000/0.00462275=5191.7kPa