1. The truss member shown has an inclination of 60 degree with respect to the x-
ID: 1845246 • Letter: 1
Question
1. The truss member shown has an inclination of 60 degree with respect to the x-axis and it is subjected to an axial compressive force of P=150 kN. The truss member is connected to the support by means of a pin as illustrated in Figures 1.a and b. The diameter of pin is given as 20 mm. The support is made of material A which has a compressive (bearing) stress-strain diagram shown in Figure 1.b. (70 p.) c. If the shear modulus (G) of material A is 4000 MPa, determine the change in length along both transverse directions (change in the dimensions of 100 mm and 150 mm).Explanation / Answer
The relationship between the stress and strainthat a particular material displays is known as that particular material's stress–strain curve. It is unique for each material and is found by recording the amount of deformation (strain) at distinct intervals of tensile or compressive loading (stress).
F=150KN
d=20mm
fy=400N/mm2
force=Anet
150*103/0.6*400
=9.615
Anet=9.615
Agr=9.615*1.514.425
ISA=150*115*15
Anet +A1+A2k
A1=(150-21.5-15/2)
A1=1815
A2=(115-15/2)
A2=1612.5
K=3A1/3A1+A2
3*1815/3*1815+1612.5
=5445/5445+1612.5=5445/70570.5=0.7715
Anet =1815+1612.5*0.771
Anet =3056>2245mm2
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