A composite rod is fixed to the wall at H. and is made of collinear rods (1), (2

Question

A composite rod is fixed to the wall at H. and is made of collinear rods (1), (2)? and (3) joined by rigid connectors B and C. Rod (1) of length L has a hollow square cross section, and is made of a material of Young's modulus 2E. whereas rod (2) of length 1.5L has a solid circular cross section and is composed of a material of Young's modulus E. Rod (3) of length L has a hollow circular cross section and is composed of a material of Young" s modulus 2 E. The cross sections of the three rods are shown in Fig. 2.2(b). Axial loads are applied to the rigid connectors at B, C, and D as shown in Fig. 2.2(a). Assuming the length of the connectors to be negligible. Calculate the axial stresses experienced by rods (1), (2), and (3). Calculate the displacements of the connectors u_B, u_c, and u_D. Assuming that all three rods have the same yielding stress sigma_y. determine the maximum P that the composite can hold with the factor of safety FS=3. Express your answers in terms of L, E, d, pi and sigma_y.

Explanation / Answer

solution:

1)here given composite rod consist of three separate rod 1,2 and 3 and by balancing forces we get that

for rod 3,

force P at point D has to be balance by same force at C so we get compressive force on rod 3 as

F3=P

for rod 2

if we take out force P from 2P at point C then only tensile force P is acting at point C and it is balanced by force of equal magnitude at pointB so tensile force on rod 2 is P

F2=P

for rod 1

if we take out P from point B then only P will act at pointB on rod and it is compressive in nature and it is supported at wall and hence compressive force on rod 1 is P

F1=P

2)so axial stress in each rod is given by

S=F/A

for rod 1

compressive stress

for square cross section A1=(3d)^2-(2d)^2

Sc1=F1/A1=P/(9d^2-4d^1)=P/5d^2

for rod 2

tensile stress

St2=F2/A2

St2=P/[(pi/4)(2d)^2]=P/pi*d^2

for rod 3

compressive stress

sc3=F3/A3

Sc3=P/[(pi/4)((2d)^2-d^2)]=4P/3*pi*d^2

3)where displacement at connector B,C and D is given by

uh=0;fixed at wall

S=e*E

e=u/L

S=F/A

ub1=-[Sc1*L/2E]=-[P*L/10*E*d^2]

ub2=-[St1*1.5L/E]=-[1.5P*L/pi*E*d^2]

uc2=[St1*1.5L/E]=[1.5P*L/pi*E*d^2]

uc3=[Sc3*L/2E]=[4*P*L/6*pi*E*d^2]

ud3=-uc3=-[Sc3*L/2E]=-[4*P*L/6*pi*E*d^2]

here resultant displacement at connector B,C and D is given by

at B

ub=ub1+ub2=-.5744[P*L/E*d^2]

at C

uc=uc2+uc3=.68967[P*L/E*d^2]

at D

ud=-uc3=-.2122[P*L/E*d^2]

4)here allowable stress is given by

Sall=Sy/FS=Sy/3

from that we must choose such value of P such that weakest rod must be able to withstand that load

for rod 1 we get force P=P1 as

Sc1=Sy/3=.2[P1/d^2]

P1=1.666*Sy*d^2

for rod 2 we get that,P=P2

St2=Sy/3=.3183*[P2/d^2]

P2=1.047*Sy*d^2

for rod 3,P=P3

Sc3=Sy/3=.4244[P3/d^2]

P3=.7854*Sy*d^2

from value of P1,P2 and P3 we get that amongest them value of P3 is very low so we choose maximum value of P as P3

P=P3=.7854*Sy*d^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.