Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 1, Problem 1.9.14P

A plane truss is subjected to loads 2P and P at joints B and C, respectively, as shown in the figure part a. The truss bars are made of two L 102 X 76 X 6.4 steel angles (see Table F-5(b): cross-sectional area or the two angles, A = 2180 mm2, and figure part b) having an ultimate stress in tension equal to 390 MPa. The angles are connected to a 12-mm-thick gusset plate at C(figure part c) with 16-mm diameter rivets; assume each rivet transfers an equal share of the member force to the gusset plate. The ultimate stresses in shear and bearing for the rivet steel are 190 MPa and 550 MPa, respectively. Determine the allowable load Pallowif a safety factor of 2.5 is desired with respect to the ultimate load that can be carried. Consider tension in the bars, shear in the rivets, bearing between the rivets and gusset plate. Disregard friction between the plates the bars, and also bearing between the rivets and the and the weight of the truss itself.

  Chapter 1, Problem 1.9.14P, A plane truss is subjected to loads 2P and P at joints B and C, respectively, as shown in the figure

Expert Solution & Answer
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To determine

The maximum load, Pallow if a safety factor of 2.5 is desired with respect to the ultimate load that can be carried.

Answer to Problem 1.9.14P

The maximum load, Pallow=45.8kN.

Explanation of Solution

Given:

  Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.9.14P , additional homework tip  1

  A=2180mm2tg=12mmdr=16mmtang=6.4mmσu=390MPaFS=2.5σa=σu2.5τa=τuFSσba=σ buFS

Member forces from truss analysis:

  Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.9.14P , additional homework tip  2

Take moment about point A as follows:

  Dy×3a=P×2a+2P×aDy=43P

Take summation of force in the vertical direction as follows:

  Dy+Ay=P+2PAy=3P43PAy=53P

Apply section method for the given truss as follows:

  Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.9.14P , additional homework tip  3

Take moment about point F as follows:

  Dy×2a=P×a+FBC×a43P×2=P+FBCFBC=53P

Take moment about point B as follows:

  Dy×2a=P×a+FFG×a43P×2=P+FFGFFG=53P

Take summation of force in the vertical direction as follows:

  Dy+FCFcos45°=P43P+12FCF=PFCF=23P

Change the direction of FCF , then the magnitude of FCF become 23P .

Consider point D as an equilibrium point which is shown below:

  Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.9.14P , additional homework tip  4

Take summation of force in the vertical direction as follows:

  Dy=FDGsin45°43P=12FDGFDG=423P

Take summation of force in the horizontal direction as follows:

  FCD=FDGcos45°FCD=12×423PFCD=43P

Consider point G as follows:

  Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.9.14P , additional homework tip  5

Take summation of force in the vertical direction as follows:

  FCG=FDGcos45°FCG=423×12PFCG=43P

Thus, all the forces are:

  FBC=53PFCD=43PFCF=23P=0.471PFCG=43P

  Pallow For tension on net section in truss bars:

  Anet=A2drtang

  Anet=1975mm2

  AnetA=0.906

  Fallow=σaAnet

The above force is less than the maximum force in a member. So, BC controls, since it has the largest member force for this loading.

  Pallow=35FBCmax

  Pallow=35(σaAnet)

  Pallow=184.879kN

  AS=2.π4dr2

The above area is less for one rivet in double shear.

  FmaxN=τaAS

Where,

N = number of rivets in a particular member

  PBC=3(35)(τaAS)PBC=55.0kN

  PCF=2(3 2 )(τaAS)PCF=129.7kN

  PCF=2(3 2 )(τaAS)PCF=129.7kN

  PCG=2(34)(τaAS)PCG=45.8kN

So, shear in rivets in CG&CD controls Pallow here,

  PCD=2(34)(τaAS)PCD=45.8kN

Now, Pallow for bearing of rivets on truss bars.

  Ab=2drtang , which is less than rivet bears on each angle in two angle pairs.

  FmaxN=σbaAb

  PBC=3(35)(σbaAb)PBC=81.101kN

  PCF=2(32)(σbaAb)PCF=191.156kN

  PCG=3(34)(σbaAb)PCG=67.584kN

  PCD=3(34)(σbaAb)PCD=67.584kN

Finally, Pallow for bearing of rivets on the gusset plate.

  Ab=drtg (Bearing area for each rivet on gusset plate)

  tg=12mm<2tang=12.8mm

So, gusset will control over angles

  PBC=3(35)(σbaAb)PBC=76.032kN

  PCF=2(32)(σbaAb)PCF=179.209kN

  PCG=2(34)(σbaAb)PCG=63.36kN

So, shear in rivets controls:

  Pallow=45.8kN

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Chapter 1 Solutions

Mechanics of Materials (MindTap Course List)

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