Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 4

Q. 4.2

Classify each of the plane trusses shown in Fig. 4.15 as unstable, statically determinate, or statically indeterminate. If the truss is statically indeterminate, then determine the degree of static indeterminacy.

Step-by-Step

Verified Solution

(a) The truss shown in Fig. 4.15(a) contains 17 members and 10 joints and is supported by 3 reactions. Thus, m + r = 2j. Since the three reactions are neither parallel nor concurrent and the members of the truss are properly arranged, it is statically determinate.                             Ans.

(b) For this truss, m = 17 , j = 10, and r = 2. Because m + r < 2j, the truss is unstable.                          Ans.

(c) For this truss, m = 21 , j = 10, and r = 3. Because m + r > 2j, the truss is statically indeterminate, with the degree of static indeterminacy i = (m + r) – 2j = 4. It should be obvious from Fig. 4.15(c) that the truss contains four more members than required for stability.                                Ans.

(d) This truss has m = 16 , j = 10, and r = 3. The truss is unstable, since m + r < 2j.                       Ans.

(e) This truss is composed of two rigid portions, AB and BC, connected by an internal hinge at B. The truss has m = 26 , j = 15, and r = 4. Thus, m + r = 2j. The four reactions are neither parallel nor concurrent and the entire truss is properly constrained, so the truss is statically determinate.

(f ) For this truss, m = 10 , j = 7, and r = 3. Because m + r < 2j, the truss is unstable.                       Ans.

(g) In Fig. 4.15(g), a member BC has been added to the truss of Fig. 4.15(f ), which prevents the relative rotation of the two portions ABE and CDE. Since m has now been increased to 11, with j and r kept constant at 7 and 3, respectively, the equation m + r = 2j is satisfied. Thus, the truss of Fig. 4.15(g) is statically determinate.                                 Ans.

(h) The truss of Fig. 4.15(f ) is stabilized by replacing the roller support at D by a hinged support, as shown in Fig. 4.15(h). Thus, the number of reactions has been increased to 4, but m and j remain constant at 10 and 7, respectively. With m + r = 2j, the truss is now statically determinate.                              Ans.

(i) For the tower truss shown in Fig. 4.15(i), m = 16 , j = 10, and r = 4. Because m + r = 2j, the truss is statically determinate.                                             Ans.

( j) This truss has m = 13 , j = 8, and r = 3. Although m + r = 2j, the truss is unstable, because it contains two rigid portions ABCD and EFGH connected by three parallel members, BF , CE, and DH, which cannot prevent the relative displacement, in the vertical direction, of one rigid part of the truss with respect to the other.                                          Ans.

(k) For the truss shown in Fig. 4.15(k), m = 19 , j = 12, and r = 5. Because m + r = 2j, the truss is statically determinate.                                      Ans.