For the connection below you may neglect any moment in the connection caused by the eccentricity of the load. You must answer the following question:Is the connection adequate? Check all limit states and other requirements. Standard Sized Holes0.6" ThickA36 Steel4"Pu = 75kPu = 75k2.0"2.0"2.25"Plan View7/8" diameter A325 BoltsNote: for the plate asa TM, the following isgiven:þPnfracture = 78.3 kipsSection View

Given: Pu=75 kBolt diameter=78 inFnv=68 ksiFu=58 ksi

Answered: Standard Sized Holes 0.6" ThickA36…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Question

For the connection below you may neglect any moment in the connection caused by the eccentricity of the load. You must answer the following question:

Is the connection adequate? Check all limit states and other requirements.

Standard Sized Holes
0.6" ThickA36 Steel
4"
Pu = 75k
Pu = 75k
2.0"
2.0"
2.25"
Plan View
7/8" diameter A325 Bolts
Note: for the plate as
a TM, the following is
given:
þPnfracture = 78.3 kips
Section View

Expert Solution

Step 1

Given:

$P_{u} = 75 k B o l t d i a m e t e r = \frac{7}{8} in F_{n v} = 68 ksi F_{u} = 58 ksi$

Step 2

Shearing of bolt.

$\begin{array}{rcl} ϕ R_{n} & = & 0.75 A_{b} F_{n v} (N o . o f b o l t s) \\ = & 0.75 \times \frac{π}{4} \times {(\frac{7}{8})}^{2} \times 68 \times 3 \\ = & 92 kips > 75 kips \end{array}$

Hence ok.

Bearing.

$\begin{array}{rcl} {(ϕ R_{n})}_{B e a r i n g} & = & 0.75 \times 2.4 d t F_{u} \\ = & 0.75 \times 2.4 \times \frac{7}{8} \times 0.6 \times 58 \\ = & 54.81 k i p s \end{array}$

Tearing

$\begin{array}{rcl} {(ϕ R_{n})}_{T e a r i n g} & = & 0.75 \times 1.2 l_{c} t F_{u} \\ = & 0.75 \times 1.2 \times (20 - (\frac{7}{8} + \frac{1}{16})) \times 0.6 \times 58 \\ = & 33.2775 k i p s \end{array}$

Step by step

Solved in 4 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions